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solmaris [256]
3 years ago
11

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the displacement of the object? the area of the rectangle under the line the area of the rectangle above the line the area of the rectangle plus the area of the triangle under the line the area of the rectangle plus the area of the triangle above the line
Physics
1 answer:
Nookie1986 [14]3 years ago
8 0

Explanation:

We have,

The initial position of an object is zero.

The starting velocity is 3 m/s and the final velocity was 10 m/s.

The object moves with constant acceleration..

The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".

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A paper airplane is thrown westward at a rate of 6 m/s. The wind is blowing at 8 m/s towards the north. Which of the following d
Andrew [12]

since airplane is thrown towards west with speed 6 m/s

while air is blowing with speed 8 m/s towards north

so here the net speed of air plane will be the resultant of airplane speed and wind speed always

SO here we can say it would be a combination of vector along west with must be of length 6 m/s and other vector is towards north with is of length 8 m/s

so correct answer must be 1st option


5 0
3 years ago
Point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergo
oksian1 [2.3K]

Answer:

e). a = 0.066 m/s^2

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

\omega_f^2 - \omega_i^2 = 2\alpha \theta

now we have

\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})

\omega = 0.177 rad/s

now the centripetal acceleration of the point P is given as

a_c = \omega^2 R

a_c = (0.177)^2(2)

a_c = 0.063 m/s^2

tangential acceleration is given as

a_t = R\alpha

a_t = 2(0.01)

a_t = 0.02 m/s^2

now net acceleration is given as

a = \sqrt{a_t^2 + a_c^2}

a = \sqrt{0.02^2 + 0.063^2}

a = 0.066 m/s^2

8 0
3 years ago
How does the temperature of water affect the speed of the sound waves?
nikitadnepr [17]

Answer:

Temperature is also a condition that affects the speed of sound. Heat, like sound, is a form of kinetic energy. Molecules at higher temperatures have more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly.

5 0
3 years ago
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3 years ago
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a car whose initial speed is 30 seconds m / s slows uniformly to a stop in 5.00 seconds what was the cars displacement ​
CaHeK987 [17]

Answer:

60

Explanation:

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3 years ago
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