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oksian1 [2.3K]
3 years ago
12

A car drives around a horizontal, circular track at constant speed. Consider the following three forces that act on the car: (1)

The upward normal force exerted on the car by the road, (2) the downward gravitational force on the car, (3) and the frictional force that is directed toward the center of the circular path.
Which of these forces does zero work on the car as the car moves along the circular path?  1, 2, and 3   3   1 and 2 
 1   2   1 and 3 
 2 and 3 
Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer: 1, 2 and 3.

Explanation:

By definition, work, is the process through which a force, applied on an object, produces a displacement of this object, and can be expressed as the dot product of the force vector and the displacement vector.

It can be also understood as the projection of the force in the direction of the displacement, so, if both vectors are perpendicular, as the projection of one vector along the other is nul, if a force and the displacement are perpendicular, no work is done.

In our case, the 3 forces mentioned, are perpendicular respect the displacement.

For normal force and gravity, as they act vertically, and the car moves along an horizontal trajectory, both are perpendicular, so no work is done.

For the friction force, as it is the only horizontal force present, it is just the centripetal force that keeps the car moving in a circular path.

It always aims to the center of the circle, along the radius at any point, so it is always perpendicular to the displacement also.

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A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the
Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

a_c =w^2r = \frac{v^2}{r}

in this case we know the speed of the runner

v =8.00\ m/s

The radius "r" will be the distance from the runner to the center of the track

r = 28.2\ m

a_c = \frac{8^2}{28.2}\ m/s^2

a_c = 2.27\ m/s^2

The answer is the last option

3 0
3 years ago
A grandfather clock has a pendulum that consists of a thin brass disk of radius r = 13.62 cm and mass 1.199 kg that is attached
Feliz [49]

Answer:

Explanation:

Expression for time period of a pendulum is as follows

T = 2\pi\sqrt{\frac{l}{g} }

l is length of pendulum from centre of bob and g is acceleration due to gravity

Given

Time period T = 1.583

g = 9.846

Substituting the values

1.583 = 2\pi\sqrt{\frac{l}{9.846} }

l = \frac{(1.583)^2\times9.846}{4\times(\frac{22}{7})^2 }

l = .6244 m

= 62.44 cm

Length of rod  = length of pendulum - radius of bob

= 62.44 - 13.62

= 48.82 cm

= .488 m

8 0
3 years ago
PLEASE HELP ME!!!! i will give brainliest to whoever gets it right
Oksanka [162]

Answer:

I believe the answer is B.)

6 0
3 years ago
Consider two carts, of masses m and 2m, at rest on an air track. if you first push one cart for 3 s and then the other for the s
Marrrta [24]
F = m₁ a₁ = m₂ a₂

if m₁ = m and m₂ = 2m :

F = ma₁ = 2m  a₂ ⇒ a₁ = 2 a₂
 
since v = at + v₀  with t = 3, v₀ = 0 ⇒ v = 3a:

v₁ = 2 v₂

since p = vm with v₁ = 2v and v₂ = v :

p₁ = v₁m₁ = 2v ⁻ m
p₂ = v₂m₂ = v ⁻ 2m

p₁ = p₂

8 0
3 years ago
Based on the data you found, about how many of the 100 aliens would become thin if the temperature were 35 C?
saw5 [17]

Answer:

Option D: More than 90

Explanation:

Thickness probably depends on temperature. The higher the temperature, the more aliens we expect to be thin.

5 0
3 years ago
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