It is very important to note down all the details that are given in the question. Based on these details the answer can be derived.
Initial velocity of the car (V1) = 0 meters/second
Final velocity of the car (V2) = 40 meters/second
Initial time (T1) = 0 Second
Final time (T2) = 60 seconds (In this case we have converted 1 minute into 60 seconds for the ease of calculation.
Then
Average acceleration = (V2 - V1)/ (T2 - T1) m/s^2
= (40 - 0)/(60 - 0) m/s^2
= 40/60 m/s^2
= 2/3 m/s^2
= 0.67 m/s^2
So the average acceleration of the car is 0.67 meters/second square.
Answer:
(a) Fₓ = 0 N
= 9.08 N
(b)
(a) Fₓ = 0 N
<u></u>
= 9.08 N
(c)
= 0 N
Fₓ = 9.08 N
Explanation:
The magnitude of the force will remain the same in each case, which is given as follows:
F = ma (Newton's Second Law)
where,
F = force = ?
m = mass = 4.54 kg
a = acceleration = 2 m/s²
Therefore,
F = (4.54 kg)(2 m/s²)
F = 9.08 N
Now, we come to each scenario:
(a)
Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:
<u>Fₓ = 0 N</u>
For upward direction the force will be positive:
<u></u>
<u> = 9.08 N</u>
<u></u>
(b)
Since the motion is in the vertical direction. Therefore the magnitude of the force in x-direction will be zero:
<u>Fₓ = 0 N</u>
For upward direction the force will be negative:
<u></u>
<u> = - 9.08 N</u>
<u></u>
(c)
Since the motion is in the horizontal direction. Therefore the magnitude of the force in y-direction will be zero:
<u></u>
<u> = 0 N</u>
<u>Fₓ = 9.08 N</u>
Answer:
t= 24080 s
Explanation:
Given that
Current in the wire ,I = 4 A
The charge ,q = 6.02 x 10²³ e C
We know that
![I=\dfrac{q}{t}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bq%7D%7Bt%7D)
I=Current
q=Charge
t=time
![t=\dfrac{q}{I}](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7Bq%7D%7BI%7D)
Now by putting the values in the above equation we get'
![t=\dfrac{6.02\times 10^{23}\times 1.6\times 10^{-19}}{4}\ s](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7B6.02%5Ctimes%2010%5E%7B23%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B4%7D%5C%20s)
t= 24080 s
Answer:
F'=708.53 N
Explanation:
We have,
The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the plane's velocity, v. It means tat,
![F=kAv^2](https://tex.z-dn.net/?f=F%3DkAv%5E2)
k is constant
If, A = 190 Ft², v = 220 mph, F = 950 pounds
Let's find k first from above data. So,
![k=\dfrac{F}{Av^2}\\\\k=\dfrac{950}{190\times 220^2}\\\\k=0.0001033](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7BF%7D%7BAv%5E2%7D%5C%5C%5C%5Ck%3D%5Cdfrac%7B950%7D%7B190%5Ctimes%20220%5E2%7D%5C%5C%5C%5Ck%3D0.0001033)
It is required to find the lifting force on the wing if the plane slows down to 190 miles per hour. Let F' is the new force. So,
![F'=0.0001033\times 190\times (190)^2\\\\F'=708.53\ \text{pounds}](https://tex.z-dn.net/?f=F%27%3D0.0001033%5Ctimes%20190%5Ctimes%20%28190%29%5E2%5C%5C%5C%5CF%27%3D708.53%5C%20%5Ctext%7Bpounds%7D)
So, the lifting force is 708.53 pounds if the plane slows down to 190 miles per hour.