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3 years ago

3. An object of mass 90 kg travels down a slide.

Physics

1 answer:

Gwar [14]3 years ago

4 0

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

$E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]$

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

$E_{k}=\frac{1}{2} *m*v^{2} \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]$

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Which object has the most gravitational potential energy?

Kipish [7]

Answer: An 8 kg book at a height of 3 m has the most gravitational potential energy.

Explanation:

Gravitational potential energy is the product of mass of object, height of object and gravitational field.

So, formula to calculate gravitational potential energy is as follows.

U = mgh

where,

m = mass of object

g = gravitational field = $9.81 m/s^{2}$

h = height of object

(A) m = 5 kg and h = 2m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 2 m\\= 98.1 J (1 J = kg m^{2}/s^{2})$

(B) m = 8 kg and h = 2 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 2 m\\= 156.96 J (1 J = kg m^{2}/s^{2})$

(C) m = 8 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 8 kg \times 9.81 m/s^{2} \times 3 m\\= 235.44 J (1 J = kg m^{2}/s^{2})$

(D) m = 5 kg and h = 3 m

Therefore, its gravitational potential energy is calculated as follows.

$U = mgh\\= 5 kg \times 9.81 m/s^{2} \times 3 m\\= 147.15 J (1 J = kg m^{2}/s^{2})$

Thus, we can conclude that an 8 kg book at a height of 3 m has the most gravitational potential energy.

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3 years ago

Galileo is heliocentric or geocentric

algol [13]

Galileo is geocentric, just like all the rest of us.

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3 years ago

Read 2 more answers

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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3 years ago

1. Which Law? When you are standing up in a subway train, and the train suddenly stops, your body continues to go forward.

riadik2000 [5.3K]

1. Law 1, since there is no other force acting on your body as you stand there, so you will continue to go forward.

2. Law 3, since the swimmer is using opposite forces to propel herself through the water. She generates a force by pushing the water which helps to push her forward.

3. Law 2, since you are giving the motorcycle more energy as a result of the gas being transformed into the energy that helps to accelerate the motorcycle's speed.

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3 years ago

A 0.50-kg red cart is moving rightward with a speed of 40 cm/s when it collides

lukranit [14]

The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.

The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.

The change in momentum of the system of the carts is 0.

<h3>Initial momentum of the carts before collision</h3>

The momentum of the carts before the collision is calculated as follows;

P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s

P(blue) = 1.5 x 0 = 0

<h3>Momentum of the carts after collision</h3>

The momentum of the carts after the collision is calculated as follows;

P(red) = 0.5 x 0.1 = 0.05 kgm/s

P(blue) = 1.5 0.1 = 0.15 kgm/s

<h3>Change in momentum of the carts</h3>

$\Delta P = P_f - P_i$

ΔP = (0.05 + 0.15) - (0.2)

ΔP = 0

Learn more about momentum here: brainly.com/question/7538238

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2 years ago