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3 years ago

The graph below shows the velocity of a car as it attempts to set a speed record. At what point is the car the fastest?

Physics

1 answer:

Maksim231197 [3]3 years ago

6 0

Answer:

B). 3.4 s

Explanation:

As we can see the graph is given between velocity and time

so here we can see that the velocity is changing here with time and initially for some time it moves with constant speed

Then it's speed decreases to next few second and then speed increases to its maximum value

The time after which velocity comes to its maximum value will reach after t = 3 s

so out of the all given options most correct option will be

$t = 3.4 s$

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On Planet Y, which has no air, a dropped object falls 9 m in 3 seconds. What is g, the acceleration due to gravity, on that plan

slavikrds [6]

Answer:

a = 2 m/s^2

which agrees with the third answer option provided.

Explanation:

Recall the kinematic formula for displacement under the action of a constant acceleration "a":

yf - yi = 1/2 a t^2

using the information provided this equation becomes:

9 = 1/2 a (3)^2

solve for a:

9 * 2 / 9 = a

then a = 2 m/s^2

which agrees with the third answer option provided.

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3 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

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3 years ago

Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up

Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

$v = \sqrt{2as}$

$v = \sqrt{2*18.9*391}$

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

$V= \sqrt{v^2 + 2aS'}$

$V = \sqrt{121.57^2+ 2 (-9.92)(332)}$

V = 90.51 m/s

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3 years ago

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