Answer:
v_{ average} = 5.57
Explanation:
The most probable value of a measure is
v_average =
∑ x_i
where N is the number of measurements
in tes case N = 3
v_{average} = ⅓ (5.63 +5.54 + 5.53)
V_{average} = 5,567
The number of significant figures must be equal to the number of figures that have the least in the readings.
v_{ average} = 5.57
Answer:
(A) She needs to move the decimal point by 3 places
Answer:
vDP = 21.7454 m/s
θ = 200.3693°
Explanation:
Given
vDE = 7.5 m/s
vPE = 20.2 m/s
Required: vDP
Assume that
vDE to be in direction of - j
vPE to be in direction of i
According to relative motion concept the velocity vDP is given by
vDP = vDE - vPE (I)
Substitute in (I) to get that
vDP = - 7.5 j - 20.2 i
The magnitude of vDP is given by
vDP = √((- 7.5)²+(- 20.2)²) m/s = 21.7454 m/s
θ = Arctan (- 7.5/- 20.2) = 20.3693°
θ is in 3rd quadrant so add 180°
θ = 20.3693° + 180° = 200.3693°
Answer:
The maximum height reached by the ball is 16.35 m.
Explanation:
Given;
initial velocity of the ball, u = 17.9 m/s
the final velocity of the ball at the maximum height, v = 0
The maximum height reached by the ball is given by;
v² = u² + 2gh
During upward motion, gravity is negative
v² = u² + 2(-g)h
v² = u² - 2gh
0 = u² - 2gh
2gh = u²
h = u² / 2g
h = (17.9)² / (2 x 9.8)
h = 16.35 m
Ttherefore, the maximum height reached by the ball is 16.35 m.
The answer is D. time really does pass more slowly in a rest frame of reference relative to a frame of reference that is moving