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Nataly_w [17]
3 years ago
15

Which statement is true? A Displacement can never be greater than distance. B Displacement can never be less than distance. C Di

splacement can never be equal to distance. D Displacement is always equal to distance.
Physics
1 answer:
qwelly [4]3 years ago
3 0

Answer:

__________________

I think the answer is A.

__________________

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How many times should the power develop by the engine of a ship increases to double is velocity i the resistance of the water to
Setler79 [48]

Answer:

If the ship speed is doubled, then the power developed is 8 times the initial value.

Explanation:

ship power is roughly proportional to the cube of the speed, so

P ∝ v³

If the speed is doubled, then the power developed becomes

P  ∝ (2)³ = 8 times

Therefore, if the ship speed is doubled, then the power developed is 8 times the initial value.

4 0
3 years ago
9) What will happen to the period of a pendulum if we change the rope of a pendulum with another which is four times with the in
Arturiano [62]

Double

Explanation:

Since the period T of a pendulum is given by

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

By increasing the length of the pendulum by 4, the period becomes

T' = 2\pi \sqrt{\dfrac{4l}{g}} = 2\left(2\pi \sqrt{\dfrac{l}{g}}\right) = 2T

You can see that the period doubles when we increase the length by a factor of 4.

5 0
3 years ago
A car travels across Texas m miles at the rate of t miles per hour. How many hours does the trip take??
Marianna [84]

Answer: The trip takes \frac{m}{t}hours

Explanation:

Velocity V is the variation of the position of a body (distance traveled d) with time T:

V=\frac{d}{T}

In this case, the car travels a distance d=m miles at a velocity V=t \frac{miles}{hour} and we need to find the time it takes the trip.

Isolating  T:

T=\frac{d}{V}=\frac{m miles}{t \frac{miles}{hour}}

Finally:

T=\frac{m}{t}hours

8 0
3 years ago
The current in a wire varies with time according to the relationship 1=55A−(0.65A/s2)t2. (a) How many coulombs of charge pass a
mario62 [17]

Answer:

(A) Q = 321.1C (B) I = 42.8A

Explanation:

(a)Given I = 55A−(0.65A/s2)t²

I = dQ/dt

dQ = I×dt

To get an expression for Q we integrate with respect to t.

So Q = ∫I×dt =∫[55−(0.65)t²]dt

Q = [55t – 0.65/3×t³]

Q between t=0 and t= 7.5s

Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]

Q = 321.1C

(b) For a constant current I in the same time interval

I = Q/t = 321.1/7.5 = 42.8A.

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