The (nonconservative) force propelling a 1.50 103-kg car up a mountain road does 5.00 106 J of work on the car. The car starts from rest at sea level and has a speed of 25.0 m/s at an altitude of 1.90 102 m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are non conservative forces.

Answers

Answer 1

Answer:

car. The car starts from rest at sea level and has a speed of 29.0 m/s at an altitude of 2.20x10^2 m above sea level. Obtain the work done on the car by the combined forces of friction and air resistance, both of which are nonconservative forces.

Explanation:

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An article in a magazine denies that Earth is warming and climate change is occurring. The writer reports that this season was the snowiest winter on record for the Northeast, with lower than average temperatures for the region. Because of the low temperatures, the writer concludes that climate change cannot possibly be occurring. What question should you ask the writer to clarify whether his stance on climate change is scientifically sound?(A) Did the snowstorms cause extensive damage?
(B) Was this the coldest month on record for this region?
(C) Were the snowfall totals of each storm larger or smaller than last year?
(D) Were long-term temperature changes investigated in addition to short-term change?

Answers

D.
This is likely to challenge the writer's stance because one piece of data cannot accurately depict a long-term trend.`

Answer:

Explanation:

The kinetic energy of the diver is 5040 J. The speed of the diver is 12 m/s. Calculate the mass of the diver.What's the mass of the diver?

Answers

Answer:

70 kg

Explanation:

We know that,

Kinetic energy = 1/2 mv²

Where m is mass and v is velocity or speed.

Putting the required values,

5040 = 1/2 × m ( 12 × 12)
5040 × 2 = m × 144
10080 = m × 144

divide both side by 144

m = 70 kg

So the mass of the diver is 70 kg respectively.

After landing from your skydiving experience, you are so excited that you throw your helmet upward. The helmet rises 5.8 m above your hands. What was the initial speed of the helmet when it left your hands? Express your answer to three significant figures and include the appropriate units How long was it moving from the time it left your hands until it returned? Express your answer to three significant figures and include the appropriate units.

Answers

Answer:

Initial velocity: approximately $10.7\; {\rm m\cdot s^(-1)}$ .

Time taken before return to initial height: approximately $2.17\; {\rm s}$ .

(Assumptions: $g = 9.81\; {\rm m\cdot s^(-2)}$ ; air resistance is negligible.)

Explanation:

Under the assumption, acceleration of the helmet would be constantly $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ .

During the interval between being thrown upward and reaching maximum height:

Initial velocity immediately after the person threw the helmet upward, $u$ , needs to be found.
Final velocity when the helmet reached maximum height is $v = 0\; {\rm m\cdot s^(-1)}$ .
Change in position over this interval is $x = 5.8\; {\rm m}$ .
Acceleration is $a = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}$ .

Apply the following SUVAT equation to find initial velocity $u$ :

$\begin{aligned} u &= \sqrt{v^(2) - 2\, a\, x} \n &= \sqrt{(0)^(2) - 2\, (-9.81)\, (5.8)} \; {\rm m\cdot s^(-1)}\n &\approx 10.668\; {\rm m\cdot s^(-1)} \n &\approx 10.7\; {\rm m\cdot s^(-1)}\end{aligned}$ .

(Round to three significant figures for the final result, but keep more significant figures for intermediary values.)

In other words, the velocity of the helmet was approximately $10.7\; {\rm m\cdot s^(-1)}$ immediately after the person threw the helmet upward.

Right before returning to the initial height, the velocity of the helmet would be the opposite of its initial velocity: $(-u) \approx (-10.7)\; {\rm m\cdot s^(-1)}$ .

The change in velocity would be:

$\begin{aligned}((-u) - u)/(a) &\approx ((-10.668) - 10.668)/((-9.81))\; {\rm s} \n &\approx 2.17\; {\rm m\cdot s^(-1)}\end{aligned}$ .

(Rounded to three significant figures.)

Final answer:

The initial speed of the helmet was 10.7 m/s and it was in the air for a total of 2.18 s.

Explanation:

This problem involves concept from physics specifically kinematics. Kinematics helps us study the motion of objects. To solve this problem, we need to use the second equation of motion: v²=u²+2as. In this case, the final speed (v) is 0 (when the helmet reaches the highest point, its velocity becomes 0), acceleration (a) is -9.8 m/s² (gravity acts downwards), and the distance (s) is 5.8 m.

Plugging in these values we get: 0 = u² - (2 * 9.8 * 5.8). Solving for u (initial velocity), we get u = √(2 * 9.8 * 5.8) = 10.7 m/s. This is the initial speed of the helmet when it left your hands.

To find out how long the helmet was in the air, we can use the first equation of motion: v = u + at. Solving for t (time), we get: t = (v - u) / a = (0 - 10.7) / -9.8 = 1.09 s going up. Because the time going up and coming down is the same, the total time the helmet was moving is 2 * 1.09 = 2.18 s.

Learn more about Kinematics here:

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A soccer player kicks a ball with a speed of 30 m/s at an angle of 10. How long does the ball stay in the air?

Answers

"10" is not a very satisfying description of an angle.
I'll assume it's 10 degrees, and that it's the angle of the initial trajectory
above the horizontal.

The vertical component of the ball's speed is 30 sin(10°) = 5.209 meters per second.

The acceleration of gravity is 9.8 meters per second² downward.

The ball continues to rise for 9.8/5.209 = 1.881 seconds after the kick.
At that time, it reaches the peak of its arc, and if nobody else has contacted it yet,
it begins to fall, and takes the same length of time to return to the ground whence
it was kicked.

So the time between the kick and return to the ground is (2 x 1.881) = 3.763 seconds.

Note: As is customary and necessary at the high school level,
we assume no air resistance.

Which of the following is an example of distance? A. A bird flies at 5 meters per second. B. A man is standing on a street corner. C. A cat travels 5 meters from the front door of a house to the edge of the street. D. A racecar increases its speed from 5 meters per second to 10 meters per second.

Answers

C.A cat travels 5 meters from the front door of a house to the edge of the street

C would be your answer unless its a trick question.

A car that travels from point A to point B in four hours, and then from point B back to point A in six hours. The road between point A and point B is perfectly straight, and the distance between the two points is 240 km. What is the car’s average velocity?

Answers

Note : Be careful with speed and velocity, even the formula is same that $v = (x)/(t)$

1. If we would like to find the velocity, we must calculate the distance from initial to final, $x_(f) - x_(i) = (delta)x$ ,

2. If we would like to find the speed, we must calculate the total distance that travelled ( $x_(total)$ .

So, in this question, because it return back to its initial point so (delta)x = 0, so average velocity is 0 m/s.

If you mean average speed, it is 480/10 = 48 kn/h

If you mean average velocity, it is 0/10 = 0 km/h

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