Answer:
4.2s
Explanation:
Given parameters:
Power = 2190W
Mass of box = 1.47 x 10⁴g
distance = 6.34 x 10⁴mm
Unknown:
Time = ?
Solution:
Power is the rate at which work is done;
Mathematically;
Power =
Time =
Work done = weight x height
convert mass to kg;
100g = 1kg;
1.47 x 10⁴g = 14.7kg
convert the height to m;
1000mm = 1m
6.34 x 10⁴mm gives 63.4m
Work done = 14.7 x 9.8 x 63.4 = 9133.4J
Time taken = = 4.2s
Answer:
1317.4 m
Explanation:
We are given that
Angle=
Initial speed =
We have to find the horizontal distance covered by the shell after 5.03 s.
Horizontal component of initial speed=
Vertical component of initial speed=
Time=t=5.03 s
Horizontal distance =
Using the formula
Horizontal distance=
Horizontal distance=1317.4 m
Hence, the horizontal distance covered by the shell=1317.4 m
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ =
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt =
Δt = 2 v₀ / g