Since you already gave us the weight of the 2.5-kg box,
we don't even need to know what the distance is, just
as long as it doesn't change.
Look at the formula for the gravitational force:
F = G m₁ m₂ / R² .
If 'G', 'm₁' (mass of the Earth), and 'R' (distance from the Earth's center)
don't change, then the Force is proportional to m₂ ... mass of the box,
and you can write a simple proportion:
(6.1 N) / (2.5 kg) = (F) / (1 kg)
Cross-multiply: (6.1 N) (1 kg) = (F) (2.5 kg)
Divide each side by (2.5 kg): F = (6.1N) x (1 kg) / (2.5 kg) = 2.44 N .
Answer:
e see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1
Explanation:
This is an ejercise in special relativity, where the speed of light is constant.
Let's carefully analyze the approach, we see the two events at the same time.
The closest event time is
c = (x₁-300) / t
t = (x₁-300) / c
The time for the other event is
t = (x₂- 600) / c
since they tell us that we see the events simultaneously, we can equalize
(x₁ -300) / c = (x₂ -600) / c
x₁ = x₂ - 300
We see that the distances are different, the only way that the two beams of light approach simultaneously is that event 2 (farthest) occurs first than event 1
Answer:
same problem is here also bro i cannot find this plz help anyone
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s