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Karolina [17]
3 years ago
8

You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s

peed) brakes and begins to slow down with the largest deceleration possible without skidding. Considering an average human reaction, you press your brakes 0.543 s later. You also brake and decelerate as rapidly as possible without skidding. Assume that the coefficient of static friction is 0.804 between both cars’ wheels and the road.
Physics
2 answers:
Greeley [361]3 years ago
8 0

there is no questions associated with this what are the questions that go with this senario


Bess [88]3 years ago
4 0

Answer:

8.1242 m/s^2

Remaining part of the question:

The acceleration of gravity is 9.8 m/s2 . Calculate the acceleration of the car in front of you when it brakes?

Explanation:

Frictional force allow as a  car to stop from motion, but it is also part of the the force required by the the car to move forward, It help keeps the tires from sliding on the road. if you are to correctly calculate the motion of  a body  across a surface, you must take into account  the interactions  between the body and the surface.

ma=μmg

The acceleration of the car can be calculated as

a=μg=(0.829)(9.8)=8.1242 m/s^2

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What do these compounds have in common H2O cl2 NH3
nasty-shy [4]

Answer:

All the compounds are covalent compounds . This means that they are formed by the sharing of pair of electrons.

7 0
3 years ago
Calculate the index of refraction for a medium in which the speed of light is 2.1x 108 m/s. The speed of light in vacuum is 3x10
strojnjashka [21]

Answer:

n = 1.42

Explanation:

The refractive index for a medium is given by the ratio of the speed of light in vacuum to the speed of light in a medium.

n=\dfrac{c}{v}\\\\n=\dfrac{3\times 10^8}{2.1\times 10^8}\\\\n = 1.42

So, the refractive index of the medium is 1.42.

5 0
3 years ago
How do you do this question? Please include free body diagrams and clear explanation, so I can understand.
vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.

∑F = ma

86.5 N − T₁ − Wa = 0

Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

ma = 3.2 kg

Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

mb × 9.8 m/s² = 55.6 N − 36.5 N

mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

mc × 9.8 m/s² = 36.5 N − 9.6 N

mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

3 0
3 years ago
An airplane flies eastward and always accelerates at a constant rate. At one position along its path it has a velocity of 34.3 m
Tomtit [17]

Explanation:

We'll need two equations.

v² = v₀² + 2a(x - x₀)

where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.

x = x₀ + ½ (v + v₀)t

where t is time.

Given:

v = 47.5 m/s

v₀ = 34.3 m/s

x - x₀ = 40100 m

Find: a and t

(47.5)² = (34.3)² + 2a(40100)

a = 0.0135 m/s²

40100 = ½ (47.5 + 34.3)t

t = 980 s

7 0
3 years ago
If a 93000 kg truck collides with a 60 kg car
algol [13]
What’s the question here?
8 0
3 years ago
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