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goblinko [34]
3 years ago
6

a particle moves a distance of 190 ft along a straight line. As it moves, it is acted upon by a constant force of magnitude 15 l

b in a direction opposite to that of the motion. What is the work done by the force
Physics
1 answer:
AleksAgata [21]3 years ago
6 0

Answer:

3868.1625 Nm

Explanation:

Given that:

Distance moved = 190 ft

Magnitude of force = 15 lb

Converting units :

1 lb = 4.45 Newton

1 ft = 0.305 meter

Hence,

Distance moved = (190 * 0.305) = 57.95 meter

Magnitude of force = (15 * 4.45) = 66.75 Newton

Workdone = Force * distance

Workdone = 66.75 N * 57.95 m

Workdone = 3868.1625 Nm

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A 20 kg crate initially at rest on a horizontal floor requires a 80 N horizontal force to set it in motion. Find the coefficient
e-lub [12.9K]

Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

N = W = m.g

The crate of m=20 Kg has a weight of:

W = 20*9.8

W = 196 N

The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

\displaystyle \mu=\frac{80}{196}

\mu=0.41

The coefficient of static friction between the crate and the floor is 0.41

7 0
2 years ago
The coefficient of cubical expansion of a substance depends upon
zzz [600]
<span>I think that the coefficient of cubical expansion of a substance depends on THE CHANGE IN VOLUME.

Cubical expansion, also known as, volumetric expansion has the following formula:

</span>Δ V = β V₁ ΔT

V₁ = initial volume of the body
ΔT = change in temperature of the body
β = coefficient of volumetric expansion.

β is defined as the <span>increase in volume per unit original volume per Kelvin rise in temperature.
</span>
With the above definition, it is safe to assume that the <span>coefficient of cubical expansion of a substance depends on the change in volume, which also changes in response to the change in temperature. </span>
7 0
3 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
erik [133]

Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

4 0
3 years ago
Two children are throwing a ball back-and-forth straight across the back seat of a car. The ball is being thrown 7 mph relative
Evgesh-ka [11]

Answer:

the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Explanation:

Given the data in the question and as illustrated in the diagram below;

Now, Lets assume line AB represent the movement of the car,

AC is the movement of the ball been thrown back and forth in the back seat

Ax is the motion of the ball it flies off the window

so from the diagram, We can see triangle ABC

where AB is 45 mph and AC = 7 mph

and angle ∠CAB = 90°

using SOH CAH TOA

TOA; tanθ = Opposite / Adjacent

tanθ = Opposite / Adjacent

tan( ∠ ABC ) = AC /  AB

we substitute

tan( ∠ ABC ) = 7 /  45

tan( ∠ ABC ) = 0.15555

( ∠ ABC ) = tan⁻¹ 0.15555

( ∠ ABC ) = 8.84°

Therefor, angle ( ∠ ABC )  is 8.84°

Meaning angle ( ∠ XAA' ) is also 8.84°

Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car

4 0
2 years ago
Would water molecules in Venus’ atmosphere, whose temperature is 740 K, escape into outer space? A water molecule has a mass tha
Akimi4 [234]

Answer:

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus.

Explanation:

The average speed of gas molecules is given by:

v_{rms}=\sqrt{\frac{3RT}{M}}

R is the gas constant, T is the temperature and M the molar mass of the gas.

We know that a water molecule has a mass that is 18 times that of a hydrogen atom:

M_H=1.01*10^{-3}\frac{kg}{mol}\\M_{H2O}=18M_H=0.02\frac{kg}{mol}

So, we have:

v_{rms}=\sqrt{\frac{3(8.314\frac{J}{mol \cdot K})740K}{0.02\frac{kg}{mol}}}\\v_{rms}=960.65\frac{m}{s}*\frac{1km}{1000m}=0.96\frac{km}{s}

The water molecule cannot escape, since the average velocity of the water molecules is less than one sixth of the escape velocity of venus:

10\frac{km}{s}*\frac{1}{6}=1.6\frac{km}{s}\\0.96\frac{km}{s}

4 0
3 years ago
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