A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V and the secondary is connected to a floodlight that draws 5A, what is the power output? Please show ALL of your work.

Answer:

The distance and average speed are 54.79 m and 10.85 m.

Explanation:

Given that,

Speed = 21.7 m/s

Time = 5.05 s

(a). We need to calculate the distance

Firstly we will find the acceleration

Using equation of motion

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

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

Where, v = final velocity

u = initial velocity

t = time

Put the value in the equation

[tex]a = \dfrac{21.7-0}{5.05}[/tex]

[tex]a = 4.297 m/s^2[/tex]

Now, using equation of motion again

For distance,

[tex]s = ut+\dfrac{1}{2}at^2[/tex]

[tex]s = 0+\dfrac{1}{2}\times4.3\times(5.05)^2[/tex]

[tex]s=54.79\ m[/tex]

The distance is 54.79 m.

(b). We need to calculate the average speed during this time

[tex]v_{avg}=\dfrac{D}{T}[/tex]

Where, D = total distance

T = time

Put the value into the formula

[tex]v_{avg}=\dfrac{54.79}{5.05}[/tex]

[tex]v_{avg}=10.85\ m/s[/tex]

Hence, The distance and average speed are 54.79 m and 10.85 m.

Final answer:

The sprinter's total time in the 100 m dash is 10 seconds.

Explanation:

In this problem, the sprinter reaches his top speed of 11.2 m/s in 2.14 seconds.

To find the total time, we need to consider two parts: the time it takes to reach the top speed and the time it takes to cover the remaining distance at that speed.

First, calculate the time it takes to reach the top speed using the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values: 11.2 = 0 + a(2.14), we can solve for a to find a = 5.23 m/s².

Now, to find the time it takes to cover the remaining distance at the top speed, we can use the equation:

t = d / v

where d is the distance and v is the velocity.

Since the total distance is 100 m, subtracting the distance covered during the acceleration phase, we get the remaining distance as 100 - (0.5)(5.23)(2.14)² = 88.2 m.

Dividing this distance by the top speed of 11.2 m/s, we find t = 88.2 / 11.2 = 7.86 s.

To find the total time, we add the time it takes to reach the top speed and the time it takes to cover the remaining distance, so the sprinter's total time is 2.14 s + 7.86 s = 10 s.

Answer:

The temperature in degrees Fahrenheit in terms of the Celsius temperature is given by .

As we know by the linear relation

[tex]\frac{^o F - 32}{212 - 32} = \frac{^o C - 0}{100 - 0}[/tex]

now we have

[tex]^o F - 32 = \frac{180}{100} ^o C[/tex]

so we have

[tex]^o F = \frac{9}{5} ^o C + 32[/tex]

The temperature in degrees Celsius in terms of the Kelvin temperature is given by

As we know by the linear relation

[tex]\frac{^o C - 0}{100 - 0} = \frac{K - 273}{373 - 273}[/tex]

now we have

[tex]C = K - 273[/tex]

the temperature in degrees Fahrenheit in terms of the Kelvin temperature .

As we know by the linear relation

[tex]\frac{^o F - 32}{212 - 32} = \frac{K - 273}{373 - 273}[/tex]

now we have

[tex]^o F - 32 = \frac{180}{100} (K - 273)[/tex]

so we have

[tex]^o F = \frac{9}{5} (K - 273) + 32[/tex]

Explanation:

Mass = density × volume

m = (0.7 g/mL) × (40 mL)

m = 28 g

Rounding to 1 significant figure, the student should weigh out 30 grams of diethylamine.

By using the formula Mass = Density x Volume, we find that the student needs to weigh out 28 grams of diethylamine for 40mL volume, as the density of diethylamine is 0.7g/mL.

The question asks for the mass of diethylamine the student should weigh out to obtain 40mL of diethylamine. Diethylamine has a density of 0.7 g/mL, according to the CRC Handbook of Chemistry and Physics. The formula to find mass when you have volume and density is:Mass = Density x Volume. Therefore, to find the mass of the diethylamine, we multiply the volume (40 mL) by the density (0.7 g/mL), which equals 28 grams. So, the student should weigh out 28 grams of diethylamine.

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The orbital period of a satellite at an altitude of 350 km above Earth's surface is approximately 1.93 hours or around 115.8 minutes. Given the options, the closest one is option D, with a period of 91 minutes.

To answer the question about the satellite orbiting Earth at an altitude of 350 km, we can utilize some physics concepts related to orbital mechanics. Note that a significant factor in determining the orbital period of a satellite (time required for a satellite to complete one orbit) is not the satellite's mass but rather its altitude or, more accurately, the total distance from the center of the Earth, considering the Earth's radius plus the satellite's altitude.

In general, the closer a satellite is to Earth, the shorter its orbital period. The information given indicates that a satellite at an altitude of 350 km will likely have an orbital period around 1.93 hours, which is roughly equivalent to 115.8 minutes. Given that, none of the available options exactly match this result. But the value closest to it would be 91 minutes, which is represented by option D.

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Answer:

81.1 m

Explanation:

X = 237 m 27.2 east of north 237 (Sin27.2 i + Cos 27.2 j)

X = 108.33 i + 210.8 j

Y = 335 east = 335 i

Displacement = X + Y = 108.33 i + 210.8 j + 335 i = 443.33 i + 210.8 j

(magnitude of displacement)^2 = 443.33^2 + 210.8^2

magnitude of displacement = 490.9 m

Distance traveled = 237 + 335 = 572 m

Difference in the magnitude of displacement and distance = 572 - 490.9

                                                                                                   = 81.1 m

Answer:

64.2 m/s

Explanation:

In the x direction:

x = x₀ + v₀ₓ t + ½ at²

400 m = 0 m + v₀ cos (π/5) t + ½ (0 m/s²) t²

t = 400 / (v₀ cos (π/5))

In the y direction:

y = y₀ + v₀ᵧ t + ½ gt²

0 m = 0 m + v₀ sin (π/5) t + ½ (-9.8 m/s²) t²

0 = v₀ sin (π/5) - 4.9 t

t = v₀ sin (π/5) / 4.9

Therefore:

400 / (v₀ cos (π/5)) = v₀ sin (π/5) / 4.9

1960 = v₀² sin (π/5) cos(π/5)

1960 = ½ v₀² sin(2π/5)

3920 / sin(2π/5) = v₀²

v₀ = 64.2 m/s

Answer:

Initial velocity = 423.08m/s

Explanation:

Using formular for Range of projectile

R =V^2 sin2theta/g

Given:

Range=0.4km= 400m

Theta=3.142/5

g= 9.8m/s^2

400= V^2×sin(2×3.142/5)/9.8

400×9.8= V^2Sin 1.26

3920= 0.0219V^2

V^2= 3920/0.0219

V^2= 178995.43

V=sqrt 178995.43

V= 423.08m/s

(a) [tex]2.56\cdot 10^4 J[/tex]

The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:

[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]

where

m = 51.0 kg is the mass of the cheetah

u = 0 is the initial speed of the cheetah (zero because it starts from rest)

u = 31.7 m/s is the final speed

Substituting, we find

[tex]W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J[/tex]

(b) 6.1 cal

The conversion between calories and Joules is

1 cal = 4186 J

Here the energy the cheetah needs is

[tex]E=2.56\cdot 10^4 J[/tex]

Therefore we can set up a simple proportion

[tex]1 cal : 4186 J = x : 2.56\cdot 10^4 J[/tex]

to find the equivalent energy in calories:

[tex]x=\frac{(1 cal)(2.56\cdot 10^4 J)}{4186 J}=6.1 cal[/tex]

The cheetah needs 25,972.35 Joules or 6.2 Calories of work to reach its top speed of 31.7 m/s.

The cheetah accelerates from a state of rest to its top speed. This involves work done on the cheetah which we can compute using the formula for kinetic energy, as work done is equal to change in kinetic energy. The initial speed of the cheetah is 0, so initial kinetic energy is also 0. The final kinetic energy when the cheetah is at its top speed is ½ m v^2, where m is the mass and v is the speed.

(a) So, the work done, W = ½ * 51.0kg * (31.7 m/s)^2 =  25972.35 J,

(b) To convert Joules into Calories, we use 1 Calorie = 4186 J, so the Calories of work done = 25972.35 J / 4186 = 6.2 Calories. Note that due to inefficiencies in the energy conversion process in bodies, the actual energy produced by the cheetah's body will be more than 6.2 Calories to reach this speed.

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The ball's position vector has components

[tex]x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ t[/tex]

[tex]y=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2[/tex]

where [tex]g=9.80\dfrac{\rm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. The ball hits the ground when [tex]y=0[/tex]:

[tex]0=1.00\,\mathrm m+\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ t-\dfrac g2t^2\implies t=1.22\,\mathrm s[/tex]

The ball's velocity vector has components

[tex]v_x=\left(8.00\dfrac{\rm m}{\rm s}\right)\cos40.0^\circ[/tex]

[tex]v_y=\left(8.00\dfrac{\rm m}{\rm s}\right)\sin40.0^\circ-gt[/tex]

so that after 1.22 s, the velocity vector is

[tex]\vec v=(6.13\,\vec\imath-6.79\,\vec\jmath)\dfrac{\rm m}{\rm s}[/tex]

and the magnitude is

[tex]\|\vec v\|=\sqrt{6.13^2+(-6.79)^2}\,\dfrac{\rm m}{\rm s}=\boxed{9.14\dfrac{\rm m}{\rm s}}[/tex]

Explanation:

It is given that,

Mass of bullet, m₁ = 0.0207 kg

Mass of wooden block, m₂ = 28 kg

Wooden block is initially at rest, u₂ = 0 m/s

After the collision the, system has speed of, v = 1.180 m/s

We need to find the initial speed (u₁) of the bullet. It can be calculated using the conservation of linear momentum as :

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

[tex]0.0207\ kg\times u_1+0=(0.0207\ kg+28\ kg)\times 1.180\ m/s[/tex]

[tex]0.0207\times u_1=33.06[/tex]

[tex]u_1=1597.10\ m/s[/tex]

So, the initial speed of the bullet is 1597.10 m/s. Hence, this is the required solution.

Answer:

Charge, q = 1.15 C

Explanation:

It is given that,

Mass of sphere, m = 441 g = 0.441 kg

Electric field, E = 5 N/C

Due to this field, the sphere suddenly acquires a horizontal acceleration of 13.0 m/s² such that,

[tex]ma=qE[/tex]

[tex]q=\dfrac{ma}{E}[/tex]

[tex]q=\dfrac{0.441\ kg\times 13\ m/s^2}{5\ N/C}[/tex]

q = 1.146 C

or

q = 1.15 C

So, the charge carried by the small sphere is 1.15 C. Hence, this is the required solution.

The electric field accounts for the unbalanced motion of the sphere. The electrostatic forces balance the linear force, for which the charge acquired by the sphere is 1.146 C.

The region or space where a charged particle experiences the electrostatic force of attraction or repulsion, is known as an electric field.

Given data:

The mass of the small sphere is, m = 441 g = 0.441 kg.

The magnitude of the electric field is, E = 5.00 N/C.

The magnitude of horizontal acceleration is, a = 13.0 m/s².

Clearly, under the influence of an electric field, the electrostatic force balances the linear force. Then,

Fe = FL

[tex]qE = ma[/tex]

Here,

q is the magnitude of charge.

Solving as,

[tex]q \times 5.0 = 0.441 \times 13\\\\q = \dfrac{0.441 \times 13}{5.0}\\\\q = 1.146 \;\rm C[/tex]

Thus, we can conclude that the magnitude of charge carried by the sphere is 1.146 C.

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Answer:

40.3 min

Explanation:

First of all, let's convert every quantity into SI units:

[tex]v_1 = 92.5 km/h = 25.7 m/s[/tex] (speed in the first part of the trip)

[tex]t_2 = 28.0 min = 1680 s[/tex] time during which the person has stopped

[tex]v=72.2 km/h = 20.1 m/s[/tex] (average speed of the whole trip)

The average speed is the ratio between the total distance covered, d, and the total time taken, t:

[tex]v=\frac{d}{t}[/tex] (1)

The total distance covered is simply

[tex]d = v_1 t_1[/tex]

where [tex]t_1[/tex] is the time during which the person has moved at 92.5 km/h.

The total time taken is

[tex]t= t_1 + t_2[/tex]

So (1) becomes

[tex]v=\frac{v_1 t_1}{t_1 + t_2}[/tex]

Solving for [tex]t_1[/tex]:

[tex]v t_1 + v t_2 = v_1 t_1\\vt_2 = (v_1+v)t_1\\t_1 = \frac{v t_2}{v_1+v}=\frac{(20.1 m/s)(1680 s)}{25.7 m/s + 20.1 m/s}=737.3 s[/tex]

which corresponds to

[tex]t_2 = 737.3 s = 12.3 min[/tex]

So the total time of the trip is

[tex]t = 28.0 min + 12.3 min = 40.3 min[/tex]

Answer:

The angle the wire make with respect to the magnetic field is 30°

Explanation:

It is given that,

Length of straight wire, L = 0.6 m

Current carrying by the wire, I = 2 A

Magnetic field, B = 0.3 T

Force experienced by the wire, F = 0.18 N

Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :

[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\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.

Answer:

6.4 N

Explanation:

Draw free body diagrams for each mass.

For the mass on the inclined surface, the sum of forces parallel to the incline are:

∑F = ma

T - Mg sin θ = Ma

For the hanging mass, the sum of the forces in the y direction are:

∑F = ma

T - mg + F = m(-a)

We're given that g = 9.8 m/s², M = 11 kg, m = 2.1 kg, F = 6.6 N, and a = 3.6 m/s².

From the second equation, we have everything we need to find the tension force.

T - (2.1)(9.8) + 6.6 = (2.1)(-3.6)

T = 6.42 N

Rounded to 2 sig figs, the tension is 6.4 N.

Answer:

option (a)

Explanation:

To make a galvanometer into voltmeter, we have to connect a high resistance in series combination.

The voltmeter is connected in parallel combination with teh resistor to find the voltage drop across it.

An ideal voltmeter has very high resistance that means it has a resistance as infinity.

Answer:

A. In series Fint

Explanation:

If you connect it to the resistor, the voltmeter's ideal internal resistance be  In series Fint.

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

[tex]I= \Delta p = m \Delta v = m (v-u)[/tex]

where

m is the mass of the object

[tex]\Delta v[/tex] is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

[tex]I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s[/tex]

(b) 155 N

The impulse can also be rewritten as

[tex]I=F \Delta t[/tex]

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

[tex]\Delta t[/tex] is the duration of the collision

In this situation, we have

[tex]\Delta t = 0.06 s[/tex]

So we can re-arrange the equation to find the magnitude of the average force:

[tex]F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N[/tex]

The light emerging from a red filter is red and less intense than the original white light. When passed through blue plastic, it would be very dim or blocked, as red light cannot pass through a blue filter.

When a beam of white light passes through a red piece of plastic, the color of the light that emerges is red. This color is seen because the red plastic acts as a filter, absorbing other colors and only allowing red light to pass through. The emerging light is less intense than the white light because some of the light energy has been absorbed by the plastic.

If the light that has already passed through a red filter is then passed through a blue piece of plastic, the emerging light would likely be very dim or completely blocked. This is because the red light, lacking blue components, will be mostly absorbed by the blue plastic, which only allows blue light to pass through.

Answer:

Option 2 is the correct answer.

Explanation:

We have equation of motion s = ut + 0.5at²

Here u = 0 m/s

So, s = 0.5at²

Distance traveled in first second = 0.5 x a x 1² = 0.5 a

Distance traveled in second second = 0.5 x a x 2² - 0.5 x a x 1²= 1.5 a

Distance traveled in third second = 0.5 x a x 3² - 0.5 x a x 2²= 2.5 a

Distance traveled in fourth second = 0.5 x a x 4² - 0.5 x a x 3²= 3.5 a

The percentage increase in its displacement during the 4th second compared to its displacement in the 3rd second

                  [tex]=\frac{3.5a-2.5a}{2.5a}=0.4=40\%[/tex]

Option 2 is the correct answer.

Explanation:

It is given that,

Magnetic field, B = 0.15 T

Charge on a proton, [tex]q=1.60218\times 10^{-19}\ C[/tex]

Mass of a proton, [tex]m=1.67262 \times 10^{-27}\ kg[/tex]

The cyclotron frequency is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

[tex]f=\dfrac{1.60218\times 10^{-19}\ C\times 0.15\ T}{2\pi \times 1.67262 \times 10^{-27}\ kg}[/tex]

f = 2286785.40 Hz

or

[tex]\omega=14368296.44\ rad/s[/tex]

[tex]\omega=1.43\times 10^7 rad/s[/tex]

Hence, this is the required solution.

Answer:

0.035 (3.5 %)

Explanation:

The thermodynamic efficiency is given by:

[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]

where

[tex]T_C[/tex] is the cold temperature

[tex]T_H[/tex] is the hot temperature

In this problem we have

[tex]T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K[/tex]

So the efficiency is

[tex]\eta = 1 - \frac{278 K}{288 K}=0.035[/tex]

The maximum thermodynamic efficiency of an engine operating with a hot reservoir at 288 K and a cold reservoir at 278 K, according to the Carnot efficiency formula, is approximately 3.47%.

The student has asked about the thermodynamic efficiency of an engine that uses the temperature gradient of the ocean. To calculate this, we can use the Carnot efficiency formula:

Carnot Efficiency Formula

η = 1 - Tc/Th, where η represents the efficiency, Tc is the cold temperature, and Th is the hot temperature. It's important to remember these temperatures need to be in Kelvin.

First, convert the given temperatures from degrees Fahrenheit to Kelvin:

Now, apply the Carnot efficiency formula:

η = 1 - (278 K / 288 K) = 1 - 0.9653 = 0.0347 or 3.47%

Therefore, the maximum thermodynamic efficiency of this oceanic temperature gradient engine would be approximately 3.47%.

Answer:

The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]

Explanation:

Given that,

Charge [tex]q= 2.00 \mu C[/tex]

Electric field E = 100 N/C i

Radius R = 10.0 cm

We need to find the flux. It can be calculate using Gauss's law

The flux of the net electric field

[tex]\phi=\dfrac{q}{\epsilon_{0}}[/tex]

[tex]\phi=\dfrac{2.00\times10^{-6}}{8.85\times10^{-12}}[/tex]

[tex]\phi=2.26\times10^{5}\ Nm^2/C[/tex]

Hence, The flux of the net electric field is [tex]2.26\times10^{5}\ Nm^2/C[/tex]

The velocity of the fish relative to water is approximately 10.44 m/s when it hits the water. The fish inherits the eagle's horizontal velocity at the start and then accelerates downward due to gravity.

The eagle's speed is given as 3.00 m/s horizontally. Therefore, when the fish falls, it also initially has a horizontal velocity of 3.00 m/s because it was brought to that speed by the eagle. The vertical component of the fish's velocity can be calculated using the second equation of motion: vf = vi + a*t, which simplifies to vf = g*t in the absence of initial vertical velocity. The time 't' it takes for the fish to hit the water can be calculated using the equation d = 0.5*a*t2, which gives t = √(2d/g). Substituting 5.00 m for 'd' and 9.81 m/s2 (approx gravitational acceleration) for 'g', we get t ≈ 1.017 seconds. Replacing 't' in the velocity equation with 1.017 seconds, which results in a vertical velocity of 9.965 m/s. The resultant velocity of the fish relative to the water when it hits would then be calculated using Pythagoras theorem since the horizontal and vertical movements are independent and at right angles to each other, giving a result of √((3.00 m/s)2 + (9.965 m/s)2) = 10.44 m/s.

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The velocity of the fish relative to the water when it hits the water is approximately 10.36 m/s at an angle of about 73.74° below the horizontal.

To calculate the velocity of the fish relative to the water when it hits the water, we need to analyze both the vertical and horizontal components of its motion separately.

Given Data:

Horizontal speed of the eagle (and the fish when dropped): [tex]v_{horizontal} = 3.00 \, \text{m/s}[/tex]

Height from which the fish falls: [tex]h = 5.00 \, \text{m}[/tex]

Step 1: Calculate the time it takes for the fish to fall 5m.

To find the time of fall, we can use the kinematic equation for vertical motion:

[tex]h = \frac{1}{2} g t^2[/tex]

where:

[tex]h[/tex] is the height (5.00 m),

[tex]g[/tex] is the acceleration due to gravity (approximately [tex]9.81 \, \text{m/s}^2[/tex]),

[tex]t[/tex] is the time in seconds.

Rearranging the equation to solve for [tex]t[/tex], we have:

[tex]t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 5.00}{9.81}} \approx \sqrt{1.02} \approx 1.01 \, \text{s}[/tex]

Step 2: Calculate the vertical velocity of the fish just before hitting the water.

The vertical velocity can be calculated using:

[tex]v_{vertical} = g t[/tex]

Substituting the time we found:

[tex]v_{vertical} = 9.81 \, \text{m/s}^2 \times 1.01 \, \text{s} \approx 9.91 \, \text{m/s}[/tex]

Step 3: Combine the horizontal and vertical velocities to find the resultant velocity.

The horizontal velocity is constant at 3.00 m/s, and the vertical velocity just before impact is approximately 9.91 m/s. We can use the Pythagorean theorem to find the magnitude of the resultant velocity:

[tex]v_{resultant} = \sqrt{v_{horizontal}^2 + v_{vertical}^2}[/tex]

Substituting the values:

[tex]v_{resultant} = \sqrt{(3.00)^2 + (9.91)^2} = \sqrt{9 + 98.20} = \sqrt{107.20} \approx 10.36 \, \text{m/s}[/tex]

Step 4: Find the direction of the velocity relative to the horizontal using tangent.

The angle [tex]\theta[/tex] can be calculated using:

[tex]\theta = \tan^{-1}\left(\frac{v_{vertical}}{v_{horizontal}}\right)[/tex]

Substituting our values:

[tex]\theta = \tan^{-1}\left(\frac{9.91}{3.00}\right) \approx \tan^{-1}(3.30) \approx 73.74^\circ[/tex]

Answer:

Heat absorbed, Q = 110110 J

Explanation:

It is given that,

The specific heat of copper is, [tex]c=385\ J/kg.K[/tex]

Mass of block, m = 2.6 kg

Initial temperature, [tex]T_1=340\ K[/tex]  

Final temperature, [tex]T_2=450\ K[/tex]

Thermal energy is given by :

[tex]Q=mc\Delta T[/tex]

[tex]Q=mc(T_2-T_1)[/tex]

[tex]Q=2.6\times 385\times (450-340)[/tex]

Q = 110110 J

So, the thermal heat of 110110 J is absorbed. Hence, this is the required solution.

Answer:

4 s

Explanation:

u = 19.6 m/s, g = 9.8 m /s^2

Let the time taken to reach the maximum height is t.

Use first equation of motion.

v = u + at

At maximum height, final velocity v is zero.

0 = 19.6 - 9.8 x t

t = 19.6 / 9.8 = 2 s

As the air resistance be negligible, is time taken to reach the ground is also 2 sec.

So, total time taken be the ball to reach at original point = 2 + 2 = 4 s

The ball takes 4 seconds to return to its original launch point.

To find the time it takes for the ball to return to its original launch point, we need to determine the total time it takes for the ball to reach its maximum height and come back down. The ball is shot straight up, which means its initial velocity is positive and its final velocity is negative (when it returns to the launch point). We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time elapsed. In this case, the final velocity is -19.6 m/s (negative because it's going down), the initial velocity is 19.6 m/s, and the acceleration is -9.8 m/s^2 (due to gravity). Plugging the values into the equation and solving for t:

v = u + at

-19.6 = 19.6 - 9.8t

-39.2 = -9.8t

t = -39.2 / -9.8

t = 4 seconds

It takes 4 seconds for the ball to return to its original launch point.

To solve this problem, we need to utilize the relationship between linear and angular motion.

Firstly, we know the speed of point A is vA = 2.7 ft/sec. In a rotating cylinder, the linear speed v of a point at distance r from the center of rotation is given by v = ω * r, where ω is the angular speed. This would allow us to find ω by dividing the speed at point A by its distance from the center.

However, in this problem, we don't know the exact distance of point A from the center. So let's denote the unknown distance as r ft. Then, the angular speed ω = vA / r.

Secondly, it's given that the magnitude of the acceleration of the point A is aA = 20.6 ft/sec². Angular acceleration α is related to linear acceleration a by the factor of the radius, i.e., a = α * r.

We don't know r itself, but we know vA and aA, so we are able to compute the ratio of α to ω or α/ω = aA/vA, which equals to 20.6/2.7. This calculation enables us to conclude that the angular velocity and angular acceleration are proportional to each other, regardless of the values of radius r and angle θ

Taken together, from this reasoning, we can conclude that the ratio of angular to linear velocity/acceleration is approximately 7.63.

Finally, considering the issue of whether the knowledge of the angle θ is necessary, we can see that the angle θ does not appear in our ratio, and it does not affect the ratios of angular to linear velocities and accelerations. Therefore, the knowledge of the angle θ is not necessary for this calculation.

Learn more about Angular velocity and angular acceleration here:

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(a) 756.9 J

The work done by a force when moving an object is given by:

[tex]W=Fd cos \theta[/tex]

where

F is the magnitude of the force

d is the displacement of the object

[tex]\theta[/tex] is the angle between the direction of the force and of the displacement

In this situation,

F = 165 N

d = 5.60 m

[tex]\theta=35.0^{\circ}[/tex]

so the work done by the tension is

[tex]W=(165 N)(5.60 m)cos 35.0^{\circ}=756.9 J[/tex]

(b) 0.646

The crate is moving at constant speed: this means that the acceleration of the crate is zero, so the net force on the crate is also zero.

There are only two forces acting on the crate along the horizontal direction:

- The horizontal component of the tension, [tex]Tcos \theta[/tex], forward

- The frictional force, [tex]-\mu N[/tex], backward, with [tex]\mu[/tex] being the coefficient of kinetic friction, N being the normal reaction of the surface on the crate

The normal reaction is the sum of the weight of the crate + the vertical component of the tension, so

[tex]N=mg-T sin \theta = (31.0 kg)(9.8 m/s^2) - (165 N)sin 35.0^{\circ}=209.2 N[/tex]

Since the net force is zero, we have

[tex]T cos \theta-\mu N =0[/tex]

where

T = 165.0 N

[tex]\theta=35.0^{\circ}[/tex]

N = 209.2 N

Solving for [tex]\mu[/tex], we find the coefficient of friction:

[tex]\mu = \frac{T cos \theta}{N}=\frac{(165.0 N)(cos 35^{\circ})}{209.2 N}=0.646[/tex]

Final answer:

In this physics problem, we determine the work done by a tension force inclining at an angle above the horizontal and find the coefficient of kinetic friction when a crate moves at a constant speed.

Explanation:

A tension force of 165 N inclined at 35.0° above the horizontal is used to pull a 31.0 kg storage crate at a distance of 5.60 m on a rough surface.

(a) Work done by the tension force: To calculate work, use the formula W = Fd cos(θ), where F is the force, d is the distance moved, and θ is the angle between the force and displacement. Here, W = 165 N * 5.60 m * cos(35.0°).

(b) Coefficient of kinetic friction: Since the crate moves at a constant speed, the work done by the tension force equals the work done against friction. Find the frictional force using the work done formula and then the coefficient of kinetic friction.

Answer:

sned me short notes of Physics of all branches (post graduation level) thanks  

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Explanation:

The rock strikes the ground after approximately 3.67 seconds. The speed of the rock just before it strikes the ground is approximately 17.0 m/s.

To find the time it takes for the rock to strike the ground, we can use the equation for vertical motion. Assuming negligible air resistance, the initial velocity is 18.0 m/s and the vertical displacement is 50.0 m.

Using the equation y = yo + vyo*t - 1/2*g*t^2, where y is the final position, yo is the initial position, vyo is the initial velocity, g is the acceleration due to gravity, and t is the time, we can solve for t. Plugging in the values, we get:

50.0 = 0 + 18.0*t - 1/2*9.8*t^2

After rearranging the equation and solving the quadratic equation, we find that t = 3.67 seconds.

To find the speed of the rock just before it strikes the ground, we can use the equation for vertical motion. The final velocity v is equal to the initial velocity vyo - g*t. Plugging in the values, we get:

v = 18.0 - 9.8*3.67

Calculating the value, we find that the speed of the rock just before it strikes the ground is approximately 17.0 m/s.

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Answer:

Magnitude of the force exerted on the handle = 66 N

Explanation:

Power is the ratio of work and time.

        [tex]P=\frac{W}{t}\\\\W=Pt=84\times 1.1=92.4J[/tex]

We have work done by rower = 92.4 J

We also have

            Work = Force x Displacement

            92.4 = Force x 1.4

            Force = 66 N

Magnitude of the force exerted on the handle = 66 N

Answer:

Displacement of the car, XY = 81.92 km

Explanation:

From the attached figure,

OX = 64.9 km (in north direction)

OY = 50 km (in west direction)

We have to find the displacement of the car. The shortest path covered by a car is called displacement of the car. Here, XY shows the displacement of the car :

Using Pythagoras equation as :

[tex]XY^2=OX^2+OY^2[/tex]

[tex]XY^2=(64.9\ km)^2+(50\ km)^2[/tex]

XY = 81.92 km

Hence, the displacement of the car is 81.92 km.

The magnitude of the car's displacement is calculated using the Pythagorean theorem by treating the westward and northward movements as the sides of a right-angled triangle. The displacement, representing the hypotenuse of that triangle, is found to be approximately 81.93 km.

The student has asked to find the magnitude of displacement of a car that moves 50.0 km West and then 64.9 km North.

Calculating Displacement

To calculate the resultant displacement, we treat the distances as vector quantities and use the Pythagorean theorem. The car's westward and northward movements are at right angles to each other, so we can draw this as a right-angled triangle where the westward distance is one side (50.0 km), the northward distance is the other side (64.9 km), and the hypotenuse will be the displacement.

Using the Pythagorean theorem:

The magnitude of the car's displacement, therefore, is approximately 81.93 km.