If you can observe, we are only given one parameter here which is the time. If you want to compute for the distance, you have to know the speed. The hint here is the 'radio transmissions'. All the information gathered by the probe from the space, is sent back to the Earth by electromagnetic waves. Hence, we must know the speed of electromagnetic waves. Since they are as fast as light, their speed is equal to 300 million meters per second. Then, we can finally determine the distance.
d = speed*time
d = (300×10⁶ m/s)(2.53 hours)
Since 1 hour = 3,600 seconds,
d = (300×10⁶ m/s)(2.53 hours)(3,600 seconds/1 hour)
d = 2.73×10⁻¹² m
"decreasing the distance of the space shuttle from Earth"
F = Gm(1)m(2)/R²
where R is the distance between the 2 objects, as it decreases, the force increases.
Answer:
0.8%
Explanation:
We are given;
Number of oscillations; n = 20
Time taken; t = 25 s
Formula for period of oscillation;
T = t/n = 25/20 = 1.25 s
We are told that the least count is 0.2 s. Thus, error is; ΔT = 0.2 s
percentage error in the measurement of time is given by;
(0.2/(20 × 1.25)) × 100% = 0.8%
Answer:
2.06 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.
Momentum before collision
Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5
Momentum after collision
The momentum after collision will be given by (9+27)*0.9=32.4
Relating the two then 9v+13.5=32.4
9v=18.5
V=2.055555555555555555555555555555555555555 m/s
Rounded off, v is approximately 2.06 m/s
