The graph below shows the speed of a car as it drives along a racetrack.

Physics

1 answer:

nordsb [41]2 years ago

4 0

Answer:

Average speed = distance/time

From 1 to 9 seconds:

Distance covered = 1 - 0.2 = 0.8 km

Time = 9 - 1 = 8 sec

Average speed = 0.8 km / 8 sec

Average speed = 0.1 km/s .

The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.

Surprise surprise ! The area under a speed/time graph is the distance covered during that time !

In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.

Explanation:

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Two cars, both of mass m, collide and stick together. Prior to the collision, one car had been traveling north at a speed 2v, wh

mel-nik [20]

Answer: $u=\frac{v}{2}\sqrt{5-4sin\phi }$

Explanation:

Given

Both cars mass is m

and solving problem in Vertical and horizontal direction

considering + y and +x to be positive and u be the final velocity of system

Conserving Momentum in Vertical direction

$m(2v)+m(-vsin\phi )=2m(ucos\theta )$

$2ucos\theta =v(2-sin\theta )$ ------1

Conserving momentum in x direction

$mvcos\phi =2musin\theta$ -----2

squaring and adding 1 &2

$(2u)^2=(2v-vsin\phi )^2+(vcos\phi )^2$

$4u^2=4v^2+v^2-4v^2sin\phi$

$4u^2=5v^2-4v^2sin\phi$

$u=\frac{v}{2}\sqrt{5-4sin\phi }$

7 0

2 years ago

(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively