Answer:
a.) the speed at the bottom is greater for the steeperhill
Explanation:
since the energy at the bottom of the steeper hilis greater

As we can see from above that v is higher when h ishigher.
Answer: Rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.
Explanation: Terminal velocity is defined as the final velocity attained by an object falling under the gravity. At this moment weight is balanced by the air resistance or drag force and body falls with zero acceleration i.e. with a constant velocity.
Case 1: Terminal velocity of a piece of tissue paper.
The weight of tissue paper is very less and it experiences an air resistance while falling downward under the effect of gravity.
Downward gravitational force, F = mg
Upward air resistance or friction or drag force will be 
So, paper will attain terminal velocity when mg =
Case 2: Rock is very heavy and require larger air resistance to balance the weight of rock relative to the tissue paper case.
Downward force on rock, F = Mg
Drag force =
Rock will attain terminal velocity when Mg =
Mg > mg
so,
>
And rock require larger drag force and to achieve it rock need to move at a very high terminal velocity.
Answer:
Explanation:
Given that,
5J work is done by stretching a spring
e = 19cm = 0.19m
Assuming the spring is ideal, then we can apply Hooke's law
F = kx
To calculate k, we can apply the Workdone by a spring formula
W=∫F.dx
Since F=kx
W = ∫kx dx from x = 0 to x = 0.19
W = ½kx² from x = 0 to x = 0.19
W = ½k (0.19²-0²)
5 = ½k(0.0361-0)
5×2 = 0.0361k
Then, k = 10/0.0361
k = 277.008 N/m
The spring constant is 277.008N/m
Then, applying Hooke's law to find the applied force
F = kx
F = 277.008 × 0.19
F = 52.63 N
The applied force is 52.63N
Answer:
Length = 2.92 m
Diameter = 0.11 mm
Explanation:
We have
, where:
is the length

We divide the first equation by the second equation to get:


Using this Area, we find the diameter of the wire:



To find the length, we multiply the two equations stated initially:


Answer:
The soda is being sucket out at a rate of 3.14 cubic inches/second.
Explanation:
R= 2in
S= π*R²= 12.56 inch²
rate= 0.25 in/sec
rate of soda sucked out= rate* S
rate of soda sucked out= 3.14 inch³/sec