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solmaris [256]
2 years ago
13

When a pendulum is at the midpoint of its oscillation, hanging straight down, which statement is true?

Physics
2 answers:
Svetach [21]2 years ago
5 0
When a pendulum is at the midpoint of its oscillation, hanging straight down ...

-- that's the fastest it's going to swing, so its kinetic energy is maximum;
and
-- that's the lowest it's going to get, so its potential energy is minimum.

'c' is your choice.
Misha Larkins [42]2 years ago
3 0

c. Kinetic energy is maximum and potential energy is minimum.

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A 480-kilogram horse runs across a field at a rate of 40 km/hr. What is the magnitude of the horse's momentum?
Andrej [43]
Momentum (p) = mass × velocity

so, 480×40 = 19,200 kg km/hr

so the answer is C !!
4 0
2 years ago
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Compressional stress on rock can cause strong and deep earthquakes, usually at _____.
valentinak56 [21]
The answer is reverse faults. 
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2 years ago
A package is dropped from a helicopter that is descending steadily at a speed v0. After t seconds have elapsed, consider the fol
qaws [65]

Answer:

Part a)

v = \sqrt{v_o^2 + g^2t^2}

Part b)

d = \frac{1}{2}gt^2

Part c)

v_f = v_o - gt

Part d)

d = \frac{1}{2}gt^2

Explanation:

Part a)

As we know that speed of package is same as that of helicopter in horizontal direction

So after time "t" the velocity in x direction will remain constant while in Y direction it will go free fall

So we have

v_y = -gt

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{v_o^2 + g^2t^2}

Part b)

Distance from helicopter is same as the distance of free fall

so we will have

d = \frac{1}{2}gt^2

Part c)

If helicopter is rising upwards with uniform speed

then final speed of the package after time t is given as

v_f = v_i + at

v_f = v_o - gt

Part d)

distance from helicopter

d = \frac{1}{2}gt^2

8 0
2 years ago
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The blade of metal cutter is shorter than that scissors of tailors<br><br>​
Naily [24]

Explanation:it is beause they are sharper and also have less surface area and therefore more pressure

8 0
2 years ago
A power station burns 75 kilograms of coal per second. Each kg of coal contains 27 million joules of energy.
Kaylis [27]

Answer:

Explanation:

a )

one kg of coal gives energy of 27 x 10⁶ J

75 kg of coal gives energy of 27 x 10⁶ x 75 J

So rate which energy is coming out of coal per second

= 27 x 10⁶ x 75 J

= 2025 x 10⁶ J /s

2025 million watts .

b ) energy output = 800 million watts

efficiency = (800 / 2025) x 100

= 39.5 % .

3 0
2 years ago
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