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2 years ago

A ball of mass 1.84 kg is dropped from a height y, =

Physics

1 answer:

DochEvi [55]2 years ago

3 0

Answer:

11.408 joules

Explanation:

soln

From the question above

Mass of ball = m = 1.84kg

Height of ball when dropped = 1.49m

Height of ball when it bounces back up = 0.87m

Recall

Potential energy (P.E) = mass × height × g

Kinetic energy(K.E) = 1/2 × mass × (velocity)²

Mechanical energy = K.E + P.E

M.E lost = M.E when ball dropped – M.E when ball bounces back up

Recall that g = acceleration due to gravity = 10ms²

P.E/K.E when ball dropped = 1.84 × 1.49 × 10

P.E/K.E when ball dropped = 27.416J

P.E/K.E when ball bounces back = 1.84 × 0.87 × 10

P.E /K.Ewhen ball bounces back = 16.008J

Since P.E = K.E

M.E when ball dropped = 27.416J

M.E when ball bounces back = 16.008J

Therefore,

M.E lost = 27.416 - 16.008

M.E lost = 11.408Joules

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7. Two people are pushing a 40.0kg table across the floor. Person 1 pushes with a force of 490N

artcher [175]

Answer:

20.4 $m/s^{2}$

Explanation:

To start doing this problem, first draw a free body diagram of the table. My teacher always tells us to do this, and I find that it is very helpful. I have attached a free body diagram to this answer- take a look at it.

First, let us see if Net force = MA. To do that, we need to determine whether the object is at equilibrium horizontally. For an object to be at equilibrium, it either needs to be moving at a constant velocity or not moving at all. Also, if an object is at equilibrium, there will not be any acceleration. But we know that there IS acceleration horizontally, so it cannot be in equilibrium. If it is not in equilibrium, we can use the formula ∑F= ma.

Let us determine the net force. Since the object is moving horizontally, we can ignore the weight and normal force, because they are vertical forces. The only horizontal forces we need to worry about are the applied force and force of friction.

Applied force = 1055 N (490 + 565)

Friction force= Unknown

To find the friction force, use the kinetic friction formula, Friction = μkN

μk is the coefficient, which the problem includes- it is 0.613.

N is the normal force, which we have to find.

*To find the normal force, we have to determine if the object is at equilibrium VERTICALLY. Since it has no acceleration vertically (it's not moving up/down), it is at equilibrium. Now, when an object is at equilibrium in one direction, it means that all the forces in that direction are equal. What are our vertical forces? Weight (mg) and Normal force (N). So it means that the Normal force is equal to the Weight.

Weight = mg = (40)(9.8) = 392 N

Normal force = 392 N

Now, plug it back into the formula (μkN): (0.613)(392) = 240.296 N

Friction = 240.296 N

Now that we know the friction, we can find the horizontal net force. Just subtract the friction force, 240.296 from the applied force, 1055 N

Horizontal Net Force: 814.704 N

Now that we know the net force, plug in the numbers for the formula

∑F= ma.

814.704 = (40.0)(a)

*Divide on both sides)

a = 20.3676 m/s^2

Round it to 3 significant figures, to get:

20.4 $m/s^{2}$

7 0

3 years ago

A 9,100-microfarad [muF] capacitor is charged to 22 volts [V]. If the capacitor is completely discharged through an iron rod

MakcuM [25]

Answer:

The temperature rises by 0.52K

Explanation:

Detailed explanation and calculation is shown in the image below

5 0

3 years ago

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

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