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MrRissso [65]
3 years ago
13

When a pendulum is swinging, the velocity is highest at which point?

Physics
1 answer:
Tems11 [23]3 years ago
3 0
At the center, when the bob is hanging straight down

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Pls help<br> What graph indicates no reference to direction of motion, and explain why.
Komok [63]

Answer:

graph A

Explanation:

the slope of the distance-time graph is speed, speed is a scalar (with magnitudes but no direction)

but the slope for the velocity time graph is acceleration, acceleration is vector quantity ( has magnitude and direction)

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A wire runs left to right and carries a current in the direction shown.
Step2247 [10]

Answer:

The direction of the magnetic field at point Z; Into the screen

Explanation:

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4 years ago
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A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally
aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m    

speed at which the ball strike = ?

vertical velocity  = 0 m/s

time at which the ball strike

s = \dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 100}{9.8}}

t = 4.53 s

vertical velocity at the time  4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = \dfrac{65}{4.53} =14.35 m/s

speed of the ball = \sqrt{44.39^2+14.35^2}

                             = 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

5 0
3 years ago
A 0.50-kg bomb is sliding along an icy pond (frictionless surface) with a velocity of 2.0 m/s to the west. The bomb explodes int
nevsk [136]

Answer:

2.667m/s to the north and 3.333 m/s to the west

Explanation:

According to law of momentum conservation, the total momentum should be conserved before and after the explosion.

Before the explosion, the momentum was

0.5*2 = 1 kg m/s to the west

Therefore the total momentum after the explosion should be the same horizontally and vertically.

Vertically speaking, it was 0 before the explosion. After the explosion:

0.2*4 + 0.3v = 0

0.3v = -0.8

v = -0.8/0.3 = -2.667 m/s

So the vertical component of the 0.3kg piece is 2.667m/s to the north

Horizontally speaking, since the 0.2kg-piece doesn't move west or east post-explosion:

0.2*0 + 0.3V = 1

0.3V = 1

V = 1/0.3 = 3.333 m/s

So the horizontal component of the 0.3kg piece is 3.333 m/s to the west

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3 years ago
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