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3 years ago

Find the average speed of a horse that traveled east for 25km in 4 hours.

Physics

1 answer:

MArishka [77]3 years ago

4 0

The direction the horse travels has nothing to do with its speed but...

The horse would be traveling 6.25 km/hr

You simply take the overall travel (25km) and divide it by the total time (4hr)

You get 6.25km as your answer, meaning that every hour the horse traveled 6.25km every hour.

4 hours later the horse has reached 25km

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A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?

AlexFokin [52]

The faster car behind is catching up/closing the gap/gaining on
the slow truck in front at the rate of (90 - 50) = 40 km/hr.

At that rate, it takes (100 m) / (40,000 m/hr) = 1/400 of an hour
to reach the truck.

(1/400 hour) x (3,600 seconds/hour) = 3600/400 = 9 seconds, exactly

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3 years ago

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Two children sit on different sides of a seesaw. The first child of mass 27 kg sits 1.5 m from the center. How far must the seco

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Explanation:

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3 years ago

Two metra trains approach each other on separate but parallel tracks. one has a speed of 90 km/hr, the other 80 km/hr. initially

Gennadij [26K]

The trains take 57.4 s to pass each other.

Two trains A and B move towards each other. Let A move along the positive x axis and B along the negative x axis.

therefore,

$v_A=90 km/h\\ v_B=-80 km/h$

The relative velocity of the train A with respect to B is given by,

$v_A_B=v_A-v_B\\ =(90km/h)-(-80km/h)\\ =170km/h$

If the train B is assumed to be at rest, the train A would appear to move towards it with a speed of 170 km/h.

The trains are a distance d = 2.71 km apart.

Since speed is the distance traveled per unit time, the time taken by the trains to cross each other is given by,

$t= \frac{d}{v_A_B}$

Substitute 2.71 km for d and 170 km/h for $v_A_B$

$t= \frac{d}{v_A_B}\\ =\frac{2.71 km}{170 km/h} \\ =0.01594 h$

Express the time in seconds.

$t=(0.01594h)(3600s/h)=57.39s$

Thus, the trains cross each other in 57.4 s.

6 0

3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

g An electric hot plate raises its own internal energy and the internal energy of a cup of water by 9000J, and there is at the s

DochEvi [55]

Answer:

11700j

Explanation:

add the two because the plate has to maintain the temp.

2700+9000=11700

3 0

2 years ago