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3 years ago

Solve for the dog's displacement during segments A and B (0-20s)?

Physics

2 answers:

Katena32 [7]3 years ago

5 0

The dog walked 5 meters backwards in the first 10 seconds, then 5 meters forward from 10-20 seconds. After 20 sec he was back where he started. Displacement (20sec) = zero.

expeople1 [14]3 years ago

3 0

Explanation:

To find displacement using velocity - time graph, we need to find area under the curve during time period 0 to 20 seconds.

Displacement [ area of Triangle (b=10 units & h=1 units) - area of trapezium (sum of parallel sides=30 units & h=1 unit)

= (1/2 * 10 * 1) - (1/2 * 30 * 1)

= - 10 m

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A mosquito and a truck have a head-on collision. Which has a larger change of momentum?

Maksim231197 [3]

Answer:

C. They have the same change of momentum.

Explanation:

Let "m" be the mass and "v" the velocity.

According to the property of conservation of momentum, when objects 1 and 2 collide:

$m_{a}v_{ai} + m_{b}v_{bi} = m_{a}v_{afinal} + m_{b}v_{bfinal}\\m_{b}v_{bfinal} - m_{b}v_{bi} = m_{a}v_{ai} - m_{a}v_{afinal}\\\Delta (m_{b}v_{b}) = -\Delta (m_{a}v_{a})$

Therefore, one can conclude that the change in momentum (Δ(mv)) has the same magnitude but opposite sign for any two objects regardless of how much heavier one might be than the other.

They have the same change of momentum.

8 0

3 years ago

Guys I really need help with these 2 questions , it's for my final plz help asap

mars1129 [50]

Answer:

(1) Initial speed, $u=0$

Final speed, $v=165.76m/s$

Average speed, $v_a_v_g=82.87m/s$

(2) Force of gravity, $F_g=12.8\times10^1^5N$

Explanation:

(1)

Given,

Distance, $S=300meter$

Time, $t=3.62second$

It is given that drag racer started at rest.

So Initial speed, $u=0$

Using Newton's second equation of motion,

$S=ut+\frac{1}{2}at^2\\300=0+\frac{a\times3.62^2}{2} \\a=45.79m/s^2$

Newton's first equation of motion,

$v=u+at\\=0+45.79\times3.62\\=165.76 m/s$

So, Final speed, $v=165.76m/s$

Average speed is defined as totle distance divided by totle time.

$v_a_v_g=\frac{S}{t}\\=\frac{300}{3.62} \\=82.87m/s$

So, Average speed, $v_a_v_g=82.87m/s$

(2)

Gravitation: It is the natural phenomenon in which two different bodies attract each other by virtue of their masses.

According to Newton's law of gravitation, the force of attraction between two bodies is directly proportional to the masses of the bodies and inversely proportional to square of distance between centers of mass of the bodies.

$F_g\propto\frac{m_1m_2}{r^2} \\F_g=G\frac{m_1m_2}{r^2}$ where $G$ is constant of proportionality and known as gravitation constant.