F = m · a
In order to accelerate 82 kg upward at the rate of 3.2 m/s², a NET upward force of (82kg · 3.2m/s²) = 262.4 Newtons is required.
But if the object is on or near the surface of the Earth, then there's a downward force of (82kg · 9.8m/s²) = 803.6 N already acting on it because of gravity.
So you need to apply (803.6N + 262.4N) = <em>1,066 Newtons UPward</em>, in order to cancel its own weight and accelerate it upward at that rate.
Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.
Sound is an example of a mechanical wave. Mechanical waves are the kinds of waves that cannot be propagated without a medium. As such, these waves cannot travel through a vacuum, just like how sound cannot travel through space, since space is a vacuum.
When driver see the child standing on road his speed is 20 m/s
So here at that instant his reaction time is 0.80 s
He will cover a total distance given by product of speed and time
now after this he will apply brakes with acceleration a = 7 m/s^2
so the distance covered before it stop is given by
so the total distance covered by it
<em>so it will cover a total distance of 44.6 m</em>