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Natali5045456 [20]
3 years ago
11

A car is traveling at 42.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0

.107, what is the minimum distance in which the car will stop?
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

Answer:

64.85 m

Explanation:

a)

μ = Coefficient of friction between road and tire on rainy day = 0.107

g = acceleration due to gravity = 9.8 m/s²

a = acceleration experienced by car due to friction = - μg = - (0.107) (9.8) = - 1.05 m/s²

v₀ = initial velocity of the car = 42 km/h = 11.67 m/s

v = final speed of the car = 0 m/s

d = minimum distance traveled before stopping

Using the equation

v² = v₀² + 2 a d

0² = 11.67² + 2 (- 1.05) d

d = 64.85 m

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A hotel elevator ascends 200m with maximum speed of 5m/s. Its acceleration and deceleration both have a magnitude of 1.0m/s2. Pa
ValentinkaMS [17]

Answer:

45 s .

Explanation:

The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .

displacement s = ?

acceleration a = 1 m /s²

Final speed v = 5 m/s

initial speed u = 0

v² = u² + 2as

5² = 0 + 2 x 1 x s

s = 12.5  m

B)  Let time of acceleration or deceleration be t

v = u + a t

5 = 0 + 1 t

t = 5 s

Similarly displacement during deceleration = 12.5 m

Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) =  175 m .

velocity of uniform motion = 5 m /s

time during which there was uniform velocity = 175 / 5 = 35 s

Total time = 5 + 35 + 5 = 45 s .

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4 0
3 years ago
A student, who weighs 720N, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is
jasenka [17]

Answer:

1) The mass of the student is approximately 73.39 kg

2) The net force on the student is approximately 947.523 N

3) The value the scale will read is approximately 96.59 kg

Explanation:

The given parameters are;

The weight of the student = 720 N

The speed at which the elevator is decreasing = 3.1 m/s²

1) The weight of the student = The mass of the student × The acceleration due to gravity

The acceleration due to gravity is a constant = 9.81 m/s²

Substituting the known values gives;

720 N = The mass of the student × 9.81 m/s²

∴ The mass of the student = 720 N/(9.81 m/s²) ≈ 73.39 kg

2) The forces acting on the student are;

i) The force of gravity which is the weight of the student acting downwards

ii) The inertia force of the slowing elevator acting downwards in the same direction as the weight of the student

The net force, F_{net} = The weight of the student + The inertia force of the slowing elevator

∴ The net force, F_{net} = 720 N + 73.39 kg × 3.1 m/s² ≈ 947.523 N

3) The scale will read the mass of the student as follows;

Mass reading of student on the scale = Force on scale/9.81

∴ Mass reading of student on the scale = 947.523/9.81 ≈ 96.59 kg

The value the scale will read = 96.59 kg.

3 0
3 years ago
Without friction, what force is needed to maintain a 1,000 kg car in uniform motion for 30 minutes?
VARVARA [1.3K]
<span>The answer is none. According to the first law of Newton, an object stays at the same speed in the same direction if there are not forces unbalancing the object. Without friction, the car would be moving forever, unless there is another force accelerating or stopping the car.</span>
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4 years ago
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