Answer:
Vf = 23 m/s
Explanation:
First we need to find the distance covered by the motorcycle 2 when it passes motorcycle 1. Using the uniform speed equation for motorcycle 1:
s₁ = v₁t₁
where,
s₁ = distance covered by motorcycle 1 = ?
v₁ = speed of motorcycle 1 = 6.5 m/s
t₁ = time = 10 s
Therefore,
s₁ = (6.5 m/s)(10 s)
s₁ = 65 m
Now, for the distance covered by motorcycle 2 at the meeting point. Since, the motorcycle started 50 m ahead of motorcycle 2. Therefore,
s₂ = s₁ + 50 m
s₂ = 65 m + 50 m
s₂ = 115 m
Now, using second equation of motion for motorcycle 2:
s₂ = Vi t + (1/2)at²
where,
Vi = initial velocity of motorcycle 2 = 0 m/s
Therefore,
115 m = (0 m/s)(10 s) + (1/2)(a)(10 s)²
a = 230 m/100 s²
a = 2.3 m/s²
Now, using 1st equation of motion:
Vf = Vi + at
Vf = 0 m/s + (2.3 m/s²)(10 s)
<u>Vf = 23 m/s</u>
Answer:
Pushing it the wrong angle causes it to be 10x harder than it is before therefore there is a certain way to move an object instead of doing it recklessly
Explanation:
According to Newton's second law. £Fext=m.a
then the acceleration of the second object will be twice as that of the first since the mass of both objects is equal(100kg) and the force F is doubled. Therefore the acceleration of object 2 will be more than that of object 1 (double)
hope this helps.
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Answer:
Explanation:
The weight of the box is
W = 253.1N
Weight is on an incline plane
θ = 39.7°
The weight of an object is always acting downward
So, the weight makes an angle of 39.7° with the vertical component
Then, it's horizontal component is
Wx = W•Sinθ
Wx = 253.1 × Sin 39.7°
Wx = 161.67 N
The horizontal component of the weight is 161.67N
This is the force acting down the plane.
Check attachment
