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Sever21 [200]
3 years ago
10

A ferris wheel car with a mass of 350 kg, travels in a

Physics
1 answer:
SpyIntel [72]3 years ago
6 0

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:

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3 years ago
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Norma-Jean [14]

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2 years ago
What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300
Vinvika [58]
The efficiency of an ideal Carnot heat engine can be written as:
\eta = 1-  \frac{T_{cold}}{T_{hot}}
where
T_{cold} is the temperature of the cold region
T_{hot} is the temperature of the hot region

For the engine in our problem, we have T_{cold}=300 K and T_{hot}=400 K, so the efficiency is
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4 0
3 years ago
A 50.0-turn circular coil of radius 5.00 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0
Blababa [14]

Answer:

The maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

Explanation:

Given that,

Number of turns in the circular coil, N = 50

Radius of coil, r = 5 cm

Magnetic field, B = 0.5 T

Current in coil, I = 25 mA

We need to find the magnitude of the maximum possible torque exerted on the coil. The magnetic torque is given by :

\tau=NIAB\ \sin\theta

For maximum torque, \theta=90^{\circ}

\tau=NIAB\\\\\tau=50\times 25\times 10^{-3}\times \pi (0.05)^2\times 0.5\\\\\tau=4.9\times 10^{-3}\ N-m

So, the maximum torque in the coil is 4.9\times 10^{-3}\ N-m.

4 0
3 years ago
A certain parallel-plate capacitor is filled with a dielectric for which κ = 5.0. The area of each plate is 0.027 m2, and the pl
alexdok [17]

Answer:

Energy will be 4.776\times 10^{-5}J  

Explanation:

We have given that dielectric constant k = 5

Area of the plate A=0.027m^2

Distance between the plates d=2mm=2\times 10^{-3}m

Electric field E = 200 kN/C

We know that capacitance is given by C=\frac{K\epsilon _0A}{d}=\frac{5\times 8.85\times 10^{-12}\times 0.027}{2\times 10^{-3}}=0.597\times 10^{-9}F

We know that electric field is given by E=\frac{V}{d}

So V=Ed=200\times 10^3\times 2\times 10^{-3}=400volt

So energy will be E=\frac{1}{2}CV^2=\frac{1}{2}\times 0.597\times 10^{-9}\times 400^2=4.776\times 10^{-5}J

6 0
3 years ago
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