I got you
Explanation:
normal force = 400 g cos 35
friction force up slope = .6 (400 g) cos 35
weight component down slope = 400 g sin 35
400 a = 400 g sin 35 - .6 (400 g cos 35)
a = g (sin 35 - .6 cos 35) = .082 g
I hope this helps you
Answer:
Explanation:
<h2><u>Given</u> :-</h2>
<h2><u>To Find</u> :-</h2>
<h2><u>Formula to be used</u> :-</h2>
Where,
- K.E. = Kinetic energy possessed by the body
- M = Mass of the body
- V = Velocity of the body
<h2><u>Solution</u> :-</h2>








- Velocity of the vehicle at the instant is

Answer:
(a) 8 V, (b) 144000 V, (c) 2 x 10^(-8) C
Explanation:
(a) charge, q = 5 μC , Work, W = 40 x 10-^(-6) J
The electric potential is given by
W = q V

(b)
charge, q = 8 x 10^(-6) C, distance, r = 50 cm = 0.5 m
Let the potential is V.

(c)
Work, W = 8 x 10^(-5) J, Potential difference, V = 4000 V
Let the charge is q.
W= q V

Use the formula dgh + p atm
d is density g is gravitational field strength h heigh or depth and p atm is atmospheric pressure
1000 x 10 x 20 + 100000 = 300000
almost 300,000 pascal
Answer:
Part a)

Part b)

Part C)

Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as



now we will have




Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have


Part C)
By momentum conservation we will have



Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.