When he prepares food, Edgar wants to use ingredients that are not high in trans fat. Based on this information, which of the following fats should he include in his recipes?a. Stick margarine made with olive oilb. Partially-hydrogenated peanut oilc. Corn oild. Shortening

Answer:b

Explanation:

Given

mass of block [tex]m=4 kg[/tex]

coefficient of static friction [tex]\mu =0.25 [/tex]

height of triangle is [tex]h=3 m[/tex]

[tex]F_{net}=mg\sin \theta -\mu _kmg\cos \theta [/tex]

[tex]a_{net}=g\sin \theta -\mu _kg\cos \theta [/tex]

[tex]a_{net}=9.8\sin 37-0.25\times 9.8\times \cos 37[/tex]

[tex]a_{net}=5.897-1.956=3.94 m/s^2[/tex]

here [tex]s=5 m[/tex]

[tex]v^2-u^2=2 a_{net}s[/tex]

[tex]v=\sqrt{2\times 3.94\times 5}[/tex]

[tex]v=6.27 m/s[/tex]

time taken to reach bottom of plane

[tex]v=u+at[/tex]

[tex]6.27=0+3.94\times t[/tex]

[tex]t=1.59 s\approx 1.6 s[/tex]                

This event is a result of interference, a physics phenomenon where waves combine to create a new wave. Changing the frequency altered the phase difference between the waves at various places in the room, leading to some students now hearing a tone because they are at points of constructive interference.

The correct answer is c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone. This situation is described by the phenomenon of interference. Interference occurs when two waves combine to form a resultant wave. When two identical sound waves from the speakers meet at a point in space, they can either constructively or destructively interfere depending on the phase difference.

If the phase difference is such that the waves reinforce each other, it results in a loud sound (constructive interference). However, if the phase difference is such that one wave cancels the other, no sound is heard (destructive interference). By doubling the frequency, the professor effectively changes the phase difference between the waves at the various points in the room. Therefore, some students' locations might now be at points of constructive interference, allowing them to hear the sound.

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The correct answer is option a. When the oscillator's frequency is doubled, the interference pattern shifts, causing some students who originally heard a loud tone to hear nothing, while others still hear a loud tone.

The student asked what happens when the professor adjusts the oscillator to produce sound waves of twice the original frequency. The correct answer is a)  Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.

The phenomenon described in the question is due to the interference of sound waves. Interference occurs when sound waves from the two speakers overlap, creating areas of constructive interference (where waves are in phase and amplify the sound) and destructive interference (where waves are out of phase and cancel each other out).

When the frequency is changed, the interference pattern shifts, causing some students who previously experienced a loud sound to now be in a zone of destructive interference, and vice versa. This adjustment results in a new distribution of loud and quiet spots within the classroom.

Answer:

Fc₂ = 4.675 N

Explanation:

r₁ = 12 cm = 0.12 m

Fc= 1.7 N

Fc = mV²/r = m w² r   ( ∴  V= r w )

Fc = m w² r   ----------------------- (1)

As angular speed (w) and mass are constant.

let    m * w² = k   ------------------(2)

Put equation (2) in equation (1).

⇒  Fc = m w² r = k * r         ( ∴ m w² = k)

⇒  Fc =k * r     ----------------------( 3)

⇒  Fc₁ =k * r₁ = 1.7 N

⇒ k = 1.7/ r  = 1.7 / 0.12  N    ( as r₁=0.12 m )

⇒ k = 14.17

Centripetal force hen the rod is fixed at 33 cm from the axis:

From equation ( 3 )

Fc₂ =  k * r₂

Fc₂ = 14.17 * 0.33 N    ( ∴ r₂ = 0.33 m)

Fc₂ = 4.675 N

To solve the problem it is necessary to apply the concept of Load on capacitors. The charge Q on the plates is proportional to the potential difference V across the two plates.

It can be mathematically defined as:

Q= CV

Where,

C = Capacitance

V = Voltage

Our values are given as,

[tex]C = 0.60pF\\V = 30V[/tex]

Substituting values in the above formula, we get

[tex]Q=CV\\Q = 0.6*30\\Q = 18pC[/tex]

Where

[tex]1pC = 10^{-12}Coulomb[/tex]

Therefore the charge must be 18pC to create a 30V Poential difference.

Answer:

77362.56 J

163730.28571 J

Explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant = [tex]3.33\times 10^{-6}[/tex]

[tex]\eta[/tex] = Heat transfer factor = 0.75

[tex]f_m[/tex] = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

[tex]V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3[/tex]

The unit energy for melting is given by

[tex]U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3[/tex]

Heat would be

[tex]Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J[/tex]

Heat required to weld is 77362.56 J

Amount of heat generation is given by

[tex]Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J[/tex]

The heat generated at the welding source is 163730.28571 J

Without the specific heat capacity and melting point of low carbon steel, the exact heat required for the weld cannot be calculated. However, given the heat transfer factor and melting factor, the heat generated at the welding source is 163,700 Joules according to the student's provided answer.

The quantity of heat required for welding low carbon steel with a fillet weld having a cross-sectional area of 25.0 mm2 and length of 300 mm depends on the specific heat capacity of the steel and the temperature change required to melt it. However, the question does not provide specific values for the specific heat capacity or the melting point of low carbon steel, which are essential to calculate the heat quantity. Normally, such calculations would also require knowledge of the latent heat of fusion for the steel.

Given the lack of necessary details to calculate the quantity of heat, we can address the part (b) of the question which relates to the heat generated at the welding source. The heat generated at the source can be calculated by dividing the actual heat needed to make the weld (which is given by the student as an unknown, hence represented by the question mark '?') by the product of the heat transfer factor and the melting factor, which are 0.75 and 0.63 respectively.

If the heat required to perform the weld ('?') is found, then the heat generated at the source can be calculated as follows: Heat at source = Heat required / (Heat transfer factor × Melting factor). According to the answer provided by the student, the heat at the source is 163,700 Joules.

Answer:

0.247

5.25×10⁻¹³ J

Explanation:

Part 1/2

Elastic collision means both momentum and energy are conserved.

Momentum before = momentum after

m v = m v₁ + 14.1m v₂

v = v₁ + 14.1 v₂

Energy before = energy after

½ m v² = ½ m v₁² + ½ (14.1m) v₂²

v² = v₁² + 14.1 v₂²

We want to find the fraction of the neutron's kinetic energy is transferred to the atomic nucleus.

KE/KE = (½ (14.1m) v₂²) / (½ m v²)

KE/KE = 14.1 v₂² / v²

KE/KE = 14.1 (v₂ / v)²

We need to find the ratio v₂ / v.  Solve for v₁ in the momentum equation and substitute into the energy equation.

v₁ = v − 14.1 v₂

v² = (v − 14.1 v₂)² + 14.1 v₂²

v² = v² − 28.2 v v₂ + 198.81 v₂² + 14.1 v₂²

0 = -28.2 v v₂ + 212.91 v₂²

0 = -28.2 v + 212.91 v₂

28.2 v = 212.91 v₂

v₂ / v = 28.2 / 212.91

v₂ / v = 0.132

Therefore, the fraction of the kinetic energy transferred is:

KE/KE = 14.1 (0.132)²

KE/KE = 0.247

Part 2/2

If a fraction of 0.247 of the initial kinetic energy is transferred to the atomic nucleus, the remaining 0.753 fraction must be in the neutron.

Therefore, the final kinetic energy is:

KE = 0.753 (6.98×10⁻¹³ J)

KE = 5.25×10⁻¹³ J

When a neutron collides elastically with the nucleus of an atom, a fraction of its kinetic energy is transferred to the nucleus. The fraction of kinetic energy transferred can be calculated using the principle of conservation of momentum and kinetic energy. For the given scenario, the fraction is 0.8636. To find the final kinetic energy of the neutron, multiply the fraction of kinetic energy transferred by the initial kinetic energy of the neutron.

When a neutron in a reactor undergoes an elastic head-on collision with the nucleus of an atom initially at rest, kinetic energy is transferred from the neutron to the atomic nucleus. The fraction of the neutron's kinetic energy transferred to the nucleus can be calculated using the principle of conservation of momentum and kinetic energy. Since the mass of the atomic nucleus is about 14.1 times the mass of the neutron, the fraction of kinetic energy transferred can be calculated as:

Fraction of kinetic energy transferred = (14.1 - 1) / (14.1 + 1) = 0.8636

For PART 2/2, to find the final kinetic energy of the neutron, we can multiply the fraction of kinetic energy transferred to the nucleus by the initial kinetic energy of the neutron:

Final kinetic energy = Fraction of kinetic energy transferred x Initial kinetic energy = 0.8636 x 6.98 × 10-13 J

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The period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].

Step 1

The period T of a simple pendulum is given by the formula:

[tex]\[ T = 2\pi \sqrt{\frac{L}{g}} \][/tex]

where:

- T is the period of the pendulum,

- L is the length of the pendulum, and

- g is the acceleration due to gravity.

To find the period [tex]\( T_{\text{Mars}} \)[/tex] on the surface of Mars, we can use the formula with the acceleration due to gravity on Mars, [tex]\( g_{\text{Mars}} = 3.71 \, \text{m/s}^2 \)[/tex], and the same length L as on Earth.

We have the period [tex]\( T_{\text{Earth}} = 2.00 \, \text{s} \)[/tex] on Earth, and we need to find [tex]\( T_{\text{Mars}} \).[/tex]

Step 2

Using the formula, we get:

[tex]\[ T_{\text{Earth}} = 2\pi \sqrt{\frac{L}{g_{\text{Earth}}}} \][/tex]

Solving for L, we find:

[tex]\[ L = \left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}} \][/tex]

Now, using this value of L, we can find [tex]\( T_{\text{Mars}} \)[/tex] using the acceleration due to gravity on Mars:

[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{L}{g_{\text{Mars}}}} \][/tex]

Substituting the known values, we get:

[tex]\[ T_{\text{Mars}} = 2\pi \sqrt{\frac{\left( \frac{T_{\text{Earth}}}{2\pi} \right)^2 g_{\text{Earth}}}{g_{\text{Mars}}}} \][/tex]

Step 3

Simplifying:

[tex]\[ T_{\text{Mars}} = T_{\text{Earth}} \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Mars}}}} \]Now, let's calculate \( T_{\text{Mars}} \):\[ T_{\text{Mars}} = 2.00 \times \sqrt{\frac{9.81}{3.71}} \]\[ T_{\text{Mars}} = 2.00 \times \sqrt{2.643} \]\[ T_{\text{Mars}} \approx 2.00 \times 1.626 \]\[ T_{\text{Mars}} \approx 3.25 \, \text{s} \][/tex]

So, the period of the simple pendulum on the surface of Mars is approximately [tex]\( 3.25 \, \text{s} \)[/tex].

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

[tex]v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2[/tex]

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

[tex]v = \omega R\\a = \alpha R[/tex]

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

[tex]v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft[/tex]

One revolution is equal to the circumference of the drum. So, total number of revolutions is

[tex]x / (2\pi R) = 6/(\pi R)[/tex]

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

[tex]a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2[/tex]

b) [tex]a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2[/tex]

The angular acceleration of drum B is 0.5 ft/s² and the number of revolutions executed by drum B during the 4-s interval is 6 ft.

To determine the angular acceleration of drum B, we can use the formula:

α = (v1 - v0) / t

Where α is the angular acceleration, v1 and v0 are the final and initial speeds respectively, and t is the time interval.

Substituting the given values, we have:

α = (4 ft/s - 2 ft/s) / 4 s = 0.5 ft/s²

To find the number of revolutions executed by drum B during the 4 s interval, we can use the formula:

θ = (ω0 + ω1) / 2 * t

Where θ is the angle in radians, ω0 and ω1 are the initial and final angular velocities respectively, and t is the time interval.

Since the tape does not slip on the drums, the angular velocity of drum B is the same as the linear velocity of the tape. Thus, ω0 = v0 and ω1 = v1.

Substituting the given values, we have:

θ = (2 ft/s + 4 ft/s) / 2 * 4 s = 6 ft

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Answer:

Normal force, N = 60000 N                      

Explanation:

It is given that,

Mass of the automobile, m = 2000 kg

Radius of the curved road, r = 20 m

Speed of the automobile, v = 20 m/s

Let N is the normal and F is the net force acting on the automobile or the centripetal force. It is given by :

[tex]N-mg=\dfrac{mv^2}{r}[/tex]

[tex]N=\dfrac{mv^2}{r}+mg[/tex]    

[tex]N=m(\dfrac{v^2}{r}+g)[/tex]                

[tex]N=2000\times (\dfrac{(20)^2}{20}+10)[/tex]                            

N = 60000 N  

So, the normal force exerted by the road on the system is 60000 Newton. Hence, this is the required solution.

Final answer:

The normal force exerted by the road on the system (car and its occupants) when moving through a curved dip at a constant speed, calculated considering both gravitational and centripetal forces, is 60,000 N.

Explanation:

To calculate the normal force exerted by the road on the system (car and its occupants), we first note that when an object is in circular motion, the net force acting on the object is directed towards the center of the circle. In this case, the net force is the centripetal force required to keep the automobile in circular motion, which can be calculated using the formula Fc = m*v2/R where m is the mass of the automobile, v is its velocity, and R is the radius of the curve.

For the given values (m = 2000 kg, v = 20 m/s, and R = 20 m), the centripetal force is calculated as Fc = 2000 kg * (20 m/s)2 / 20 m = 2000 kg * 400 m2/s2 / 20 m = 40,000 N. This force is provided by the component of the normal force that acts towards the center of the circular path. Additionally, the normal force must counteract the gravitational force acting on the automobile (Fg = m*g), which is 2000 kg * 10 m/s2 = 20,000 N.

However, in the scenario of a car moving through a curved dip, the normal force also provides the centripetal force. The total normal force exerted by the road must therefore support the weight of the car and provide the centripetal force needed for circular motion. Thus, the total normal force is N = Fg + Fc = 20,000 N + 40,000 N = 60,000 N.

The work done by vector field F on a particle moving along curve C can be computed using the line integral ∫F.dr. To compute the line integral, we need to parameterize curve C and evaluate the dot product of the vector field and the parameterized curve.

To compute the work done by a vector field F on a particle moving along a curve C, we can use the line integral. The line integral of a vector field F along a curve C can be computed using the formula: ∫F.dr. In this case, F(x, y, z) = xi + yj + zk and C is given by r(t) = (1 + 2sin(t))i + (1 + 3sin(2t))j + (1 + sin(3t))k. We need to parameterize C to compute the line integral. Let's rewrite r(t) as:

r(t) = i + 2sin(t)i + j + 3sin(2t)j + k + sin(3t)k

We can then calculate the line integral using the given parameterization. Substituting r(t) into F(x, y, z), we get:

F(r(t)) = (1 + 2sin(t))i + (1 + 3sin(2t))j + (1 + sin(3t))k

Now, we can compute the line integral by evaluating ∫F(r(t)).dr over the given interval 0 ≤ t ≤ π/2. This involves evaluating the dot product of F(r(t)) and r'(t). The work done by F on the particle moving along C is the value of the line integral.

Final answer:

To compute the work done by a vector field on a particle moving along a curve, we can use the line integral formula. In this case, the curve C is given by r(t) = (1+2sin(t))i + (1+3sin(2t))j + (1+sin(3t))k, where 0 ≤ t ≤ π/2. The vector field F(x, y, z) = xi + yj + zk. The work done is equal to the line integral of F along C.

Explanation:

To compute the work done by a vector field on a particle moving along a curve, we can use the line integral formula:

Work = ∫C F · dr

In this case, the curve C is given by r(t) = (1+2sin(t))i + (1+3sin(2t))j + (1+sin(3t))k, where 0 ≤ t ≤ π/2. The vector field F(x, y, z) = xi + yj + zk.

The work done is equal to the line integral of F along C, so:

Work = ∫0^π/2 F(r(t)) · r'(t) dt

Now, we can substitute the given expressions for F and r(t) and evaluate the integral to find the work done.

To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.

Since the forces are balanced the Spring force is equal to the force of the weight that is

[tex]F_s = F_g[/tex]

[tex]kx = mg[/tex]

Where,

k = Spring constant

x = Displacement

m = Mass

g = Gravitational Acceleration

Re-arrange to find the spring constant

[tex]k = \frac{mg}{x}[/tex]

[tex]k = \frac{8*9.8}{0.1}[/tex]

[tex]k = 784N/m[/tex]

Just before launch the compression is 40cm, then from Potential Elastic Energy definition

[tex]PE = \frac{1}{2} kx^2[/tex]

[tex]PE =\frac{1}{2} 784*0.4^2[/tex]

[tex]PE = 63.72J[/tex]

Therefore the energy stored in the spring is 63.72J

Answer:

Some of the Forces that are related to to the gg dropping are Grvaity, air resistance/ Air friction! hope this helps . Reley on Newtons third Law!! Also use The words Compresion and Velocity in your report!! :) LMK IF THIS HELPED>

Explanation:  

When an egg is dropped off a bridge, the main forces acting on it are gravity, air resistance, and, if it falls into water, the buoyant force.

When an egg is dropped off a bridge, several forces act on it as it falls. The main force acting on the egg is gravity, which pulls the egg downwards towards the ground. Another force is air resistance, which opposes the motion of the falling egg and slows it down. Additionally, there may be a buoyant force acting on the egg if it falls into water, which pushes the egg upwards. Gravity is the force that gives weight to objects and pulls them towards the center of the Earth. It is responsible for the downward motion of the egg.

Air resistance is the frictional force exerted by the air on the falling egg. It increases with the speed and surface area of the falling object. Buoyant force, on the other hand, is an upward force exerted by a fluid, such as water, on an object partially or fully submerged in it. If the egg falls into water, the buoyant force would act on it, partially counteracting the force of gravity. Overall, the forces acting on the egg as it falls off a bridge are gravity, air resistance, and the buoyant force (if it falls into water).

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Answer:

1421 seconds or 23.67 minutes

Explanation:

Q = Heat = 1200 W

c = Specific heat of human body = 3480 J/kgK

[tex]\Delta T[/tex] = Change in temperature = (44-37)°C

t = Time taken

[tex]Q=\dfrac{mc\Delta T}{t}\\\Rightarrow t=\dfrac{mc\Delta T}{Q}\\\Rightarrow t=\dfrac{70\times 3480\times (44-37)}{1200}\\\Rightarrow t=1421\ s[/tex]

The time student jogs before irreversible body damage occurs is 1421 seconds or 23.67 minutes

The body maintains a stable temperature through thermoregulation, with mechanisms like sweating playing an important role in removing heat. During physical activity, energy production increases and this heat needs to be dissipated to maintain body temperature. Failure of these mechanisms can lead to dangerous overheating.

The phenomenon mentioned in the question deals with the concept of thermoregulation, which is the process through which the body maintains its core temperature within certain boundaries. When a person exercises, the body produces more heat as a byproduct of the energy utilised during physical activity. This heat is then removed by perspiration and other mechanisms to ensure body temperature remains within a safe range.

The energy generated (in this case, 1200 W) can be converted into heat and then removed from the body to maintain a constant body temperature. Sweat, as part of the body's thermoregulation system, plays a crucial role in this process. As sweat (which is water) evaporates from the skin, it takes a significant amount of heat energy with it, effectively cooling the body.

If these cooling mechanisms were ineffective and the body could not release this heat, it could cause a dangerous increase in body temperature, potentially leading to irreversible damage to proteins in the body if the temperature reaches 44°C or above. That's why adequate fluid intake is essential to replace the liquid lost through sweat and to prevent dehydration.

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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

Explanation:

1 )  tire of radius 0.381 m rotating at 12.2 rpm

12.2 rpm = 12.2 /60 rps

n = .20333 rps

angular speed

= 2πn

= 2 x 3.14 x .20333

= 1.277 rad / s

2 ) a bowling ball of radius 12.4 cm rotating at 0.456 rad/s

angular speed = .456 rad/s

3 ) a top with a diameter of 5.09 cm spinning at 18.7∘ per second

18.7° per second = (18.7 / 180) x 3.14 rad/s

= .326  rad/s

4 )

a rock on a string being swung in a circle of radius 0.587 m with

a centripetal acceleration of 4.53 m/s2

centripetal acceleration = ω²R

ω is angular velocity and R is radius

4.53 = ω² x .587

ω  = 2.78 rad / s

5 )a square, with sides 0.123 m long, rotating about its center with corners moving at a tangential speed of 0.287 m / s

The radius of the circle in which corner is moving

= .123 x √2

=.174 m

angular velocity = linear velocity / radius

.287 / .174

1.649 rad / s

The perfect order is

4 ) > 5> 1 >2>3.

Explanation:

To arrange the listed objects according to their angular speeds from largest to smallest, we must first calculate the angular speed ω (in radians per second) for each object, as this unit provides a common ground for comparison. The angular speed can be found using the formula ω = v/r where v is the linear velocity and r is the radius.

Arranging these angular speeds from largest to smallest:

Answer:

a. closer to 20∘C

Explanation:

[tex]m_{p}[/tex] = mass of pallet = 50 g = 0.050 kg

[tex]c_{p}[/tex] = specific heat of pallet = specific heat of iron

[tex]T_{pi}[/tex] = Initial temperature of pellet = 200 C

[tex]m_{w}[/tex] = mass of water = 50 g = 0.050 kg

[tex]c_{w}[/tex] = specific heat of water

[tex]T_{wi}[/tex] = Initial temperature of water = 20 C

[tex]T_{e}[/tex] = Final equilibrium temperature

Also given that

[tex]c_{w} = 9 c_{p} [/tex]

Using conservation of energy

Energy gained by water = Energy lost by pellet

[tex]m_{w} c_{w} (T_{e} - T_{wi}) = m_{p} c_{p} (T_{pi} - T_{e})\\(0.050) (9) c_{p} (T_{e} - 20) = (0.050) c_{p} (200 - T_{e})\\ (9) (T_{e} - 20) =  (200 - T_{e})\\T_{e} = 38 C[/tex]

hence the correct choice is

a. closer to 20∘C

Final answer:

When a 50-g iron pellet at 200°C is dropped into 50 g of water at 20°C, the final temperature of the system upon reaching thermal equilibrium will be closer to 20°C. This is due to water's higher specific heat capacity, causing it to undergo a smaller temperature change during heat transfer.

Explanation:

The question involves a calorimetry problem, where two objects at different temperatures - a 50-g pellet of iron at 200°C and 50 g of water at 20°C - are combined and allowed to reach thermal equilibrium. Given that water has a specific heat capacity that is significantly higher than that of iron, the thermal energy will transfer from the iron to the water until both reach the same temperature.

As the water has a higher specific heat capacity, it will undergo a smaller change in temperature for a given amount of heat transfer. Hence, when the iron and water achieve thermal equilibrium, the final temperature will be closer to 20°C than to 200°C. This is because the 50 g of water will absorb more heat from the iron pellet with a less significant rise in temperature, compared to the temperature drop that iron experiences.

Based on the principles of heat transfer and specific heat capacity, the correct answer to the student's question is: a. closer to 20°C.

Answer:

D, using a spring scale to exert a force on the block. Measure the acceleration of the block and the applied force

Explanation:

For this you would use the net force equation acceleration=net force * mass however you will want to isolate mass so it would be acceleration/ net force to get mass. Then process of elimination comes to play.

The experiment that could be used to determine the inertial mass of a block is ; ( D ) Using a spring scale to exert a force on the block.Measure the acceleration of the block and the applied force.

Inertial mass is the mass of a body that poses an inertial resistance to the acceleration of a body when forces are applied to it. before you inertial mass can be determined, force must be applied to the body at rest .

The experiment that is suitable to determine the inertial mass is using a spring scale to apply a force on the block, then take measurement of the acceleration experienced by the block and amount of force applied.

Inertial mass =  acceleration / net force applied

Hence we can conclude that The experiment that could be used to determine the inertial mass of a block is  Using a spring scale to exert a force on the block.Measure the acceleration of the block and the applied force.

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Answer:

h = 0.362 m

Explanation:

The pressure equation with depth is

      P₂ = [tex]P_{atm}[/tex] +ρ g h

The gauge pressure is

       P2 -  [tex]P_{atm}[/tex] = ρ g h

This is the pressure that muscles can create

       P₂ -  [tex]P_{atm}[/tex]= 3740 Pa

But still the person needs a small pressure for the transfer of gases, so

      P₂ -  [tex]P_{atm}[/tex] = 3740 - 188 = 3552 Pa

This is the maximum pressure difference, where the person can still breathe,

Let's clear the height

      h = 3552 / ρ g

      h = 3552 / (1000 9.8)

      h = 0.362 m

This is the maximum depth where the person can still breathe normally.

Final answer:

The maximum depth that a person could swim to and still breathe with the snorkel is approximately 1.9 cm.

Explanation:

To find the depth that a person could swim to and still breathe with the snorkel, we need to consider the pressure difference needed for inhalation. The maximum reduction in air pressure that the chest muscles can produce in the lungs is 3740 Pa. The minimum pressure difference needed for inhalation is 188 Pa. Therefore, the maximum depth h that a person could swim to and still breathe with the snorkel is determined by the difference in atmospheric pressure at the surface and the pressure at that depth.

Using the equation for gauge pressure, we can calculate the maximum depth:

Pgauge = ρgh

Where:

Pgauge is the gauge pressure

ρ is the density of the fluid (water)

g is the acceleration due to gravity

h is the depth

Given that the minimum pressure difference is 188 Pa, we can rearrange the equation to solve for h:

h = Pgauge / (ρg)

Using the values for the density of water (1000 kg/m³) and acceleration due to gravity (9.8 m/s²), we can calculate the maximum depth:

h = 188 Pa / (1000 kg/m³ * 9.8 m/s²) = 0.019 m = 1.9 cm

Therefore, a person could swim to a maximum depth of approximately 1.9 cm and still breathe with the snorkel.

The correct step for obtaining a common denominator in the expression 'a = a1 + F/m' is to multiply each term by m/m, which results in '(m/m * a1/1) + (m/m * F/m)'. This leads to the acceleration, a, equalling 4.71 m/s2 when substituting the given values into the expression.

The student is trying to solve for the acceleration (a) of an object using the equation a = a1 + F/m, where a1 is given as 3.00 meters/second2, F as 12.0 kilogram.meter/second², and m as 7.00 kilograms. To obtain a common denominator for the two terms a1 and F/m, we look for an expression that allows us to combine these two fractions.

The correct step for obtaining a common denominator is option d. This is written as:

(m/m * a1/1) + (m/m * F/m)

This is because multiplying by m/m is equivalent to multiplying by 1, which does not change the value of the expression. For the term a1, since there is no denominator, multiplying by m/m effectively gives it a common denominator with F/m. The calculation becomes:

a = (m * a1 + F) / m

Substituting the given values:

a = (7.00 kg * 3.00 m/s2 + 12.0 kg*m/s2) / 7.00 kg

= (21.00 + 12.00) kg*m/s2 / 7.00 kg

= 33.00 kg*m/s2 / 7.00 kg

= 4.71 m/s2

Thus, the newton's second law, a = F/m, can be used to calculate the acceleration of the object.

Answer:

a. K = 1080.61 N/m

b. mₐ = 96.28 kg

Explanation:

T = 2π * √ m / K

T₁ = 2π * √m₁ / K , T₂ = 2π * √m₂ / K

m₁ = 36.2 kg , m₂ = m₁ + mₐ

T₁ = 1.15 s , T₂ = 2.2 s

Equal period to determine the force constant of the spring

a.

K = 4π²* m₁ / T₁²

K = 4π² * 36.2 kg / 1.15² s

K = 1080.61 N/m

So replacing and solve can find m₂

b.

m₂ = T₂² * m₁ / T₁²   ⇒  m₂ = 2.20² s * 36.2 kg  / 1.15 ² s

m₂ = 132.48

mₐ = m₂ - m₁

mₐ = ( 132.48 - 36.2 ) kg = 96.28 kg

Answer:

(a) Velocity at time t will be  [tex]v(t)=t^2+4t-32[/tex]

(B) Distance will be -48 m

Explanation:

We have given [tex]a(t)=2t+4[/tex]

And [tex]v(0)=-32[/tex]

(a) We know that [tex]v(t)=\int a(t)dt[/tex]

So [tex]v(t)=\int (2t+4)dt[/tex]

[tex]v(t)=t^2+4t+c[/tex]

As [tex]v(0)=-32[/tex]

So [tex]-32=0^2+4\times 0+c[/tex]

c = -32

So [tex]v(t)=t^2+4t-32[/tex]

(b) We have to find the distance traveled

So [tex]s(t)=\int_{0}^{6}v(t)dt[/tex]

[tex]s(t)=\int_{0}^{6}(t^2+4t-32)dt[/tex]

[tex]s(t)=\int_{0}^{6}(\frac{t^3}{3}+2t^2-32t)[/tex]

[tex]s=(\frac{6^3}{3}+2\times 6^2-32\times 6)-0=72+72-192=-48m[/tex]

Final answer:

To find the velocity at time t, integrate the acceleration function and add the initial velocity. To find the distance traveled, integrate the velocity function over the given time interval.

Explanation:

To find the velocity at time t, we can integrate the acceleration function over the given time interval and add the initial velocity. The integral of 2t + 4 is t^2 + 4t, so the velocity function is v(t) = t^2 + 4t - 32.

To find the distance traveled, we can integrate the velocity function over the given time interval. The integral of t^2 + 4t - 32 is (1/3)t^3 + 2t^2 - 32t. Evaluating this integral from 0 to 6 gives us the distance traveled during the given time interval.

Answer:

2258.4 m

Explanation:

Distance covered is a product of speed and time hence

s=vt where s is the displacement/distance covered, v is the speed and t is the time taken

s=24*94.1=2258.4 m

Therefore, the distance covered is 2258.4 m

Answer:

t = 17199 years

Explanation:

given,

mass of sample = 1.09 Kg

Activity of living material   = 15 decays / min /g

Activity of living material   = 15 x 1000 decays /min /kg

Activity of living material per 1.09 kg A = 1.09 x 15 x 1000 decays / min

Activity of after time t is A ' = 2020

half life = 57300 years

desegregation constant

λ = 0.693 / 5700

[tex]A'= A e^{-\lambda\ t}[/tex]

[tex]A'= 1.09 \times 15 \times 1000 e^{-\lambda\ t}[/tex]

[tex]2020= 1.09 \times 15 \times 1000 e^{-\dfrac{0.693}{5700}\times t}[/tex]

[tex]0.124=e^{-\dfrac{0.693}{5700}\times t}[/tex]

taking ln both side

[tex]\dfrac{0.693}{5700}\times t = 2.09[/tex]

t = 17199 years

Answer:

B. Balanced Forces

Explanation:

The net force is defined as the sum of all the forces acting on an object. Therefore, if the forces are balanced, they will counteract each other, causing the net force to be zero, then the object will continue at rest or moving with constant velocity.

The term below best describes the forces on an object with a a net force of zero B. Balanced Force

Balanced Forces refer to the situation where the forces acting on an object are equal in magnitude and opposite in direction, resulting in a net force of zero. When balanced forces act on an object, they cancel each other out, leading to no change in the object's state of motion.

This concept is tied to Newton's First Law of Motion, which states that an object at rest remains at rest, and an object in motion continues to move with a constant velocity, unless acted upon by an unbalanced force.

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Answer:

The pressure inside the hose 7000 Pa to the nearest 1000 Pa.

[tex]h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}[/tex]

[tex]V_{1} =V_{2}[/tex]

The pressure at the site of the puncture is [tex]P₂ =7161.3 Pa[/tex]

Explanation:

According to Poiseuille's law, [tex]P_{1} - P_{2}  = \frac{128μLQ}{πD^{4} }[/tex]

Where [tex]P_{1}[/tex] is the pressure at a point [tex]1[/tex] before the leak, [tex]P_{2}[/tex] is the pressure at the point of  the leak [tex]2[/tex], μ = dynamic viscosity, L = the distance between points [tex]1[/tex] and [tex]2[/tex], Q = flow rate, D = the diameter of the garden hose.

Also,  from the equation [tex]P =ρgh[/tex], the equations [tex]h₁ = \frac{P₁} {ρg}[/tex] and [tex]h₂ = \frac{P₂} {ρg}[/tex] can be derived.

Combining Poseuille's law with the above, we get [tex]h₁ - ρgh₂ =  \frac{128μLQ}{πD^{4} }[/tex]

[tex]h₁ = \frac{\frac{128μLQ}{πD^{4} } + ρgh₂}{ρg}[/tex]

[tex]V =\frac{Q}{A}[/tex]

Since the hose has a uniform diameter, the nozzle at the end is closed and neither point [tex]1[/tex] nor [tex]2[/tex] lie after the puncture,

[tex]V_{1} =V_{2}[/tex]

The pressure at the site of the puncture [tex]P₂ =ρgh₂[/tex]

[tex]P₂ =1kgm⁻³ ×9.81ms⁻¹×0.73m[/tex]

[tex]P₂ =7161.3 Pa[/tex]

Answer: Option (c) is the correct answer.

Explanation:

It is known that the relation between resistance, length and cross-sectional area is as follows.

          R = [tex]\rho \frac{l}{A}[/tex]

Let the resistance of resistor A is denoted by R and the resistance of resistor B is denoted by R'.

Hence, for resistor A the expression for resistance according to the given data is as follows.

                 R = [tex]\rho \frac{2l}{2A}[/tex]

On cancelling the common terms we get the expression as follows.

             R = [tex]\rho \frac{l}{A}[/tex]

Now, the resistance for resistor B is as follows.

           R' = [tex]\rho \frac{l'}{A'}[/tex]

Thus, we can conclude that the statement, Wire A has the same resistance as wire B, accurately compares the resistances of wire resistors A and B.  

The question pertains to calculating the displacement at end A of a steel rod with a spring attached when subject to a specific force, using concepts of mechanical engineering such as Hooke's law and elastic deformation.

The student's question involves calculating the displacement of end A when a 60-kN force is applied to a 20-mm-diameter rod made of A-36 steel which has a spring attached with a stiffness k = 55 MN/m and taking into account that the modulus of elasticity E = 200 GPa. This problem is a straightforward application of Hooke's law combined with the elastic deformation formula (stress = force/area, strain = change in length/original length, and Hooke's law: F = kx for springs), which are part of mechanical engineering and physics topics on material strength and deformation.

To find the displacement, we should consider the deformation of the rod under the applied force and the compression of the spring separately. The rod's deformation can be found using the modulus of elasticity E and the cross-sectional area derived from the diameter, while the spring's compression is directly related to the force applied and the spring constant k.

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a) 12.95*10^20 molecules

b) 5.84*10^20 molecules

Explanation:

a)Tylenol

Mass of Tylenol = 325 mg

(m)=325*10^-3 g

Molecular weight of Tylenol(M) = 151.16256 g/mol

the number of moles present  

n = m/M

= 325*10^-3 g /151.1656 g/mol

= 2.15 *10^-3 mol

the number of molecules present

N= nNa

=(2.15*10^ -3 mol)(6.023*10^23 molecules/mol)

N=12.95*10 ^20 molecules

b)Advil

Mass of Advil = 200 mg

m=200*10^-3 g

Molecular weight of Advil(M) =206.28082 g/mol

the number of moles

n=m/M

=200*10^ -3 g/ 206.28082 g/mol

=0.96955*10^ -3 mol

the number of molecules resent

N=nNa

=(0.96955*10^ -3)(6.023*10^ 23 molecules/ mol)

N=5.84*10^20 molecules

Tylenol contains more molecules of active ingredient hence it is a more effective drug.

We have to find the number of molecules in each of the drugs. In the drug, tylenol, the molar mass of the compound C8H9NO2 is;

8(12) + 9(1) + 14 + 2(16) = 96 + 9 + 14 + 32 =151 g/mol

Number of moles = 325 × 10^-3 g/151 g/mol = 0.0022 moles

1 mole contains 6.02 × 10^23 molecules

0.0022 moles contains  0.0022 moles ×  6.02 × 10^23 molecules /1 mole = 1.32 × 10^21 molecules

For Advil;

Molecular mass= 13(12) + 18(1) + 2(16) = 156 + 18 + 32 = 206 g/mol

Number of moles = 0.2 g/ 206 g/mol = 0.00097 moles

1 mole contains 6.02 × 10^23 molecules of advil

0.00097 moles contains  0.00097 moles × 6.02 × 10^23 molecules /1 mole = 5.8  × 10^20

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Newton's second law for linear and rotational motion allows us to find the minimum angle so that the ladder does not slide is:

                 θ= 54.2º

Newton's second law for stable rotational motion that torque is equal to the product of the moment of inertia times the angular acceleration, when the angular acceleration is zero we have an equilibrium condition.

         Σ τ = 0

         τ = F r sin θ

Where θ is the torque, F the force, (r sin θ) the perpendicular distance.

In the attached we have a free-body diagram of the forces.

x- axis

      R - fr = 0

y -axis

      N - W = 0

      N = W

The friction force has the expression.

       fr = μ N

we substitute

       R = μ W          (1)

We use Newton's second rotational law where the pivot point is the base of the ladder.

      R y - W x = 0

Let's use trigonometry.

      sin θ = y / L

      cos θ = x / L

      y = L sin θ

     x = L cos θ

Let's  substitute.

      R L sin θ = [tex]W \frac{L}{2} cos \theta[/tex]  

      R sin θ = [tex]\frac{1}{2} W cos \theta[/tex]  

We use equation 1.

     μ W sin θ =[tex]\frac{1}{2}[/tex]  W cos θ

     tan θ = [tex]\frac{1}{2 \mu }[/tex]  

      θ = tan⁻¹ [tex]\frac{1}{2 \mu}[/tex]

We calculate.

     θ = tan⁻¹  [tex]\frac{1 }{2 \ 0.36}[/tex]

     θ = 54.2º

In conclusion, using Newton's second law we can find the minimum angle for the ladder not to slide is:

      tae 54.2º

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The minimum angle between the ladder and the floor at which the ladder will not slip is approximately 21.8 degrees.

To determine the minimum angle (θ) between the ladder and the floor at which the ladder will not slip, we need to consider the forces acting on the ladder. These forces include the normal reaction force (N) from the floor, the static friction force (f) between the ladder and the ground, and the weight of the ladder (w). The ladder will not slip as long as the net torque acting on it is zero, which occurs when the static friction force is greater than or equal to the force component that wants to make the ladder slip.

In this case, the force component that wants to make the ladder slip is the horizontal component of the ladder's weight, which is w*sin(θ). The static friction force can be calculated using the equation f = μsN, where μs is the coefficient of static friction and N is the normal reaction force.

We can set up an inequality to find the minimum angle: μsN ≥ w*sin(θ). Substituting the values given in the problem, we have 0.36*N ≥ w*sin(θ). Dividing both sides by N gives us 0.36 ≥ sin(θ). Taking the inverse sine of both sides gives us θ ≥ sin-1(0.36).

Therefore, the minimum angle between the ladder and the floor at which the ladder will not slip is approximately 21.8 degrees.

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Answer:

19

Explanation:

d = Length of cell = 2.02 cm

[tex]\lambda[/tex] = Wavelength = 600 nm

n = Refractive index of air = 1.00028

In the Michelson interferometer the relation of the number of the bright-dark-bright fringe is given by

[tex]\Delta m=\dfrac{2d}{\lambda}(n-1)\\\Rightarrow \Delta m=\dfrac{2\times 2.02\times 10^{-2}}{600\times 10^{-9}}\times (1.00028-1)\\\Rightarrow \Delta m=18.853\approx 19[/tex]

The number of bright-dark-bright fringe shifts that are observed as the cell fills with air are 19