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Lorico [155]
3 years ago
6

Suppose a bicycle was coasting on a level surface, and there was no friction. What would happen to the bicycle?

Physics
2 answers:
andrew-mc [135]3 years ago
8 0

Answer: B. It would keep going at a constant speed

Explanation: I got this correct in Study Island :)

balandron [24]3 years ago
3 0
It will be ccccccccccccccccccccccccccc
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The x vector component of a displacement vector has a magnitude of 146 m and points along the negative x axis. The y vector comp
larisa86 [58]

Answer:  

a) the magnitude of r is  184.62

b) the direction is 37.74° south of the negative x-axis

   

Explanation:

Given the data in the question;

as illustrated in the image blow;

To find the the magnitude of r, we will use the Pythagoras theorem

r² = y² + x²

r = √( y² + x²)

we substitute

r = √((-113)² + (-146)²)

r = √(12769 + 21316 )

r = √(34085 )

r = 184.62

Therefore, the magnitude of r is  184.62

To find its direction, we need to find ∅

from SOH CAH TOA

tan = opposite / adjacent

tan∅ = -113 / -146

tan∅ = 0.77397

∅ = tan⁻¹( 0.77397 )

∅ = 37.74°

Therefore, the direction is 37.74° south of the negative x-axis

7 0
3 years ago
Which oftthe following is the least reliable source of background information for a scientific project? General internet site, g
kenny6666 [7]
General internet site
7 0
3 years ago
a 5.0 kg ball is dropped from a 2.5 m high window. what is the velocity of the ball just before it hits the ground?
julsineya [31]

Answer:

Approximately 7.0\; \rm m \cdot s^{-1}.

Assumption: the ball dropped with no initial velocity, and that the air resistance on this ball is negligible.

Explanation:

Assume the air resistance on the ball is negligible. Because of gravity, the ball should accelerate downwards at a constant g = 9.81 \; \rm m \cdot s^{-2} near the surface of the earth.

For an object that is accelerating constantly,

v^2 - u^2 = 2\, a \, x,

where

  • u is the initial velocity of the object,
  • v is the final velocity of the object.
  • a is its acceleration, and
  • x is its displacement.

In this case, x is the same as the change in the ball's height: x = 2.5\; \rm m. By assumption, this ball was dropped with no initial velocity. As a result, u = 0. Since the ball is accelerating due to gravity, a = 9.81\; \rm m \cdot s^{-2}.

v^2 - u^2 = 2\, g \cdot h.

In this case, v would be the velocity of the ball just before it hits the ground. Solve for

v^2 = 2\, a\, x + u^2.

\begin{aligned}v &= \sqrt{2\, a\, x + u^2} \\ &= \sqrt{2\times 9.81 \times 2.5 + 0} \\ &\approx 7.0\; \rm m\cdot s^{-1}\end{aligned}.

3 0
3 years ago
A 6 kg bowling ball is accelerated at a rate of 2.3 m/s² down the lane. How much force was necessary to produce this acceleratio
Maurinko [17]
The answer is D hoped this helped
4 0
3 years ago
What is the sentence that we use to<br> remember how to convert metric<br> prefixes?
Marrrta [24]

Answer:

The mnemonic I can use to memorize the metric prefixes in this order is: Gigantic Monsters Killed One Million Men Napping Peacefully. All right, so again, gigantic monsters killed one million men napping peacefully.

8 0
3 years ago
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