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Alla [95]
3 years ago
8

The coefficients of friction between a race cars tyres and the track surface are

Physics
1 answer:
Luba_88 [7]3 years ago
5 0

the question is about tyres of a race car, which are made of rubber and will be in contact with a race track, which is generally made from asphalt, the static coefficient of friction is in the range of (0.5–0.8), in dry conditions (Source: Friction and Friction Coefficients ).

Explanation:

please mark me as a brainlieast

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From the window of a house that is placed 15 m
kow [346]

Answer:

a) 52.915 m

b) The vertical velocity is approximately 21.092 m/s

The resultant velocity is approximately 26.5 m/s

Explanation:

a) The height of the window in the house from which the water was thrown = 15 m

The speed of the stream of water thrown = 20 m/s

The angle at which the water was thrown = 37° over the horizontal

The acceleration due to gravity, g = 10 m/s²

a) The distance from the base of the house at which the water will fall is given as follows;

y = y₀ + u·t·sin(θ) + 1/2·g·t²

Where;

y = The vertical height reached    

u = The initial velocity

t = Time of flight

From the point the steam of water is thrown, we get;

y₀ = 15 m

Therefore;

y = 15 + 20 × t × sin(37°) - 1/2 × 10 × t²

y = 15 + 20 × t × sin(37°) - 5 × t²

When y = 0, Ground level, we get

0 = 15 + 20 × t × sin(37°) - 5 × t²

5·t² - 20×sin(37°)×t -15 = 0

∴ t = (20 ×sin(37°) ± √((-20 × ·sin(37°))² - 4 × (5) × (-15)))/(2 × 5)

t ≈ 3.3128302, or t ≈ 0.906

Therefore, the time of flight of the water, t ≈ 3.3128302 seconds

The distance from the base of the house at which the water will fall = The horizontal distance travelled by the water, x

x = u·cos(θ)×t

∴ x = 20 × cos(37°) × 3.3128302 ≈ 52.915

The distance from the base of the house at which the water will fall = x ≈ 52.915 m

b) The velocity at which the water will reach the ground, 'v', is given as follows;

The vertical velocity, v_y = u·sin(θ)·t - g·t

At the ground, t ≈ 3.3128302 seconds

∴ v_y = 20 × sin(37) - 10 × 3.3128302 ≈ -21.092

The vertical velocity at which the water will reach the ground, v_y ≈ 21.092 m/s (downwards)

The resultant velocity, v = √(v_y² + vₓ²)

∴ v = √(21.092² + (0 × cos(37°))²) ≈ 26.5

The resultant velocity at which the water will reach the ground, v ≈ 26.5 m/s.

5 0
2 years ago
When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can
andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

F=m\frac{v^2}{r}

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

3 0
2 years ago
If a hypothesis is falsifiable, _____.
san4es73 [151]

Answer:

it must be possible to prove it wrong

Explanation:

4 0
2 years ago
Read 2 more answers
1. The planet Jupiter completes a revolution of the sun in 11.5 years. Express it in seconds. Given that one year= 3.154 × 10^7
xenn [34]

Answer:

The planet Jupiter completes one revolution of the sun in 362710000 seconds. Long time, right?

Explanation:

3.154x10^7=3.154x10000000=31540000

11.5x31540000=362710000

7 0
1 year ago
_______ measures an object's change in position per unit time.
Deffense [45]

I believe your answer should be C. Speed.

6 0
3 years ago
Read 2 more answers
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