Answer:
Explanation:
Assuming school is at the end of the 20 mile route, then
20 mi / 35 mi/hr = 0.57142...hr
which is about 34 minutes 17 seconds
Explanation:
The given data is as follows.


Electric potential energy (
) = ?
Formula to calculate electric potential energy is as follows.
= 
= 
Thus, we can conclude that the electric potential energy of the pair of charges when the second charge is at point b is
.
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:






we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:



Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:







t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:



So once we got the two distances we can now find the difference between them:

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Answer:
alcohol thermometers are used rather than Mercury thermometers in very cold regions because alcohol has a lower freezing point than Mercury.
Explanation:
While riding in a hot air balloon,
which is steadily at a speed of 1.01 m/s, and your phone accidentally falls.
<span>(a)
</span>The
speed of your phone after 4 s is:
V= u +
at
V= 1.01
+ (9.8)(4)
V=
40.21 m/s
<span>(b)
</span>The balloon
is ____ far:
V = u +
at
V= 1.01
+ (9.8)(1)
V=10.81
–distance at 1 one second
V= u +
at
V= 1.01
+ (9.8)(2)
V= 20.61-distance
at 2 seconds
V= u+ at
V=
30.41- distance at 3 seconds
V=
40.21- distance at 4 seconds
D=
102.04 m
<span>(c)
</span>If the
balloon is rising steadily at 1.01 m/s:
V= -1.1
m/s
<span> </span>