James weighs 120 pounds. He ran up the steps which rose 12 feet. John weighs 91 pounds. He ran up the same steps as James. If both boys reached the top of the 12-foot stairs at the same time, which boy had the greatest horsepower?

Answers

Answer 1

Answer: Mass of James = 54.431kg
Distance ran by James = 3.6576m
Mass of John= 41.277kg
Distance relan by John= 3.6576m

Work done by James = 54.431kg*10N/kg*3.6576m= 1990.87J

Work done by John= 41.277kg*10N/kg*3.6576m=1509.75J

Horsepower = work done / time since time is same, James has larger horsepower

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You have devised an experiment to measure the kinetic coefficient of friction between a ramp and block. You place the block on the ramp at an angle high enough that it starts sliding. You measure the time it takes to fall down a known distance. The time it takes to fall down the ramp starting from a standstill is 0.5 sec, ???? = 1 kg, θ = 45o, and the distance it falls, L, is 0.5 m. What is µk? (8 pts)
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What is the frequency of an xray with a wavelength of 2x10-10m?a) 1.5x1018hz
b) 6.67x10-19hz
c) 3x108hz
d) 1.5hz

Show calculation

Answers

Answer:

1.5 x 10¹⁸hz

Explanation:

Given parameters:

Wavelength = 2 x 10⁻¹⁰m

Unknown:

Frequency = ?

Solution:

To find the frequency, use the expression below;

V = f x wavelength

V is the speed of light = 3 x 10⁸m/s

f is the frequency

Now;

Insert the parameters

3 x 10⁸ = 2 x 10⁻¹⁰ x frequency

Wavelength = $(3 x 10^(8) )/(2 x 10^(-10) )$ = 1.5 x 10¹⁸hz

What is the difference between V(peak voltage) and Vrms (root-mean-square) of AC voltage source?

Answers

Answer:

V(peak voltage) is the highest voltage that the waveform will ever attain and the Vrms(root-mean-square) is the effective voltage of the total waveform representing the AC source.

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the other pulls in the same direction with a force of 290 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answers

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_(x) = P+T-\mu_(k)\cdot N = 0$ (Ec. 1)

$\Sigma F_(y) = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_(k)$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_(k)\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_(k) =(P+T)/(m\cdot g)$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,(m)/(s^(2))$ , then:

$\mu_(k) = (400\,N+290\,N)/((300\,kg)\cdot \left(9.807\,(m)/(s^(2)) \right))$

$\mu_(k) = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

Final answer:

The calculation of the coefficient of kinetic friction involves setting the total force exerted by the workers equal to the force of friction, as the crate moves at a constant speed. The coefficient of kinetic friction is then calculated by dividing the force of friction by the normal force, which is the weight of the crate. The coefficient of kinetic friction for the crate on the floor is approximately 0.235.

Explanation:

To calculate the coefficient of kinetic friction, we first must understand that the crate moves at a constant velocity, indicating that the net force acting on it is zero. Thus, the total force exerted by the workers (400 N + 290 N = 690 N) is equal to the force of friction acting in the opposite direction.

Since the frictional force (F) equals the normal force (N) times the coefficient of kinetic friction (μk), we can write the equation as F = μkN. Here, the normal force is the weight of the crate, determined by multiplying the mass (m) of the crate by gravity (g), i.e., N = mg = 300 kg * 9.8 m/s² = 2940 N.

Next, we rearrange the equation to solve for the coefficient of kinetic friction: μk = F / N. Substituting the known values (F=690 N, N=2940 N), we find: μk = 690 N / 2940 N = 0.2347. Thus, the coefficient of kinetic friction for the crate on the floor is approximately 0.235.

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At an intersection of hospital hallways, a convex spherical mirror is mounted high on a wall to help people avoid collisions. The magnitude of the mirror's radius of curvature is 0.560 m.A) Locate the image of a patient10.6m from the mirror. B) Indicate whether the image is upright or inverted.C) Determine the magnification of the image.

Answers

Answer:

Explanation:

For a convex mirror, the value of its image distance and its focal length are negative.

using the mirror formula 1/f = 1/u+1/v

f is the focal length = Radius of curvature/2 = 0.560/2

f= 0.28m

u is the object distance = 10.6m

v is the position of the image = ?

On substitution;

1/0.28 = 1/10.6 + 1/-v

3.57 = 0.094 - 1/v

3.57 - 0.094 = -1/v

3.476 = -1/v

v = -1/3.476

v = -0.2877m

B) Since the image distance is negative, this means that the image is an upright and a virtual image. All Upright images has their image distance to be negative.

C) Magnification = Image distance/object distance

Magnification = 0.2877/10.6

Magnification = 0.0271

A block sliding on ground where μk = .193 experiences a 14.7 N friction force. What is the mass of the block

Answers

Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77

What is Friction?

According to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab), it is not treated as a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Scientists began piecing together the laws governing friction in the 1400s, according to the book Soil Mechanics(opens in new tab), but because the interactions are so complex.

F=μ*m, n=w which also means n=mg, 14.7=0.193*n, n=76.2, 76.2=m*9.8, m=7.77.

Therefore, Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77.

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Answer:

7.77

Explanation:

F=μ*m

n=w which also means n=mg

14.7=0.193*n

n=76.2

76.2=m*9.8

m=7.77

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. For the water, determine the heat transfer, in kJ per kg of water. Kinetic and potential energy effects can be ignored.

Answers

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

$u_(1) = u_(g)$ = 2553.6 kJ/kg

$v_(1) = v_(g) = 0.4625 m^(3)/kg$

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, $v_(2) = v_(g) = 0.4625 m^(3)/kg$

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at $v_(2) = v_(g) = 0.4625 m^(3)/kg$ and temperature $T_(2) = 360^(o)C$ so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

$u_(2) = u_{\text{at 5 bar, 400^(o)C}} + (\frac{v_(2) - v_{\text{at 5 bar, 400^(o)C}}}{v_{\text{at 7 bar, 400^(o)C - v_(at 5 bar, 400^(o)C)}}})(u_{at 7 bar, 400^(o)C - u_(at 5 bar, 400^(o)C)})$

$u_(2) = 2963.2 + ((0.4625 - 0.6173)/(0.4397 - 0.6173))(2960.9 - 2963.2)$

= 2963.2 - 2.005

= 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$ ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

$\Delta K.E = \Delta P.E$ = 0

Now, equation will be as follows.

Q - W = $\Delta U + \Delta K.E + \Delta P.E$

Q - 0 = $\Delta U + 0 + 0$

Q = $\Delta U$

Now, we will obtain the heat transfer per unit mass as follows.

$(Q)/(m) = \Delta u$

$(Q)/(m) = u_(2) - u_(1)$

= (2961.195 - 2553.6)

= 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

Final answer:

The heat transfer is 227.4 kJ per kg of water.

Explanation:

Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C. To determine the heat transfer in kJ per kg of water, we need to calculate the heat absorbed by the water as it reaches 360°C.

Using the specific heat capacity of water (4,186 J/kg°C) and the change in temperature (360°C - 100°C), we can calculate the heat transfer:

Qw = mw * cw * AT = (1 kg) * (4186 J/kg°C) * (360°C - 100°C) = 227,440 J = 227.4 kJ

Therefore, the heat transfer is 227.4 kJ per kg of water.

Heat transfer is the process by which thermal energy moves from one object or substance to another due to a difference in temperature. This fundamental phenomenon occurs through three main mechanisms: conduction, convection, and radiation. Conduction involves the direct transfer of heat through a material, such as metal. Convection is the transfer of heat through the movement of fluids (liquids or gases). Radiation is the emission of electromagnetic waves carrying heat energy. Understanding heat transfer is essential in various fields, including physics, engineering, and environmental science, as it governs temperature regulation, climate dynamics, and the functioning of countless technological devices.

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