Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,
v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.
Hope this helps :)
When describing linear motion, you need only one graph representing each of the three terms, while projectile motion requires a graph of the x and y axes. Graphs of simple harmonic motion are sine curves. Circular motion is different from other forms of motion because the speed of the object is constant.
Answer:
<em>2.753*10^-11N</em>
Explanation:
According to Newton's law of gravitation, the force between the masses is expressed as;
F = GMm/d²
M and m are the distances
d is the distance between the masses
Given
M = 3.71 x 10 kg
m = 1.88 x 10^4 kg
d = 1300m
G = 6.67 x 10-11 Nm²/kg
Substitute into the formula
F = 6.67 x 10-11* (3.71 x 10)*(1.88 x 10^4)/1300²
F = 46.52*10^(-6)/1.69 * 10^6
F = 27.53 * 10^{-6-6}
F = 27.53*10^{-12}
F = 2.753*10^-11
<em>Hence the gravitational force between the asteroid is 2.753*10^-11N</em>
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Answer:
film is at distance of 3.07 cm from lens
Explanation:
Given data
focal length = 3.06 cm
distance = 10.4 m = 1040 cm
to find out
How far must the lens
solution
we apply here lens formula that is
1/f = 1/p + 1/q
here f = 3.06 and p = 1040 so we find q
1/f = 1/p + 1/q
1/3.06 = 1/1040 + 1/q
1/ q = 0.3258
q = 3.0690 cm
so film is at distance of 3.07 cm from lens
