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4 years ago

12

the time required for one cycle, a complete motion that returns to its starting point, it called the_____. period medium frequen

cy periodic motion

1 answer:

Salsk061 [2.6K]4 years ago

3 0

Answer:

The time required for one cycle, a complete motion that returns to its starting point,it is called periodic motion

Explanation:

I hope this will help you:)

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A stone is dropped from rest from the top of a cliff into a pond below. If its initial height is 10 m, what is its speed when it

Brut [27]

Answer:

14 m/s

Explanation:

The motion of the stone is a free fall motion, so an accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground. So, we can use the following SUVAT equation:

$v^2 -u^2 = 2gh$

where

v is the final speed of the stone as it reaches the water

u = 0 is the initial speed

g = 9.8 m/s^2 is the acceleration

h = 10 m is the distance covered by the stone

Solving for v, we find

$v=\sqrt{u^2+2gh}=\sqrt{0+2(9.8 m/s^2)(10 m)}=14 m/s$

8 0

3 years ago

Which vector goes from (-1, -3) to (-4, -1)

Alisiya [41]

Its this (couldn’t write it down on here properly so i had to ss it)
Basically, it’s just the difference between the x values at the top and the difference between the Y values at the bottom.

6 0

2 years ago

Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving

saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

$\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh$

$v_2^2=v_1^2-2gh$

$v_2^2=(7.7)^2-2\times 10\times 1$

$v_2=6.26\ m/s$

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0

4 years ago

Read 2 more answers

A 75Kg man jumps from a window 1.0m above the sidewalk. If the man falls with his knees and ankles locked, the only cushion for

liq [111]

Answer:

150153.06122 N

Explanation:

m = Mass of person = 75 kg

h = Height of fall = 1 m

g = Acceleration due to gravity = 9.81 m/s²

F = Force

s = Displacement = 0.49 cm

Potential energy is given by

$P=mgh\\\Rightarrow P=75\times 9.81\times 1\\\Rightarrow P=735.75\ J$

Work is given by

$W=Fs\\\Rightarrow F=\frac{W}{s}\\\Rightarrow F=\frac{735.75}{0.0049}\\\Rightarrow F=150153.06122\ N$

The average force exerted is 150153.06122 N

8 0

3 years ago

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0

3 years ago