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Ulleksa [173]
2 years ago
12

Which one of the statements below is true about mechanical waves?

Physics
1 answer:
TEA [102]2 years ago
5 0
They require a medium to travel through
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A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12 m. The sand is initially at rest and is dis
Marianna [84]

Answer:

<h2>E. 3.95kW</h2>

Explanation:

Power is defined as the rate of workdone.

Power = Workdone/time taken

Given Workdone = Force * distance

Power = Force * distance/time taken

Power = mgd/t (F = mg)

m = mass of the sand in kg

g = acceleration due to gravity in m/s²

d = vertical distance covered in metres

t = time taken in seconds

Given m = 2000kg, d = 12m, t = 1min = 60secs, g = 9.8m/s²

Power = 2000*9.8*12/60

Power = 3920Watts

Minimum rate of power that must be supplied to this machine is 3920Watts or 3.92kW

5 0
3 years ago
Does electric and magnetic objects both have north and south poles
Aleksandr [31]

Explanation:

hope this answer was helpful

6 0
3 years ago
How would the period of a simple pendulum be affected if it were located on the moon instead of the earth?
OlgaM077 [116]

Answer:

On moon time period will become 2.45 times of the time period on earth

Explanation:

Time period of simple pendulum is equal to T=2\pi \sqrt{\frac{l}{g}} ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth

As when we go to moon, acceleration due to gravity on moon is \frac{1}{6} times os acceleration due to gravity on earth

So time period of pendulum on moon is equal to

T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}} --------eqn 2

Dividing eqn 2 by eqn 1

\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}

T_{moon}=\sqrt{6}T=2.45T

So on moon time period will become 2.45 times of the time period on earth

5 0
3 years ago
What does the dashes line represent
Yuki888 [10]
 dashed lines<span> represent bonds away from the viewer.</span>
3 0
3 years ago
A ball is thrown at a 60.0° angle above the horizontal across level ground. It is thrown from a height of 2.00 m above the groun
klio [65]

Answer:

Option E is correct.

Time the ball remains in the air before striking the ground is closest to 3.64 s

Explanation:

yբ = yᵢ + uᵧt + gt²/2

yբ = 0

yᵢ = 2 m

uᵧ = u sinθ = 20 sin 60 = 17.32 m/s

g = -9.8 m/s², t = ?

0 = 2 + 17.32t - 4.9t²

4.9t² - 17.32t - 2 = 0

Solving the quadratic equation,

t = 3.647 s or t = -0.1112 s

time is a positive variable, hence, t = 3.647 s. Option E.

7 0
3 years ago
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