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bearhunter [10]
3 years ago
13

Which type of motion does a spinning ball have as it rolls along a curved path?

Physics
1 answer:
finlep [7]3 years ago
5 0

Answer:

The ball is going to be spinning in a clockwise direction.

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Which color is in the middle of the light spectrum?
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3 0
2 years ago
The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper e
Mashcka [7]

Answer:

Explanation:

When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.

This force is provided by a component of   T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,

T cos θ will provide the centripetal force . So

Tcosθ = mw²R

Tsinθ component will balance the weight .

Tsinθ = mg

Dividing the two equation

Tanθ = \frac{g}{\omega^2R}

Hence for a given w , θ depends upon g or weight .

8 0
3 years ago
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A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult
Dvinal [7]

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

5 0
3 years ago
A car is traveling at a constant speed of 33 m/s on a highway. At the instant this car passes an entrance ramp, a second car ent
Paha777 [63]

Answer:

0.8712 m/s²

Explanation:

We are given;

Velocity of first car; v1 = 33 m/s

Distance; d = 2.5 km = 2500 m

Acceleration of first car; a1 = 0 m/s² (constant acceleration)

Velocity of second car; v2 = 0 m/s (since the second car starts from rest)

From Newton's equation of motion, we know that;

d = ut + ½at²

Thus,for first car, we have;

d = v1•t + ½(a1)t²

Plugging in the relevant values, we have;

d = 33t + 0

d = 33t

For second car, we have;

d = v2•t + ½(a2)•t²

Plugging in the relevant values, we have;

d = 0 + ½(a2)t²

d = ½(a2)t²

Since they meet at the next exit, then;

33t = ½(a2)t²

simplifying to get;

33 = ½(a2)t

Now, we also know that;

t = distance/speed = d/v1 = 2500/33

Thus;

33 = ½ × (a2) × (2500/33)

Rearranging, we have;

a2 = (33 × 33 × 2)/2500

a2 = 0.8712 m/s²

3 0
2 years ago
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