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3 years ago

5

What would be the path of an object thrown in the air if there was no gravity?

2 answers:

alex41 [277]3 years ago

8 0

Answer:D) in a straight line.

Hey

Newton's first law says that if an object is at rest it will stay at rest. But if it is moving it will continue moving in a straight line if there is no external force. If there where no gravity the object that you throw will keep going in a straight line according to Newton's first law.

Svetllana [295]3 years ago

7 0

Answer:

What would be the path of an object thrown in the air if there was no gravity?

the answer is D

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You drop a rock down a well that is 5.4 m deep. How long does it take the rock to hit the bottom of the well?

Natali5045456 [20]

Equation of motion:

$y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}$

Since initial velocity is zero, the second term goes away:

$y_{f}=y_{o}+0+ \frac{1}{2} at^{2}$

$y_{f}=y_{o}+\frac{1}{2} at^{2}$

$y_{f}-y_{o}= \frac{1}{2} at^{2}$

$y_{f}-y_{o}=5.4m$

$5.4m= \frac{1}{2} at^{2}$

$\frac{2(5.4m) }{a} = t^{2}$

$a = g = 9.81 \frac{m}{ s^{2}}$

$\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}$

$1.1 s^{2} = t^{2}$ $\sqrt{1.1 s^{2}} = \sqrt{t^{2}}$

<u><em> $t = 1.05s$ </em></u>

4 0

3 years ago

Read 2 more answers

Match these items.

horrorfan [7]

Iron oxide = small region within a magnet
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4 0

3 years ago

A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i

tresset_1 [31]

Answer:

a) Acceleration of the car is given as

$a_{car} = -21 m/s^2$

b) Acceleration of the truck is given as

$a_{truck} = 10.15 m/s^2$

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

$m_1v_1 + m_2v_2 = (m_1 + m_2)v$

now we have

$1400 (6.32) + 2900(0) = (1400 + 2900) v$

$v = 2.06 m/s$

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

$a_{car} = \frac{v_f - v_i}{t}$

$a_{car} = \frac{2.06 - 6.32}{0.203}$

$a_{car} = -21 m/s^2$

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

$a_{truck} = \frac{v_f - v_i}{t}$

$a_{truck} = \frac{2.06 - 0}{0.203}$

$a_{truck} = 10.15 m/s^2$

5 0

3 years ago

A boy weighs 50 kg and is running with 225 J of energy, what is his velocity?

Oksanka [162]

0.22

Explanation:

50 ÷ 225 = 0.22

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8 0

3 years ago

Read 2 more answers

A 16000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5400 kg additional load

Alisiya [41]

Answer:

The speed of the car when load is dropped in it is 17.19 m/s.

Explanation:

It is given that,

Mass of the railroad car, m₁ = 16000 kg

Speed of the railroad car, v₁ = 23 m/s

Mass of additional load, m₂ = 5400 kg

The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :

$m_1v_1=(m_1+m_2)v$

$v=\dfrac{m_1v_1}{m_1+m_2}$

$v=\dfrac{16000\ kg\times 23\ m/s}{(16000+5400)\ kg}$

v = 17.19 m/s

So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.

6 0

3 years ago