Answer:
Ok, her position changed over time since the road is curved.
The velocity changed since the direction is changing
AccelerTion: theres a change in velocity since the direction changes
Direction: changed over time
Displacement : difference between 1st and final position
Explanation:
Answer:
1.176 N
Explanation:
= mass of the bottle = 0.30 kg
= initial speed of the bottle = 2.8 m/s
= final speed of the bottle = 0 m/s
= stopping distance traveled = 1.0 m
= magnitude of frictional force acting on bottle
Using work-change in kinetic energy theorem
N
direction :
frictional force acts in opposite direction of motion.
That particular strike was very roughly 2.4 km (1.5 miles) away from them.
That's if you use 340 m/s (1120 ft/sec) for the speed of sound.
But the air in the region for several thousand feet around a thunderstorm
is doing weird things to sounds that pass through it, so you can't use any
exact number for the speed of sound in a stormy area.
The only thing you can be absolutely sure of is that Johnny and his friends
need to round up their equipment and get in the house. NOW !
1) Current in the wire: 0.0875 A
The current in the wire is given by:

where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is

2) Drift velocity of the electrons: 
The drift velocity of the electrons in the wire is given by:

where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is

So the cross-sectional area is

So, the drift velocity is

Answer:
Explanation:
Given that,
Basket ball is drop from height
H=10m
It is dropped on planet mass
And the acceleration due to gravity on Mars is given as
g= 3.7m/s²
Time taken for the ball to reach the ground
Initial velocity of the body is zero
u=0m/s
Using equation of motion: free fall
H = ut + ½gt²
10 = 0•t + ½ × 3.7 ×t²
10 = 0 + 1.85t²
10 = 1.85t²
Then, t² =10/1.85
t² = 5.405
t = √ 5.405
t = 2.325seconds
So the time the ball spend on the air before reaching the ground is 2.325 seconds