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Nezavi [6.7K]
3 years ago
13

A student is standing in an elevator that is

Physics
1 answer:
daser333 [38]3 years ago
4 0
The force that the student exerts on the floor of the elevator must be "(1) less than the weight of the student when at <span>rest" since the elevator is "beating" him downward.</span>
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If you exert a force of 100.0 N to lift a box a distance of 0.5 m, how much work do you do? 200 J 400 J 50 J 26 J
jeyben [28]
Work = force x distance
= 100N (force) x 0.5m (distance)
=  50J
4 0
3 years ago
A 94.7 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.75 r
Marta_Voda [28]

Answer:

The angular velocity of the platform is 1.114 rad/s.

Explanation:

Step 1:  Given data

Mass of the horizontal circular platform = 94.7 kg

Mass of the monkey = 21.1 kg

Initial angular velocity = 1.75 rad/s

A monkey drops a 9.25 kg bunch of bananas

They hit the platform at 4/5 of its radius from the center

Model the platform as a disk of radius 1.63 m

Step 2: Calculate the moment of inertia of the disk

I = ½ * m * r² = ½ * 94.7 * 1.63² = 125.80

Step 3: Calculate the initial angular momentum

I = 125.80 * 1.75 = 220.15

Step 4: Calculate the moment of inertia for the bananas

For the bananas, r = 4/5 * 1.63 = 1.304 m

I = 9.25 * 1.304² = 15.73

Step 5: Calculate Moment of inertia for the monkey

I = 21.1 * 1.63² = 56.06

Step 6: Total moment of inertia = 125.80 + 15.73 + 56.06 = 197.59

Step 7: Calculate final angular momentum = 197.59 * ω

197.59 * ω = 220.15

ω = 220.15 / 197.59

This is approximately 1.114 rad/s.

3 0
3 years ago
A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point
Mandarinka [93]

Answer:

F=9.09\times 10^{-9} N

Explanation:

We are given that

Mass of spherical shell,m_1=1900 kg

Mass=m_2=1.80 kg

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

F=G\frac{m_1m_2}{r^2}

G=6.67\times 10^{-11} Nm^2/kg^2

Using the formula

F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}

F=9.09\times 10^{-9} N

Hence,the gravitational force =F=9.09\times 10^{-9} N

6 0
3 years ago
A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag
Kobotan [32]
Do you still need help with this question??
5 0
3 years ago
scott and jafar are ready to work a problem. in this simulation, assume the coordinates of the points are as follows. a (0 s, 30
max2010maxim [7]

Answer:

Explanation:

point a represents time 0 and position coordinate 30

point b represents time 10 s , and position coordinate 50 m .

time elapsed = 10 - 0 = 10 s .

displacement = 50 m - 30 m

= 20 m

average velocity = displacement / time elapsed

= 20 / 10

= 2 m /s .

7 0
3 years ago
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