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seraphim [82]
3 years ago
10

Explain how tangential speed depends on distance.

Physics
1 answer:
Alinara [238K]3 years ago
6 0
The equation v=rω says that the tangential speed v is proportional to the distance r from the center of rotation. ... This makes sense because a point farther out from the center has to cover a longer arc length in the same amount of time as a point closer to the center.
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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
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The answer would be 2.8m height on earth takes 
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You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time
irinina [24]
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8 0
3 years ago
Which statement best describes the movement of electrical current when a capacitor is used in a circuit?
Masteriza [31]

Answer: B.

Capacitors prevent current from moving through a circuit 

Explanation:

If a direct voltage is applied on the capacitor, no conduction current flows through the capacitor if its insulating medium is perfect insulator. This is due to the fact that there are no free charge carriers in such medium. Basically the real insulator contains very few charge carriers and therefore a very small leakage current passes in the capacitor depending on the conductivity of the insulator.

If an alternating voltage is applied on the capacitor, a displacement current passes through the capacitor irrespective of the insulating medium. This current is termed also the capacitive current. It flows because of changing electric displacement with time.

4 0
3 years ago
Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr
Illusion [34]

Answer:

F_{sprinter}=110.4N

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

W=F_{total}.d

So

W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N  

7 0
3 years ago
Read 2 more answers
A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0
3 years ago
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