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3 years ago

How much potential energy does a 50-N box have when lifted at a height of 1.5M?

Physics

1 answer:

nikitadnepr [17]3 years ago

3 0

The correct answer is: Option (A) 75 J

Explanation:

First, be careful with the units here. As you can see it is mentioned that there is a 50N box. It means that the weight (mg) of the box is given as the unit is Newton, not its mass (which is in kg).

As,

Potential-energy = mass * acceleration-due-to-gravity * height

PE = m*g*h --- (A)

In equation (A), mg is actually the weight of the box, which is given.

mg = 50N

h = height = 1.5m

Plug the values in equation (A):

PE = 50 * 1.5 = 75 J (Option A)

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The stiffness of a particular spring is 42 N/m. One end of the spring is attached to a wall. A force of 2 N is required to hold

Liula [17]

Answer:

$L_{o}=0.1224m$

Explanation:

Given data

Force F=2 N

Length L=17 cm = 0.17 m

Spring Constant k=42 N/m

To find

Relaxed length of the spring

Solution

From Hooke's Law we know that

$F_{spring}=k_{s}|s|\\F_{spring}=k_{s}(L-L_{o})\\ 2N=(42N/m)(0.17m-L_{o})\\2=7.14-42L_{o}\\-42L_{o}=2-7.14\\42L_{o}=5.14\\L_{o}=(5.14/42)\\L_{o}=0.1224m$

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3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

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3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

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. As we increase the quantum number of an electron in a one-dimensional, infinite potential well, what happens to the number of

Natalija [7]

Answer:

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Explanation:

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E ∝ n^2

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Aleks [24]

Answer:

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Good luck!

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2 years ago