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Sav [38]
2 years ago
8

A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?​

Physics
1 answer:
harina [27]2 years ago
6 0

Answer:

impulse = 8820 kg·\frac{m}{s} or 8820 N·s

Explanation:

Impulse J is equal to the average force F_{av} multiplied by the elapsed time Δt or in equation form, J = F_{av}Δt

As long as your force of 450 N is constant then that value is your average force F_{av} and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg·\frac{m}{s} or 8820 N·s.

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Do objects tend to stay moving because of a force called Interia
ahrayia [7]

The first law of Newton’s law state an object in motion will stay in motion and an object at rest will stay at rest unless acted upon with an outside force. Other know as the law of inertia so yes it does

6 0
3 years ago
Sometimes, I have to take my cat, Gigi, in the car. She doesn’t like this, and tends to hide under the couch when she knows she
Reika [66]

Answer:

F - fr = ma ,    N - W = 0

Explanation:

In this exercise we are asked to identify the forces that act on the jack, for this we will use Newton's second law

On the x axis

We have two forces: the friction and the force of the girl who pulls the cat

           F - fr = ma

On the y axis

There are two forces: normal and weight

         N - W = 0

A diagram of these forces can be seen in the attachment

8 0
2 years ago
Two steel guitar strings have the same length. String A has a diameter of 0.513 mm and is under 403 N of tension. String B has a
Mnenie [13.5K]

Answer:

\frac{v_{A}}{v_{B}} = 1.785

Explanation:

T_{A} = Tension force in string A = 403 N

T_{B} = Tension force in string B = 800 N

d_{A} = diameter of string A = 0.513 mm

d_{B} = diameter of string B = 1.29 mm

v_{A} = wave speed of string A

v_{B} = wave speed of string B

Ratio of the wave speeds is given as

\frac{v_{A}}{v_{B}} = \sqrt{\frac{T_{A}}{T_{B}}} \left ( \frac{d_{B}}{d_{A}} \right )

\frac{v_{A}}{v_{B}} = \sqrt{\frac{403}{800}} \left ( \frac{1.29}{0.513} \right )

\frac{v_{A}}{v_{B}} = 1.785

5 0
3 years ago
When a metal bonds with a nonmetal, they form a(n)
MArishka [77]
I believe it would be B.
5 0
3 years ago
If it has enough kinetic energy, a molecule at the surface of the Earth can "escape the Earth's gravitation", in the sense that
user100 [1]

Answer:

K_E=mgr_E

Explanation:

By conservation of energy, the sum of the kinetic and gravitational potential energies at the surface of the Earth must be equal than their sum at infinity, so we have:

K_E+U_E=K_\infty+U_\infty

\frac{mv_E^2}{2}-\frac{GM_Em}{r_E}=\frac{mv_\infty^2}{2}-\frac{GM_Em}{r_\infty}

Where G=6.67\times10^{-11}Nm^2/kg^2 is the gravitational constant,M_E=5.97\times10^{24}kg and r_E=6371000m are the mass and radius of the Earth, <em>m </em>is the mass of the particle, v_E its velocity at the surface of the Earth (which would be its escape velocity) and v_\infty and r_\infty are the velocities and distance at infinity, which would be null and infinity respectively, so the right hand side of our equation is 0J, which leaves us with:

\frac{GM_Em}{r_E}=\frac{mv_E^2}{2}=K_E

Also, since the force the molecule experiments is the force of gravity (disregarding drag), we can write its weight in terms of Newton's Law of Gravitation:

F=mg=\frac{GM_Em}{r_E^2}

Which means that:

\frac{GM_Em}{r_E}=mgr_E

So finally putting all together we can write:

K_E=\frac{GM_Em}{r_E}=mgr_E

4 0
3 years ago
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