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soldi70 [24.7K]
3 years ago
10

A 700 kg car makes a turn going at 30 m/s with radius of

Physics
1 answer:
Gnoma [55]3 years ago
5 0

Answer:5250 N

Explanation: ig:iihoop.vince

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You pull with a force of 295 N on a rope that is attached to a block of mass 22 kg, and the block slides across the floor at a c
Sergeeva-Olga [200]

Answer:

Fnet = 0

Explanation:

  • Since the block slides across the floor at constant speed, this means that it's not accelerated.
  • According Newton's 2nd Law, if the acceleration is zero, the net force on the sliding mass must be zero.
  • This means that there must be a friction force opposing to the horizontal component of the applied force, equal in magnitude to it:

       F_{appx} = F_{app} * cos \theta = 295 N * cos 35 = 242 N  (1)

  • In the vertical direction, the block is not accelerated either, so the sum of the normal force and the vertical component of the applied force, must be equal in magnitude to the force of gravity on the block:

      F_{appy} = F_{app} * cos \theta = 295 N * sin 35 = 169 N  (2)

⇒    169 N + Fn = Fg = 216 N  (3)

  • This means that there must be a normal force equal to the difference between Fappy and Fg, as follows:
  • Fn = 216 N - 169 N = 47  N (4)

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2 years ago
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natali 33 [55]
The layers form from sand dunes
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3 years ago
What is the most abundant element in the universe?<br> A. helium<br> B. hydrogen<br> C. carbon
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The answer is Carbon
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Select the correct answer. The motion of a car on a position-time graph is represented with a horizontal line. What does this in
enyata [817]

If the car's motion appears as a horizontal line on a <u><em>position-time </em></u>graph, it shows that as time changes, the car's position doesn't change.

This is just a complicated way to say that the car is <em>not moving</em>.<em> (A)</em>

7 0
3 years ago
A cyclist is coasting at 15 m/s when she starts down a 450 m long slope that is 30 m high. The cyclist and her bicycle have a co
Flura [38]

Answer:

Her speed at the bottom of the slope is 25.665 m/s

Explanation:

Here we have

Initial velocity, v₁= 15 m/s

Final velocity = v₂

The energy balance present in the system can be represented as

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = W

Where:

m = Mass of the cyclist = 70 kg

W = work done by the drag force = -F_Dd

Where:

d = Distance traveled = 450 m

Therefore,

\frac{1}{2}mv_2^2 -\frac{1}{2}mv_1^2 - mgh = -F_Dd and

v_2^2 =\frac{ \frac{1}{2}mv_1^2 + mgh  -F_Dd}{ \frac{1}{2}m}  = v_1^2 + 2gh -\frac{   2F_Dd}{ m} = 15^2 + 2\times 9.8\times 30 - \frac{2\times 12\times 450}{70}

= 658.714 m²/s²

v₂ = 25.665 m/s

Her speed at the bottom of the slope = 25.665 m/s.

6 0
3 years ago
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