If the net force acting on an object is 100 N to the left, Are the forces unbalanced or balanced?

The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di

drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

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3 0

1 year ago

Identify the following physical quantities as scalars or vectors.

34kurt

Position displacement velocity acceleration are vectors and the rest are scalars

6 0

4 years ago

Read 2 more answers

Is the us warmer than antarctica?

Likurg_2 [28]

Answer:

Yes

Explanation:

Antarctica is usually always cold.

7 0

3 years ago

7. What is the acceleration of the box?<br> a. 2.5 m/s2<br> b. 4 m/s2<br> c. 6 m/s2<br> d. 10 m/s2

faust18 [17]

This is your answer
b) 4m/s2
Here’s my work

6 0

3 years ago

A 22-g bullet traveling 265 m/s penetrates a 1.9 kg block of wood and emerges going 125 m/s .

Usimov [2.4K]

The body moves at a velocity of 1.62m/s after the bullet emerges.

<h3>Given:</h3>

Mass of bullet, $m_1$ = 22g

= 0.022 kg

Mass of the block, $m_2$ = 1.9 kg

Velocity of bullet , $v_1$ = 265 m/s

$v_2 = 0$

According to the law of collision which states that the momentum of the body before the collision is equal to the momentum of the body after the collision.

After penetration;

$v^{'}_1 =125 m/s$

$v^{'}_2=?$

The formula for calculating the collision of a body is expressed as:

p = mv

m is the mass of the body

v is the velocity of the body

∴ Momentum before = Momentum after

Substitute the given parameters into the formula as shown:

$m_1v_1+ m_2v_2 = m_1v^{'}_1+ m_2v^{'}_2\\0.022* 265 + 0 = 0.022*125+1.9*v^{'}_2\\5.83 = 1.9 v^{'}_2\\v^{'}_2 = 1.62 m/s$

Therefore, It moves with a velocity of 1.62 m/s.

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6 0

2 years ago