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liq [111]
3 years ago
13

If a 1200kg car travels 20 m/s and slows down for a stop, what is its momentum?​

Physics
1 answer:
frosja888 [35]3 years ago
3 0

Answer:

24000kgm/s

Explanation:

Given parameters:

Mass of car  = 1200kg

Velocity = 20m/s

Unknown:

Momentum  = ?

Solution:

Momentum is the amount of motion a body possesses.

   Momentum  = mass x velocity

 Now insert the parameters and solve;

 Momentum  = 1200 x 20  = 24000kgm/s

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Please help which one is correct
Natalija [7]

Answer:

option (i) is correct

Explanation:

as there is no air resistance, no force is acting on the object horizontally, but gravitational acceleration will obviously act, regardless of the air resistance... option (i) is correct

8 0
3 years ago
A spring is 6.0cm long when it is not stretched, and 10cm long when a 7.0N force is applied. What force is needed to make it 20c
Artist 52 [7]

Answer:

Approximately 25\; {\rm N} (assuming that this spring is ideal.)

Explanation:

The displacement of a spring is the new length of the spring relative to the original length.

For example:

  • When the 6.0\; {\rm cm}-spring in this question is stretched to 10\; {\rm cm}, the displacement is x = (10\; {\rm cm} - 6.0\; {\rm cm}).
  • Likewise, if this spring is stretched to 20\; {\rm cm}, the displacement would be (20\; {\rm cm} - 6\; {\rm cm}).

If this spring is ideal, the force on the spring would be proportional to the displacement of the spring. In other words, if a force of F_{\text{a}} displaces this spring by x_{\text{a}}, while a force of F_{\text{b}} displaces this spring by x_{\text{b}}, then:

\displaystyle \frac{F_{\text{a}}}{x_{\text{a}}} = \frac{F_{\text{b}}}{x_{\text{b}}}.

In this question, it is given that a force of F_{\text{a}} = 7.0 \; {\rm N} would stretch this spring by x_{\text{a}} = (10\; {\rm cm} - 6.0\; {\rm cm}). Thus, the force F_{\text{b}} required to stretch this spring by x_{\text{a}} = (20\; {\rm cm} - 6.0\; {\rm cm}) would satisfy:

\displaystyle \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}}= \frac{F_{\text{b}}}{20\; {\rm cm} - 6.0\; {\rm cm}}.

Rearrange and solve for F_{\text{b}}:

\begin{aligned} F_{\text{b}} &= \frac{7.0\; {\rm N}}{10\; {\rm cm} - 6.0\; {\rm cm}} \, (20\; {\rm cm} - 6.0\; {\rm cm}) \\ &\approx 25\; {\rm N}\end{aligned}.

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2 years ago
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podryga [215]
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3NaOH + FeCl3 → 3NaCl + Fe(OH)3
8 0
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