An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g 10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

Explanation:

It is given that,

Charge acquired on rubbing a balloon on your hair, q = -1.93 C

If you move the balloon 25 cm away from your hair, r = 25 cm = 0.25 m

Electric field acting between the balloon and your hair is given by :

[tex]E=\dfrac{kq}{r^2}[/tex]

k = electrostatic constant

[tex]E=\dfrac{9\times 10^9\ Nm^2/C^2\times 1.93\ C}{(0.25\ m)^2}[/tex]

[tex]E=2.779\times 10^{11}\ N/C[/tex]

or

[tex]E=2.78\times 10^{11}\ N/C[/tex]

Hence, this is the required solution.

Answer:

i) [tex]v_1 = 0.40 m/s[/tex]

ii) [tex]v_2 = 3.60 m/s[/tex]

Explanation:

Part A)

As we know that diameter of the hose pipe is 2.71 cm

Now the area of crossection of the pipe will be

[tex]A = \pi (\frac{D}{2})^2[/tex]

[tex]A = \pi (\frac{0.0271}{2})^2 = 5.77 \times 10^{-4} m^2[/tex]

Now the flow rate is defined as the rate of volume

It is given as

[tex]Q = \frac{Volume}{time} = Area \times speed[/tex]

[tex]\frac{20 L\times \frac{10^{-3} m^3}{1L}}{1.45 \times 60 seconds} = 5.77 \times 10^{-4} \times v[/tex]

[tex]v = 0.40 m/s[/tex]

Part b)

As per equation of continuity we know

[tex]A_1 v_1 = A_2 v_2[/tex]

now we have

[tex]\pi (\frac{d_1}{2})^2 v_1 = \pi (\frac{d_2}{2})^2v_2[/tex]

[tex](2.71)^2 (0.40) = (\frac{2.71}{3})^2 v_2[/tex]

[tex]v_2 = 3.60 m/s[/tex]

Answer:

option (D)

Explanation:

If we plot a graph between the velocity of the object and the time taken, the slope of graph gives the value of acceleration of the object and the area under the graph gives the product of velocity and time taken that means it is displacement

Answer:

D

Explanation:

Calculate v = (v + u) / 2. ...

Average velocity (v) of an object is equal to its final velocity (v) plus initial velocity (u), divided by two.

The average velocity calculator solves for the average velocity using the same method as finding the average of any two numbers. ...

Given v and u, calculate v. ...

Given v and v calculate u.

Answer is D

Answer:

71.8 N

Explanation:

T = Tension force in the strap

W = net work done = 752 J

f = force of friction = 8 N

d = displacement = 15 m

θ = angle between tension force and horizontal displacement = 36 deg

work done by frictional force is given as

W' = - f d

Work done by the tension force is given as

W'' = T d Cos36

Net work done is given as

W = W' + W''

W = T d Cos36 - f d

752 = T (15) Cos36 - (8) (15)

T = 71.8 N

The rock has height [tex]y[/tex] at time [tex]t[/tex] according to

[tex]y=v_0t-\dfrac g2t^2[/tex]

where [tex]v_0[/tex] is the velocity with which it was thrown, and g = 9.8 m/s^2 is the acceleration due to gravity.

Complete the square to get

[tex]y=\dfrac{{v_0}^2}{2g}-\dfrac g2\left(t-\dfrac{v_0}g\right)^2[/tex]

which indicates a maximum height of [tex]\dfrac{{v_0}^2}{2g}[/tex] occurs when [tex]t=\dfrac{v_0}g[/tex]. We're told this time is 2.6 s after the rock is thrown:

[tex]2.6\,\mathrm s=\dfrac{v_0}{9.8\frac{\rm m}{\mathrm s^2}}\implies v_0=25.48\dfrac{\rm m}{\rm s}[/tex]

So when t = 1.6 s, the rock reaches the tower's height of

[tex]y=v_0(1.6\,\mathrm s)-\dfrac g2(1.6\,\mathrm s)^2\approx\boxed{28\,\mathrm m}[/tex]

The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock.

The height of the tower can be calculated by finding the maximum height reached by the rock and then subtracting the initial height of the rock. Since the rock reaches its maximum height 1.0 s after passing the top of the tower, we can use the formula:

Max height = y1 - y0 - (1/2) * g * t^2

where y1 is the maximum height, y0 is the initial height, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time it takes to reach the maximum height.

Since the rock is 8.10 m above its starting point at t = 1.00 s, we can plug in the values:

Max height = 8.10 m - 0 m - (1/2) * (9.8 m/s^2) * (1.0 s)^2

Simplifying this equation, we find that the maximum height reached by the rock is 3.15 m. Therefore, the height of the tower is 8.10 m + 3.15 m = 11.25 m.

Explanation:

It is given that,

Frequency of diagnostic ultrasound, f = 3.82 MHz = 3820 Hz

The speed of the sound in air, v = 343 m/s

(a) We need to find the wavelength in air of such a sound wave. Let it is given by λ₁

i.e. [tex]\lambda=\dfrac{v}{\nu}[/tex]

[tex]\lambda_1=\dfrac{343\ m/s}{3820\ Hz}[/tex]

[tex]\lambda_1=0.089\ m[/tex]

(b) If the speed of sound in tissue is 1650 m/s .

[tex]\lambda_2=\dfrac{v}{\nu}[/tex]

[tex]\lambda_2=\dfrac{1650\ m/s}{3820\ Hz}[/tex]

[tex]\lambda_2=0.43\ m[/tex]

Hence, this is the required solution.

Answer:

Work done, W = 0 J

Explanation:

It is given that,

Weight of the bucket, W = F = 35 N

It is lifted vertically 3 m and then returned to its original position. We need to find the work gravity do on the bucket during this process.

Work done when the bucket is lifted vertically, W₁ = -mgh

Work done when the bucket returned to its original position, W₂ = +mgh

Net work done, W = W₁ + W₂

W = 0 J

So, the work done on the bucket is zero. Hence, this is the required solution.

Final answer:

The total work done by gravity on a bucket of water that is lifted and then returned to its original position is zero joules (0 J), because gravity does equal amounts of positive and negative work on the bucket during the lifting and lowering phases, respectively.

Explanation:

The question examines the concept of work done by a force, which in physics is defined as the product of the force applied to an object and the distance over which that force is applied, provided the force is applied in the direction of motion. When a bucket of water is lifted vertically and then returned to its original position, gravity does work on the bucket on the way down, but because the bucket returns to its starting position, the total work done by gravity over the entire journey is zero joules (d). This is because gravity does positive work as the bucket is lowered and an equal amount of negative work as the bucket is lifted, resulting in a net work of zero.

Concretely, when the bucket is lifted, work is done against gravity and when it is lowered, gravity does work on the bucket. However, since the starting and ending points are the same, the net work done by gravity over the entire process is zero. It's important to notice that this is true regardless of the path taken; as long as the initial and final positions are the same, the work done by a conservative force such as gravity will be zero.

Answer:

(a) 446.88 N

(b) 241.08 N

Explanation:

m = 60 kg, μs = 0.760, μk = 0.410

(a) To just start the crate moving: the coefficient of friction is static.

F = μs m g

F = 0.760 x 60 x 9.8

F = 446.88 N

(b) To slide the crate with constant speed: the coefficient of friction is kinetic.

F = μk m g

F = 0.410 x 60 x 9.8

F = 241.08 N

Static friction is the friction that occurs in the body when the body is just about to move. The static friction force will be 446.88 N.While the kinetic friction force will be 241.08 N.

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).

Mathematically it is defined as the product of the coefficient of friction and normal reaction.

m is the mass of crate  = 60 kg

μs is the coefficient of static friction = 0.760

μk  is the coefficient of kinetic friction= 0.410

(a) Static friction is the friction that occurs in the body when the body is just about to move.

[tex]\rm F_s = \mu_s m g\\\\\rm F_s = 0.760\times60\times9.8\\\\\rm F_s = 446.88 N[/tex]

Hence static friction force will be 446.8 N.

(b) When the body is moving in a straight and inclined plane the value of friction force acting on the body is known as kinetic friction.

[tex]\rm F_k = \mu_km g\\\\\rm F_k = 0.410\times60\times9.8\\\\\rm F_k= 241.08\; N[/tex]

Hence kinetic friction force will be 241.08 N.

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Answer:

9.62 kg

Explanation:

From a z-score table, P(z<-2.33) ≈ 0.01.  So 9.5 should be 2.33 standard deviations below the mean.

z = (x − μ) / σ

-2.33 = (9.5 − μ) / 0.05

-0.1165 = 9.5 − μ

μ = 9.6165

Rounding to 2 decimal places, the mean should be set to 9.62 kg.

Answer:

12.3 m/s

Explanation:

m = mass of the box sliding on frictionless flat surface = 12.5 kg

x = compression of the spring = 14.1 cm = 0.141 m

k = spring constant = 94.5 kN/m = 94500 N/m

v = initial speed of the box

Using conservation of energy

Kinetic energy of the box = Spring potential energy

(0.5) m v² = (0.5) k x²

m v² = k x²

Inserting the values

(12.5) v² = (94500) (0.141)²

v = 12.3 m/s

Answer:

Acceleration, [tex]a=\dfrac{1}{8}(-i+9j)\ m/s^2[/tex]

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}[/tex]    

[tex]a=\dfrac{(-i+9j)}{8\ s}\ m/s^2[/tex]    

or

[tex]a=\dfrac{1}{8}(-i+9j)\ m/s^2[/tex]

Hence, the value of acceleration vector is solved.

The electric field is [tex]-0.225N/C[/tex]  towards the negative x-axis.

The electric force (F) acting on a charged object in an electric field (E) is given by the following formula:

[tex]F=qE[/tex]

Here, q is the charge of the object and E is the electric field.

Given:

Acceleration, [tex]a=2.4 \times 10^3 m/s^2[/tex]

Mass, [tex]m=3.0 \times 10^{-3} kg[/tex]

Charge, [tex]q=-32C[/tex]

The force on the charge is computed as:

[tex]qE=ma\\E=\frac{3.0 \times 10^{-3} \times 2.4 \times 10^3}{-32}\\E=-0.225N/C[/tex]

The electric field is towards the negative x-axis.

Therefore, the electric field is [tex]-0.225N/C[/tex]  towards the negative x-axis.

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The electric field that causes the object with a mass of 3.0 × 10⁻³kg and a charge of -32C to accelerate at 2.4 × 10³m/s^2 is -7.5 × 10² N/C, and it is in the +x axis direction.

The student's question involves determining the electric field that causes an object to experience a specific acceleration. The object has a mass of 3.0 × 10⁻³ kg and a charge of -32 C. To find the electric field E, we can use Newton's second law of motion and the definition of the electric force: F = ma = qE, where F is the force, m is the mass, a is the acceleration, q is the charge, and E is the electric field.

To calculate the electric field, we set F equal to the product of mass and acceleration and then solve for E:

E = F / q
= (m × a) / q
= (3.0 × 10^⁻³ kg × 2.4 × 10³ m/s²) / (-32 C)
= -7.5 × 10²N/C

The negative sign indicates that the electric field's direction is opposite to the charge's motion. Since the charge is negative and it accelerates in the direction of the +x axis, the electric field must be in the +x axis direction.

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

Answer:

Distance travelled, d = 0.21 m

Explanation:

It is given that,

Initial velocity of electron, u = 500,000 m/s

Acceleration of the electron, a = 500,000,000,000 m/s²

Final velocity of the electron, v = 675,000 m/s

We need to find the distance travelled by the electron. Let distance travelled is s. Using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{(675000\ m/s)^2-(500000\ m/s^2)^2}{2\times 500000000000\ m\s^2}[/tex]

s = 0.205 m

or

s = 0.21 m

So, the electron will travel a distance of 0.21 meters. Hence, this is the required solution.

(a) [tex]y(t)=250 - 4.9 t^2[/tex]

For an object in free-fall, the vertical position at time t is given by:

[tex]y(t) = h + ut - \frac{1}{2}gt^2[/tex]

where

h is the initial vertical position

u is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

In this problem,

h = 250 m

u = 0 (the stone starts from rest)

So, the vertical position of the stone is given by

[tex]y(t) = 250 - \frac{1}{2}(9.8) t^2 = 250 - 4.9 t^2[/tex]

(b) 7.14 s

The time it takes for the stone to reach the ground is the time t at which the vertical position of the stone becomes zero:

y(t) = 0

Which means

[tex]y(t) = h - \frac{1}{2}gt^2=0[/tex]

So for the stone in the problem, we have

[tex]250 - 4.9 t^2 = 0[/tex]

Solving for t, we find:

[tex]t=\sqrt{\frac{250}{4.9}}=7.14 s[/tex]

(c) -70.0 m/s (downward)

The velocity of an object in free fall is given by the equation

[tex]v(t) = u - gt[/tex]

where

u is the initial velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Here we have

u = 0

So if we substitute t = 7.14 s, we find the velocity of the stone at the time it reaches the ground:

[tex]v=0-(9.8 m/s^2)(7.14 s)=-70.0 m/s[/tex]

The negative sign means the direction of the velocity is downward.

(d) 6.94 s

In this situation, the stone is thrown downward with an initial speed of 2 m/s, so its initial velocity is

u = -2 m/s

So the equation of the vertical position of the stone in this case is

[tex]y(t) = h + ut - \frac{1}{2}gt^2=250 - 2t - 4.9 t^2[/tex]

By solving the equation, we find the time t at which the stone reaches the ground.

We find two solutions:

t = -7.35 s

t = 6.94 s

The first solution is negative, so it has no physical meaning, therefore we discard it. So, the time it takes for the stone to reach the ground is:

t = 6.94 s

An object in free fall, like a stone dropped from a tower, has its motion governed by the acceleration due to gravity. Using the physics of motion, we can find the height of the stone at any given time, the time it takes to reach the ground, the velocity it strikes the ground, and the time taken if it is initially thrown downward.

To solve these types of questions, we need to use the physics of motion. In the case of an object in free fall like a stone dropped from a tower, the only force acting on it is gravity, which pulls it downwards.

(a) The formula h(t) = 250 - 1/2gt^2 represents the height of the stone above ground level at any time t. Here, g is the acceleration due to gravity (9.8 m/s^2).

(b) The stone will reach the ground when h(t) = 0. Solving the equation 250 - 1/2*9.8*t^2 = 0 gives t ≈ 7.18 seconds (rounded to two decimal places).

(c) The velocity v with which the stone strikes the ground can be found using v = gt. Substituting g = 9.8 m/s^2 and t = 7.18 s gives v ≈ 70.4 m/s (rounded to one decimal place).

(d) If the stone is thrown downward with initial velocity of 2 m/s, the equation for time becomes 250 -2t- 1/2*9.8*t^2 = 0. Solving this gives t ≈ 7.04 seconds (rounded to two decimal places).

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Answer:

Amount of radioactive particles left after 60 years = 100 particles.

Explanation:

Amount of radioactive particles before 60 years = 400

Amount of radioactive particles present today = 200

That is radio active particles reduced to half. That is 60 years is half life of this radio active material.

After 60 years this 200 radio active particles will reduce to half.

Amount of radioactive particles left after 60 years = 0.5 x 200 = 100 particles.

Final answer:

The sample will have 100 radioactive particles remaining 60 years from today, based on the half-life of the material being 60 years.

Explanation:

The question concerns the concept of radioactive decay and specifically the half-life of a radioactive sample. In this case, the sample's quantity of radioactive particles was observed to decrease from 400 to 200 over a span of 60 years. Thus, the half-life of the material is 60 years, which is the time it takes for half of the radioactive atoms (parent nuclei) to decay into their decay products (daughter elements).

Given that the sample has 200 particles today, we can predict that in another 60 years, the number of radioactive particles will again be halved. Therefore, after 60 years from today, we expect there to be 100 radioactive particles remaining in the sample.

Answer:

(a) 426.8 J

(b) 20.5 m/s

(c)  23.727 °C

Explanation:

(a)

E = Energy of medium-sized pear = 102 cal

we know that , 1 cal = 4.184 J

hence

E = 102 (4.184 J)

E = 426.8 J

(b)

KE = kinetic energy of the object = E = 426.8 J

m = mass of the object = 2.03 kg

v = speed of the object = ?

Kinetic energy of the object is given as

KE = (0.5) m v²

inserting the values

426.8 = (0.5) (2.03) v²

v = 20.5 m/s

(c)

Q = Amount of heat added to water = E = 426.8 J

m = mass of water = 3.79 kg

c = specific heat of water = 4186 J/(Kg °C)

T₀ = initial temperature = 23.7 °C

T = Final temperature = ?

Using the equation

Q = m c (T - T₀)

426.8 = (3.79) (4186) (T - 23.7)

T = 23.727 °C

We first convert Calories to Joules and get 426,768 J. Then, using the kinetic energy equation and the given mass, we find the final speed of the object to be approximately 460 m/s. Using the heat formula to calculate the temperature change, we find that the final temperature of water would be 49.4°C.

Let's tackle these parts one by one:

(a) Converting Calories to Joules: The conversion factor is given as 1 Calorie = 4184 Joules (noting that a Calorie in food is actually a kilocalorie). When we multiply 102 Cal by this conversion factor, we get 102 Cal x 4184 J/Cal = 426,768 J.

(b) Kinetic Energy: The formula for kinetic energy is KE = 1/2 mv², where m is mass and v is speed. If all the energy from the pear is transformed to kinetic energy, we can set this equal to 426,768 J, then solve for v. We get: √((2*426768 J) / 2.03 kg) ≈ 460 m/s as the final speed of the object.

(c) Temperature Change in Water: The formula to calculate temperature change is Q = mcΔT, where Q is the heat added (in Joules), m is the mass of the water, c is the specific heat capacity, and ΔT is the temperature change. We know that Q = 426,768 J, m = 3.79 kg, and c = 4186 J/kg°C. We can rearrange the formula to solve for ΔT: ΔT = Q / (m*c). So, ΔT = 426768 J / (3.79 kg * 4186 J/Kg°C) ≈ 25.7°C. So, the final temperature would be the initial temperature plus this change, or 23.7°C + 25.7°C = 49.4°C.

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Answer:

Peak current = 16.9 A

Explanation:

Given that

RMS voltage = 120 Volts

[tex]V_{rms} = 120 V[/tex]

AC is connected across resistance

[tex]R = 10 ohm[/tex]

now by ohm's law

[tex]V = i R[/tex]

[tex]120 = i (10)[/tex]

[tex]i_{rms} = \frac{120}{10} = 12 A[/tex]

now peak value of current will be given as

[tex]i_{peak} = \sqrt{2} i_{rms}[/tex]

[tex]i_{peak} = \sqrt2 (12) = 16.9 A[/tex]

Answer:

A) 0.0 kJ

Explanation:

Change in the internal energy of the gas is a state function

which means it will not depends on the process but it will depends on the initial and final state

Also we know that internal energy is a function of temperature only

so here the process is given as isothermal process in which temperature will remain constant always

here we know that

[tex]\Delta U = \frac{3}{2}nR\Delta T[/tex]

now for isothermal process since temperature change is zero

so change in internal energy must be ZERO

Answer:

0.84μF

Explanation:

Charge is same through both the capacitors since they are in series. Total voltage is the sum of the voltages of the individual capacitors.. So voltage across the 2nd capacitor is 120- 90 =30 V.

Charge across first capacitor is Q = C₁V₁ = 90 x0.28 = 25.2μC

Therefore capacitance of 2nd capacitor =

C₂ = Q÷V₂ = 25.2÷30 = 0.84 μF

The decibel level at 1.20 m is approximately 93 dB. The sound intensity at 36.0 m is approximately 1.58 × 10^-13 W/m². The decibel level at 36.0 m is approximately 83 dB.

(a) To find the decibel level, we can use the formula B(dB) = 10 log10(I/Io), where I is the sound intensity and Io is the reference intensity. Plugging in the given values, we get B = 10 log10(5.50 ✕ 10^-3/10^-12). After evaluating this expression, we find that the decibel level is approximately 93 dB.

(b) To find the sound intensity at a distance of 36.0 m, we can use the inverse square law for sound, which states that the intensity is inversely proportional to the square of the distance. Using the formula I2 = I1(d1^2/d2^2), where I1 is the initial intensity, I2 is the final intensity, d1 is the initial distance, and d2 is the final distance, we find that the sound intensity at 36.0 m is approximately 1.58 × 10^-13 W/m².

(c) To find the decibel level at a distance of 36.0 m, we can use the same formula as in part (a), but with the intensity at 36.0 m as the new value for I. Plugging in the values, we get B = 10 log10(1.58 × 10^-13/10^-12). After evaluating this expression, we find that the decibel level at a distance of 36.0 m is approximately 83 dB.

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We calculated the decibel level at [tex]1.20 m[/tex] to be approximately [tex]97.4 dB[/tex] . The sound intensity at [tex]36.0 m[/tex]   is about [tex]6.11 \times 10^{-6} W/m^2[/tex] , and its corresponding decibel level is approximately [tex]67.9 dB[/tex].

(a) Finding the decibel level

The decibel (dB) level can be found using the formula:

[tex]\[\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)\][/tex]

where [tex]I[/tex]  is the sound intensity, and I0 is the reference intensity ([tex]10^{-12} W/m^2[/tex]). Given, I[tex]= 5.50 \times 10^{-3} W/m^2[/tex],

[tex]\[\beta = 10 \log_{10} \left( \frac{5.50 \times 10^{-3}}{10^{-12}} \right)\]\beta = 10 \log_{10} \left( 5.50 \times 10^{9} \right)\]\beta \approx 10 \times 9.74 = 97.4 \, \text{dB}\][/tex]

So, the corresponding decibel level is approximately [tex]97.4 dB[/tex].

(b) Finding the sound intensity at a distance of [tex]36.0 m[/tex]

Sound intensity decreases with the square of the distance from the source. We can use the inverse square law:

[tex]\[I_2 = I_1 \left( \frac{r_1^2}{r_2^2} \right)\][/tex]

Given, [tex]\[I_1 = 5.50 \times 10^{-3} \, \text{W/m}^2, \quad r_1 = 1.20 \, \text{m}, \quad r_2 = 36.0 \, \text{m}\][/tex],

[tex]\[I_2 = 5.50 \times 10^{-3} \left( \frac{1.20^2}{36.0^2} \right)\]I_2 = 5.50 \times 10^{-3} \left( \frac{1.44}{1296} \right)\]I_2 \approx 6.11 \times 10^{-6} \, \text{W/m}^2\][/tex]

The sound intensity at 36.0 m is approximately [tex]6.11 \times 10^{-6} W/m^2[/tex].

(c) Finding the decibel level at a distance of [tex]36.0 m[/tex]

Using the decibel formula again:

[tex]\[\beta = 10 \log_{10} \left( \frac{I}{I_0} \right)\][/tex]

where

[tex]I = 6.11 \times 10^{-6} \, \text{W/m}^2\]\beta = 10 \log_{10} \left( \frac{6.11 \times 10^{-6}}{10^{-12}} \right)\]\beta = 10 \log_{10} \left( 6.11 \times 10^{6} \right)\]\beta \approx 10 \times 6.79 = 67.9 \, \text{dB}\][/tex]

The corresponding decibel level at [tex]36.0 m[/tex] is approximately [tex]67.9 dB[/tex].

Answer:

[tex]-0.25 rad/s^2[/tex]

Explanation:

The equivalent of Newton's second law for rotational motions is:

[tex]\tau = I \alpha[/tex]

where

[tex]\tau[/tex] is the net torque applied to the object

I is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

In this problem we have:

[tex]\tau = -12.5 Nm[/tex] (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

[tex]I=50.0 kg m^2[/tex] is the moment of inertia

Solving for [tex]\alpha[/tex], we find the angular acceleration:

[tex]\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2[/tex]

Answer:

T = 50 + 80e^(-0.2t)

Explanation:

Newton's law of cooling says the rate of change of temperature with respect to time is proportional to the temperature difference:

dT/dt = k (T − Tₐ)

Separating the variables and integrating:

dT / (T − Tₐ) = k dt

ln (T − Tₐ) = kt + C

T − Tₐ = Ce^(kt)

T = Tₐ + Ce^(kt)

Given that Tₐ = 50 and k = -0.2:

T = 50 + Ce^(-0.2t)

At t = 0, T = 130.

130 = 50 + Ce^(0)

130 = 50 + C

C = 80

Therefore:

T = 50 + 80e^(-0.2t)

The formula for the object cooling is [tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex], where [tex]t[/tex] is in minutes.

The object is cooled by heat mechanism of Convection, Convection is a Heat Transfer mechanism in which is a solid object is cooled due to a fluid in motion and is described by the Newton's Law of Cooling, whose Differential Equation is:

[tex]\frac{dT}{dt} = -r\cdot (T-T_{m})[/tex] (1)

Where:

[tex]T[/tex] - Temperature of the solid, in degrees Fahrenheit.

[tex]r[/tex] - Cooling rate, in [tex]\frac{1}{min}[/tex].

[tex]T_{m}[/tex] - Water temperature, in degrees Fahrenheit.

The solution of this Differential Equation is:

[tex]T(t) = T_{m} + (T_{o}-T_{m})\cdot e^{-r\cdot t}[/tex] (2)

Where [tex]T_{o}[/tex] is the initial temperature of the solid, in degrees Fahrenheit.

If we know that [tex]T_{m} = 50\,^{\circ}F[/tex], [tex]T_{o} = 130\,^{\circ}F[/tex] and [tex]r = 0.2[/tex], then the formula for the object cooling is:

[tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex]

The formula for the object cooling is [tex]T(t) = 50 + 80\cdot e^{-0.2\cdot t}[/tex], where [tex]t[/tex] is in minutes.

Here is a question related to the Newton's Law of Cooling: https://brainly.com/question/13724658

Answer:

This is because copper is a conductor, and adding more copper spreads electricity more, and thus the magnetic field.

Please mark for Brainliest!!  Thanks! :D

For more information or if you have any questions, please comment below and I'll respond ASAP!! :D

Answer: 80.384 cubic cm /min

Explanation:

Let V denote the volume and r denotes the radius of the spherical snowball .

Given : [tex]\dfrac{dr}{dt}=-0.1\text{cm/min}[/tex]

We know that the volume of a sphere is given by :-

[tex]V=\dfrac{4}{3}\pi r^3[/tex]

Differentiating on the both sides w.r.t. t (time) ,w e get

[tex]\dfrac{dV}{dt}=\dfrac{4}{3}\pi(3r^2)\dfrac{dr}{dt}\\\\\Rightarrow\ \dfrac{dV}{dt}=4\pi r^2 (-0.1)=-0.4\pi r^2[/tex]

When r= 8 cm

[tex]\dfrac{dV}{dt}=-0.4(3.14)(8)^2=-80.384[/tex]

Hence, the volume of the snowball decreasing at the rate of 80.384 cubic cm /min.

The volume of the snowball is decreasing at a rate of 8.03 cubic cm per minute when the radius of the snowball is 8 cm.

The rate at which the volume of the spherical snowball is decreasing significantly depends on the rate of decrease in the sphere's radius. The volume formula of a sphere is V = 4/3πR³. With differentiation, volume change in the sphere over time, or dV/dt, can be represented as dV/dt = 4πR² * dR/dt. Plugging in the given values, dR/dt = -0.1 cm/min and R = 8cm, we find that dV/dt = -8.03 cm³/min. This indicates that the volume of the snowball is decreasing at a rate of 8.03 cm³/min. Remember, the answer is given as a positive number, i.e., without the negative sign, which represents a decrease.

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Answer:

Angle the wire make with respect to the magnetic field is 30°.

Explanation:

It is given that,

Length of wire, L = 0.6 m

Current flowing in the wire, I = 2 A

Magnetic field strength, B = 0.3 T

It is placed in the magnetic field. It will experience a force of, F = 0.18 N. We need to find the angle the wire make with respect to the magnetic field. The force acting on the wire is given by :

[tex]F=I(L\times B)[/tex]

[tex]F=ILB\ sin\theta[/tex]

[tex]\theta=sin^{-1}(\dfrac{F}{ILB})[/tex]

[tex]\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\ m\times 0.3\ T})[/tex]

[tex]\theta=30^{\circ}[/tex]

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.

By substituting the given values and solving for sin(θ), we determined that the wire makes a 30° angle with the magnetic field. Therefore, the wire experiences the magnetic force at this angle.

To determine the angle between a wire carrying a current and a uniform magnetic field, we can use the formula for the magnetic force on a current-carrying wire:

F = I * L * B * sin(θ)

Where:
F is the magnetic force (0.18 N)
I is the current (2.0 A)
L is the length of the wire (0.60 m)
B is the magnetic field strength (0.30 T)
θ is the angle between the wire and the magnetic field

Rearrange the formula to solve for sin(θ):

sin(θ) = F / (I * L * B)

Substitute the given values:

sin(θ) = 0.18 N / (2.0 A * 0.60 m * 0.30 T)

Simplify the expression:

sin(θ) = 0.18 / 0.36 = 0.50

Take the sin (inverse sine) to find θ:

θ = sin⁻¹(0.50) = 30°

Therefore, the wire makes an angle of 30° with respect to the magnetic field.

Answer:

173.2 Newtons

Explanation:

Since the scale is a reading of the normal force, the force of gravity and the normal force are in opposite directions, which means the forces can be subtracted from each other. In this situation, we can use Newtons 2nd Law: Net Force =(mass)( acceleration). The net force in this situation would be the Force of Gravity minus the Normal Force. So the equation you will begin with is Fg(which is mg) -  Normal Force = ma. We want the normal Force, so our new equation derived for Fnormal will be mg - ma = Fnormal.

The force that a scale in a rising and decelerating elevator reads can be calculated with the equation F= m*(g-a). Given that the mass, acceleration due to gravity, and elevator deceleration are 59.1 kg, 9.80 m/s2, and 6.87 m/s2 respectively, the force would be 172.8 Newtons.

The subject of this question is Physics, and it seeks to investigate the reading on a scale in an elevator when it is rising and decelerating. The force that the scale reads can be represented by the equation F= m*(g-a), where 'F' is the force, 'm' is the mass, 'g' is acceleration due to gravity, and 'a' is the rate at which the elevator is decelerating. In your case:

So, the force the scale reads is F = 59.1 kg * (9.80 m/s2 - 6.87 m/s2). This is equal to 172.8 Newtons.

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Answer:

(a) 0.0178 Ω

(b) 3.4 A

(c) 6.4 x 10⁵ A/m²

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

[tex]R=\frac{L}{A\sigma }[/tex]

[tex]R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }[/tex]

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

[tex]i = \frac{V}{R}[/tex]

[tex]i = \frac{0.060}{0.0178}[/tex]

i = 3.4 A

(c)

Current density is given as

[tex]J = \frac{i}{A}[/tex]

[tex]J = \frac{3.4}{5.3\times10^{-6}}[/tex]

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

[tex]E = \frac{J}{\sigma }[/tex]

[tex]E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}[/tex]

E = 9.01 x 10⁻³ V/m

Answer:

180 m

Explanation:

Case 1.

U = 40 km/h = 11.1 m/s, V = 0, s = 20 m

Let a be the acceleration.

Use third equation of motion

V^2 = u^2 + 2 as

0 = 11.1 × 11.1 - 2 × a × 20

a = 3.08 m/s^2

Case 2.

U = 220 km/h = 33.3 m/s, V = 0

a = 3.08 m/s^2

Let the stopping distance be x.

Again use third equation of motion

0 = 33.3 × 33.3 - 2 × 3.08 × x

X = 180 m

Answer:

696.83 m/s

Explanation:

m = mass of water = 565 g = 0.565 kg

c = specific heat of water = 4186 J/(kg⁰C)

ΔT = Change in temperature = T₂ - T₁ = 80 - 22 = 58 ⁰C

v = speed gained by water

Using conservation of energy

Kinetic energy gained by water = heat required to warm water

(0.5) m v² = m c ΔT

(0.5) v² = c ΔT

(0.5) v² = (4186) (58)

v = 696.83 m/s