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ki77a [65]
4 years ago
10

a 2,000 pound car is driving at 60 miles/hour along a straight, level road. what is the net force acting on the car?

Physics
1 answer:
SVETLANKA909090 [29]4 years ago
5 0

Answer:

0

Explanation:

According to Newton's second law, the net force is equal to the mass times the acceleration.  Since the car is not accelerating, the net force is 0.

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Found throughout the nervous system, __________ aid and support the function of neurons.
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A ferris wheel car with a mass of 350 kg, travels in a
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Answer:Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F = ma. For uniform circular motion, the acceleration is the centripetal acceleration—a = ac. Thus, the magnitude of centripetal force Fc is Fc = mac.

By using the expressions for centripetal acceleration ac from  

a

c

=

v

2

r

;

a

c

=

r

ω

2

, we get two expressions for the centripetal force Fc in terms of mass, velocity, angular velocity, and radius of curvature:  

F

c

=

m

v

2

r

;

F

c

=

m

r

ω

2

.

You may use whichever expression for centripetal force is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the center of curvature, because ac is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r, you get  

r

=

m

v

2

F

c

.

This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

The given figure consists of two semicircles, one over the other. The top semicircle is bigger and the one below is smaller. In both the figures, the direction of the path is given along the semicircle in the counter-clockwise direction. A point is shown on the path, where the radius from the circle, r, is shown with an arrow from the center of the circle. At the same point, the centripetal force is shown in the opposite direction to that of radius arrow. The velocity, v, is shown along this point in the left upward direction and is perpendicular to the force. In both the figures, the velocity is same, but the radius is smaller and centripetal force is larger in the lower figure.

Figure 1. The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the Fc, the smaller the radius of curvature r and the sharper the curve. The second curve has the same v, but a larger Fc produces a smaller r′.

Explanation:

6 0
3 years ago
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professor190 [17]

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3 years ago
1) [25 pts] A 90-kg merry-go-round of radius 2.0 m is spinning at a constant speed of 20 revolutions per minute. A kid standing
Ray Of Light [21]

Answer:

(A) 180 kg·m²

(B) 0.111 rad/s²

(C) The number of revolutions the merry-go-round will complete until it finally stops is 3.142 or π rev

Explanation:

The equation of the moment of inertia of a solid cylinder is presented as follows;

I = \frac{1}{2}MR^2

Where:

I = Moment of inertia of the merry-go-round

M = Mass of the merry-go-round

R = Radius of the merry-go-round

Therefore, I = 1/2×90×2² = 180 kg·m²

(B) For the angular acceleration we have;

Therefore, since the force × radius = The torque, we have, angular acceleration is found as follows

F × R = τ

10.0 × 2.0 = 20 = I×α = 180×α

α = 20/180 = 0.111 rad/s².

angular acceleration = 0.111 rad/s².

(C) Here we have ω₀ = 20 rev/ min = 20×2×π rad/min = π·40/60 rad/s

2/3·π rad/s

ω = ω₀ - α×t

∴ t = ω₀/α = (2/3·π rad/s)/(0.111 rad/s²) = 18.85 s

Hence we have

θ = ω₀·t + 1/2·α·t², plugging in the values, we have;

θ = 2/3·π×18.85 - 1/2·0.111·18.85²

θ = 19.74 rad

Therefore, since 2·π radian = 1 revolution

The number of revolutions the merry-go-round will complete until it stops is 19.74/(2·π) = 3.142 or π revolutions.

5 0
3 years ago
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