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2 years ago

A car at a position 150 km [W] of Toronto travels to a position 400 km [W] of

Physics

1 answer:

tigry1 [53]2 years ago

6 0

Answer:

The total displacement of the car is 250 km.

Explanation:

Given that,

A car at a position 150 km [W] of Toronto travels to a position 400 km [W] of Toronto.

Total displacement = final position - initial position

= 400 km - 150 km

= 250 km

So, the total displacement of the car is 250 km.

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Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway

SashulF [63]

Answer:

v = 12.12 m/s

Explanation:

Given that,

The mass of the cart, m = 75 kg

The roller coaster begins 15 m above the ground.

We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

$mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s$

So, the velocity of the cart is 12.12 m/s.

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2 years ago

What three steps can you take to ensure your measurments are both accurate and precise?

Ivan

First, use a high-quality measurement tool. Next, measure carefully. Finally, repeat the measurement a few times. Hope it helps!

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2 years ago

A elephant kicks a 5.0\,\text {kg}5.0kg5, point, 0, start text, k, g, end text stone with 150\,\text J150J150, start text, J, en

S_A_V [24]

The speed of the stone is 7.7 m/s

Explanation:

The kinetic energy of a body is the energy possessed by the body due to its motion. Mathematically,

$K=\frac{1}{2}mv^2$

where

m is the mass of the body

v is its speed

For the stone in this problem, we have:

K = 150 J is its kinetic energy

m = 5.0 kg is its mass

Re-arranging the equation for v, we find the speed of the stone:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(150)}{5.0}}=7.7 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

3 years ago

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed of a

Pie

Answer:

Part a)

$v = 7407.1 m/s$

Part b)

$v_{rel} = 1.05 \times 10^4 m/s$

Explanation:

Part a)

As we know that orbital velocity at certain height from the surface of Earth is given as

$v = \sqrt{\frac{GM}{R+h}}$

here we know that

$M = 5.98 \times 10^{24} kg$

$R = 6.37 \times 10^6 m$

$h = 900 km = 9.0 \times 10^5 m$

now we have

$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{6.37 \times 10^6 + 9.0 \times 10^5}}$

$v = 7407.1 m/s$

Part b)

When a loose rivet is moving in same orbit but at 90 degree with the previous orbit path then in that case the relative speed of the rivet with respect to the satellite is given as

$v_{rel} = \sqrt{2} v$

$v_{rel} = 1.05 \times 10^4 m/s$

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2 years ago

an object on a planet has a mass of 243 Kg. what is the acceleration of the object, if the radius of the planet is 2.32 x 10^7m

Ksenya-84 [330]

Answer:

7.87x10^5m/s^2

Explanation:

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2 years ago