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user100 [1]
2 years ago
10

A truck skids for a distance of 25 m with the road pushing on its tires with force of 1500 N as its brakes

Physics
2 answers:
Ray Of Light [21]2 years ago
8 0

Answer: -38000 J

Explanation:

Khan approved

MakcuM [25]2 years ago
6 0

Answer:-6800J

Explanation: 8.0m x 850N = 6800

Rewritten as a negative when brake/stop -6800

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Which of the following is a major ocean on Earth?
yarga [219]
The Indian Ocean. 
The rest of your options are not oceans, they are bodies of water. (e.g: lakes,rivers,gulfs) 
3 0
3 years ago
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A train starting from rest picks up a speed of 20 m/s in 20 s while travelling on a straight path. It continues to move at the s
Andrej [43]

Answer:

1300 m

Explanation:

As the path is straight, so the speed is equivalent to velocity. Now. assuming that the acceleration and deceleration of the train are constant. So, change of velocity with respect to time for acceleration as well as deceleration is constant. Hence, the slope of the speed-time graph is constant for the time of acceleration as well as deceleration. The speed for the time from 20 s to 70 s is constant, so slope for this interval of time is zero. The speed-time graph is shown in the figure.

The total distance covered by the train during the entire journey is the area of the speed-time graph.

Area=\frac{1}{2}(20\times 20)+ 50\times 20+\frac{1}{2}(20\times 10)

=200+1000+100

=1300

As velocity is in m/s and time is in s so the unit of area is m

Hence, the total distance is  1300m.

8 0
3 years ago
A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the
liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

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3 years ago
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Cs would be most reactive.
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3 years ago
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What is the speed of the sun as it orbits the center of the galaxy ?
FrozenT [24]
800,000 km/hr is the answr
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