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Aloiza [94]
2 years ago
3

You’re exploring a cave. Standing in a low cavern, you steady yourself by pushing straight up on the rock ceiling. Identify the

forces acting on you

Physics
1 answer:
Debora [2.8K]2 years ago
4 0

Explanation:

The attached figure shows the free diagram of the person standing in a low cavern.

The forces acting on the person are as follows :

1. Its weight, mg (in downward direction)

2. F₁ = Force exerted by the Earth on the person (in upward direction)

3. F₂ = Force exerted by the ceiling on the person (in downward direction)

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Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
2 years ago
When the balloon hits the ground, it rebounds slightly. What is the source of the energy for this rebound? A. When the balloon h
nevsk [136]

Answer:

The correct answer is c.    When the balloon hits the ground, the rubber envelope stretches, storing elastic potential energy; this elastic potential energy is converted to the gravitational potentiaL

Explanation:

Let's analyze the situation of the globe

When it touches the ground, the part that is in contact decreases its velocity to zero, but the upper part of the ball continues to move, which creates that the molecules approach slightly, if we approximate the spring links, a repulsive force is created that after all the particles reach zero speed. The force of the springs moves the ball up until the force decreases to zero.

We can relate this force of Hooke with an elastic energy

This energy can be stored in the deformation of the system, as elastic potential energy, which is subsequently transformed into gravitational potential energy when the balloon is lifted.

The correct answer is c

7 0
2 years ago
¿Alguien me puede ayudar? Problema: Un niño le pide gastada a su papá y éste le contesta que le dará los $120 que tiene en su bo
VLD [36.1K]

Answer: there are 15 coins of $2 and 18 coins of $5

Explanation:

I will answer in English.

X is the number of $5 coins.

Y is the number of $2 coins.

We have the system of equations:

Y + X = 33

Y*2 + X*5 = 120

first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:

Y = 33 - X.

Then we can replace it in the other equation:

(33 - X)*2 + X*5 = 120

66 - X*2 + X*5 = 120

X*3 = 54

X = 54/3 = 18

and using the equation for Y.

Y = 33 - X = 33 - 18 = 15

So there are 15 coins of $2 and 18 coins of $5

3 0
2 years ago
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
Serggg [28]

a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

L=I\omega

where

I is the moment of inertia

\omega is the angular speed

Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

\omega_1 = 5.00 rad/s is the initial angular speed

I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

\omega is the angular speed

Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

7 0
2 years ago
The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.
Alisiya [41]

Answer:

Explanation:

The pail is rotated at a constant rate in vertical circular path  so it has the minimum speed at all points along its circular path . That means at top position the velocity is almost zero. In that case the centripetal force at top position will be provided by its weight or

mg = mv² / r ( r is radius of  vertical circular path )

v = √ rg

At the bottom position its velocity will be increased due to loss of potential energy

so 1/2 m V² = 1/2 m v² + mg x 2r  

V =√ 5 gr

If R be the reaction force at the bottom by bottom of pail

R - mg = mV² / r

R = mg +mV² / r

= mg + m x 5gr / r

R = 6mg

This is the magnitude of the force exerted by the water on the bottom of the pail .

7 0
2 years ago
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