Answer:
Approximately 
(assuming that the projectile was launched at angle of 
above the horizon.)
Explanation:
Initial vertical component of velocity:
.
The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing 
is the same as the altitude 
at which this projectile was launched: 
.
Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 
(upwards,) the vertical velocity right before landing would be 
(downwards.) The change in vertical velocity is:
.
Since there is no drag on this projectile, the vertical acceleration of this projectile would be 
. In other words, 
.
Hence, the time it takes to achieve a (vertical) velocity change of 
would be:
.
Hence, this projectile would be in the air for approximately .
Solution:
Let the slope of the best fit line be represented by '
'
and the slope of the worst fit line be represented by '
'
Given that:
= 1.35 m/s
= 1.29 m/s
Then the uncertainity in the slope of the line is given by the formula:
(1)
Substituting values in eqn (1), we get
= 0.03 m/s
Answer:
270 mi/h
Explanation:
Given that,
To the south,
v₁ = 300 mi/h, t₁ = 2 h
We can find distance, d₁
To the north,
v₂ = 250 mi/h, d₂ = 750 miles
We can find time, t₂
Now,
Average speed = total distance/total time
Hence, the average speed for the trip is 270 mi/h.
#1. A. Waxing crescent.
#2. 1.
#3. C.
#4. C.
Answer:
Volume strain is 0.02
Explanation:
Volume strain is defined as the change in volume to the original volume.
It is given that,
Initial volume of the plastic box is 2 m³
It is then submerged below the surface of a liquid and its volume decreases to 1.96 m³
We need to find the volume strain on the box. It is defined as the change in volume divided by the original volume. So,
So, the volume strain on the box is 0.02.
