Answer:It increases confidence because the more times you conduct the same experiment over and over should either prove your hypothesis right and wrong and eliminate any random occurrences that might affect your results.
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Explanation:
Answer:
160.75 N
Explanation:
The downward velocity has no effect on the force situation, it is only changes in velocity (plus, of course, gravity, which is always there) that require a force. At constant velocity, the bottom spring s_3 is supporting its mass m_3 to balance gravity.
As the elevator slows, though, it also ends up slowing down the spring arrangement, too. However, because the stretching takes time, it means that some damped harmonic motion will be set up in the spring chain.
When the motion has finally damped out, the net force the bottom spring s3 exerts on m3 has two components--that of gravity and of the deceleration of the elevator:
F_3net = m3 * (g + a) = 10.5×(9.81+5.5)= 10.5×15.31= 160.75 N
Answer:
A 1.0 min
Explanation:
The half-life of a radioisotope is defined as the time it takes for the mass of the isotope to halve compared to the initial value.
From the graph in the problem, we see that the initial mass of the isotope at time t=0 is
The half-life of the isotope is the time it takes for half the mass of the sample to decay, so it is the time t at which the mass will be halved:
We see that this occurs at t = 1.0 min, so the half-life of the isotope is exactly 1.0 min.
Answer:
Explanation:
(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.
Hence, his average velocity is 300m/150s=2ms^−1
(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.
Therefore, his average speed for this journey is 400m210s=1.9ms−1.
For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.
Hence, his average velocity for this case is 200m/210s=0.95ms^−1
Answer:
T = 1.2 s
T = 15.1 m = 15 m
Explanation:
This is a case of projectile motion:
TOTAL TIME OF FLIGHT:
The formula for total time of flight in projectile motion is:
T = 2 V₀ Sinθ/g
where,
T = Total Time of Flight = ?
V₀ = Launch Speed = 13.9 m/s
θ = Launch Angle = 25°
g = 9.8 m/s²
Therefore,
T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)
<u>T = 1.2 s</u>
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RANGE OF BALL:
The formula for range in projectile motion is:
R = V₀² Sin2θ/g
where,
R = Horizontal Distance Covered by ball = ?
Therefore,
T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)
<u>T = 15.1 m = 15 m</u>
