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3 years ago

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As part of a physics experiment, you carry a bathroom scale calibrated in newtons onto an elevator and stand on it. At rest, you

check the scale and it reads 588 N. Then the elevator starts accelerating upward at 2.0 m/s2 and you check the reading again. Now what does the scale show

1 answer:

mafiozo [28]3 years ago

5 0

Answer: 708 N

Explanation:

Given

At rest, Elevator reads 588 N

When it starts moving upward at $2\ m/s^2$ , apparent weight changes

i.e. weight can be given by

$\Rightarrow W'=m(g+a)\\\Rightarrow W'=mg+mg\cdot \dfrac{a}{g}\\\\\Rightarrow W'=W\left(1+\dfrac{a}{g}\right)\\\\\Rightarrow W'=588\left(1+\dfrac{2}{9.8}\right)\\\\\Rightarrow W'=707.99\approx 708\ N$

The apparent weight is 708 N

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Suppose a baseball pitcher throws the ball to his catcher.

amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

6 0

3 years ago

What are two ways in which the suns energy can be captured and used?

My name is Ann [436]

The oldest way ... the way we've been using as long as we've been
walking on the Earth ... has been to use plants. Plants sit out in the
sun all day, capturing its energy and using it to make chemical compounds.
Then we come along, cut the plants down, and eat them. Our bodies
rip the chemical compounds apart and suck the solar energy out of them,
and then we use the energy to walk around, sing, and play video games.

Another way to capture the sun's energy is to build a dam across a creek
or a river, so that the water can't flow past it. You see, it was the sun's
energy that evaporated the water from the ocean and lifted it high into
the sky, giving it a lot of potential energy. The rain falls on high ground,
up in the mountains, so the water still has most of that potential energy
as it drizzles down the river to the ocean. If we catch it on its way, we
can use some of that potential energy to turn wheels, grind our grain,
turn our hydroelectric turbines to get electrical energy ... all kinds of jobs.

A modern, recent new way to capture some of the sun's energy is to use
photovoltaic cells. Those are the flat blue things that you see on roofs
everywhere. When the sun shines on them, they convert some of its
energy into electrical energy. We use some of what they produce, and
we store the rest in giant batteries, to use when the sun is not there.

7 0

3 years ago

Read 2 more answers

1. A stationary 50 tonne submarine fires a missile of mass 40kg at

White raven [17]

The velocity of the submarine immediately after firing the missile is 0.0104 m/s

Explanation:

Mass of the submarine M=50 tonne= $50\times 1000=50000kg$

Mass of the missile m=40 kg

velocity of the missile v= 13m/s

we have to calculate the velocity of the submarine after firing

This is the recoil velocity and its expression is derived from the law of conservation of momentum

recoil velocity of the submarine

$V=-mv/M\\=-40\times 13/50000\\=0.0104 m/s$

8 0

3 years ago

Choose the true statement concerning the mall or subatomic particles:

solniwko [45]

C is true, and just one of those has as much mass as about 1,840 electrons.

4 0

3 years ago

Read 2 more answers

An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and

PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

$E_i+K_i+W_f=K_f+U_f$

The work done by the friction force is given by:

$W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]$

so:

$\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m$

3 0

3 years ago