What is the metric unit for time?

An elevator carrying a person of mass m is moving upward and slowing down. How does the magnitude f of the force exerted on the

V125BC [204]

The force f of the elevator on the man keeps reducing as the elevator keeps going up while the gravitational force mg keeps increasing moving upwards.

<h3>What is an elevator?</h3>

An elevator is an electrical device that lifts people up and down a tall building or structure.

for the elevator to go up, f > mg.

for the elevator to come down mg > f.

Analysis

since the force on the man is f = ma

where a is the acceleration of the elevator, then it means when a increases, f will increase and when it decreases, f would decrease. slowing down means a, is decreasing going up and this reduces the force as the elevator keeps going up.

on the other hand, gravity acts faster on bodies that are slower in motion so since g, increases going up, mg would also increase.

Learn more about forces in an elevator : brainly.com/question/13526583

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6 0

1 year ago

Rahul sits in a chair reading a book. Which force is equal to the force Rahul exerts on Earth?

Mademuasel [1]

The first opiton is the answer A)<span>Rahul’s weight
</span>

8 0

2 years ago

Read 2 more answers

How much does 1 liter of water weigh

Zinaida [17]

Weight = (mass) x (gravity)

On Earth ...

Weight = (1 kg) x (9.8 m/s^2)

Weight = 9.8 Newtons

8 0

2 years ago

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. T

motikmotik

Answer:

V' = 0.84 m/s

Explanation:

given,

Linear speed of the ball, v = 2.85 m/s

rise of the ball, h = 0.53 m

Linear speed of the ball, v' = ?

rotation kinetic energy of the ball

$KE_r = \dfrac{1}{2}I\omega^2$

I of the moment of inertia of the sphere

$I = \dfrac{2}{5}MR^2$

v = R ω

using conservation of energy

$KE_r = \dfrac{1}{2}( \dfrac{2}{5}MR^2)(\dfrac{V}{R})2$

$KE_r = \dfrac{1}{5}MV^2$

Applying conservation of energy

Initial Linear KE + Initial roational KE = Final Linear KE + Final roational KE + Potential energy

$\dfrac{1}{2}MV^2 + \dfrac{1}{5}MV^2 = \dfrac{1}{2}MV'^2 + \dfrac{1}{5}MV'^2 + M g h$

$0.7 V^2 = 0.7 V'^2 + gh$

$0.7\times 2.85^2 = 0.7\times V'^2 +9.8\times 0.53$

V'² = 0.7025

V' = 0.84 m/s

the linear speed of the ball at the top of ramp is equal to 0.84 m/s

6 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

2 years ago