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Dahasolnce [82]
3 years ago
5

You throw a rock horizontally off a cliff with a speed of 20 m/s. after two seconds, the magnitude of the velocity of the rock i

s closest to
Physics
1 answer:
Ksivusya [100]3 years ago
7 0
So the magnitude of the resultant vector of the vertical and horizontal together is a right triangle with 20 on one side and 19.62 going down. 
<span>A^2 + B^2 = C^2 </span>
<span>20^2 + 19.62^2 = C^2 </span>
<span>400 + 385 = C^2 </span>
<span>C = 28.0 m/s magnitude </span>
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In your daily life you come across a range of motion in which acceleration is in the direction of motion
irakobra [83]
That makes no sense to me somehow
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3 years ago
A diver makes 1.0 revolutions on the way from a 9.2-m-high platform to the water. Assuming zero initial vertical velocity, find
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Answer:

Average angular velocity ≈ 4.59 rad/s

Explanation:

Using the equation of motion,

H = ut + (1/2)t² ............................ equation 1.

Where H= height, u = initial velocity(m/s), g = acceleration due to gravity(m/s²), t = time(s)  u= 0 ∴ ut =0

H =(1/2)gt².................................... equation 2.

making t² the subject of the relation in equation 2,

∴ t² = 2H/g

Where H = 9.2 m, g= 9.8 m/s

∴ t² = ( 2×9.2)/9.8

t = √(2 × 9.2/9.8) = √(18.4/9.8)

 t = 1.37 s.

The average angular velocity = θ/t

Where θ = is the number of revolution that the diver makes, t  = time

           θ = 1 rev.

Since 1 rev = 2π (rad)

           t = 1.37 s

 Average angular velocity = 2π/t

π = 3.143

 Average angular velocity = (2×3.143)/1.37 = 6.286/1.37

   Average angular velocity ≈ 4.59 rad/s

8 0
3 years ago
When a person steps forward out of a small boat onto a dock, the boat recoils backward in the water. why does this occur??
bixtya [17]
Here is the correct answer of the given question above. When a person steps forward out of a small boat onto a dock, the boat recoils backward in the water and this occurs because the total momentum of the system is conserved. Hope this helps.
4 0
3 years ago
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a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit
ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)

We have

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\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}

Then the total work is

W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right)  \approx \boxed{36.8\,\rm J}

5 0
2 years ago
In the rectangle to the left, if
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perimeter of a rectangle = 2(L+B)

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7 0
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