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san4es73 [151]
3 years ago
11

Give an example of a situation in which there is a force and a displacement, but the force does no work. explain why it does no

work
Physics
2 answers:
makkiz [27]3 years ago
6 0
Work is equal to force times distance times cosine of the angle between the force and the path vector making it a scalar or dot product. Work is independent of the path taken. It refers to positive change in KE or negative change in PE.
 So for example if you lift a box from ground to waist height then, you did positive work over the distance you lifted it. If you pick it up to that same height, then put it back done on the ground at same place, then no work is done. 

Hope this makes sense. Any questions please ask. Thank You!
harkovskaia [24]3 years ago
6 0
Say you have an object such as a block on a flat surface with no friction to resist it while being displaced. The block moves but no work is done.
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ball is dropped from rest and hits the ground 1.1 seconds later. What height was the ball released from? ​
Aleonysh [2.5K]
Height = 1/2 * 9.8 * (1.1^2)
5 0
3 years ago
Distance travelled can be found from the
meriva

Answer:

A

Explanation:

This is because distance traveled (i.e. displacement) is the integral of the velocity function, and velocity is the first derivative of the displacement function. To put this in perspective, the area bounded by a curve can be found by taking the integral of the equation of the curve, taking values on the x-axis as limits.

8 0
3 years ago
a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related
nasty-shy [4]

Answer:

32km per hour

Explanation:

Explanation:

In first case v = a t

==> a t = 40 km p h

Now distance covered S1 + S2 + S3

S1 = 1/2 a t^2 and S3 = 1/2 a t^2

But S2 = 3t * 40 = 120 t km

Hence total distance = at^2 + 120 t

Time taken (total) = t + 3t + t = 5 t

Hence average speed = at^2 + 120 t / 5 t

Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour

8 0
3 years ago
A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the
Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\

\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s

a)

i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A

b)

q=q_o*e^{-\frac{t}{\tau}}

q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC

c) the maximum current occurs when t=0

i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A

6 0
3 years ago
An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu
Doss [256]

Answer:

400000\ \text{N/C}

Explanation:

q_1 = Charge at 3000 m = 40 C

q_2 = Charge at 1000 m = -40 C

r_1 = 3000 m

r_2 = 1000 m

k = Coulomb constant = 9\times10^9\ \text{Nm}^2/\text{C}^2

Electric field due to the charge at 3000 m

E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}

Electric field due to the charge at 1000 m

E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}

Electric field at the aircraft is E_1+E_2=40000+360000=400000\ \text{N/C}.

7 0
3 years ago
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