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dexar [7]
3 years ago
14

Two cars approach a street corner at right angles to each

Physics
1 answer:
irakobra [83]3 years ago
4 0

Answer:

the angle is given by

Tan theta = 35/59 = 0.59

so theta = Tan ^-1 ( 0.59 )

theta = 30.54 deg.

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A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

<u>F = 51.3°</u>

8 0
2 years ago
How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?
Anestetic [448]

Answer:

21 m

Explanation:

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d=vt

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v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

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3 0
3 years ago
A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>
8 0
3 years ago
If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre
Dmitrij [34]
M <span>represent mass in kg
</span><span>v represents speed in m/s
</span><span>r represents radius in m

Now, just substitute these into the formula:
</span>F =  \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }<span>

</span>
3 0
3 years ago
Which most accurately describes the path that sound​ travels??
emmainna [20.7K]
What are the options ?
6 0
3 years ago
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