Answer:
A.) 8 m/s
B.) 7.0 m
Explanation:
Given that a block is given an initial velocity of 8.0 m/s up a frictionless 28° inclined plane.
(a) What is its velocity when it reaches the top of the plane?
Since the plane is frictionless, the final velocity V will be the same as 8 m/s
The velocity will be 8 m/s as it reaches the top of the plane.
(b) How far horizontally does it land after it leaves the plane?
For frictionless plane,
a = gsinø
Acceleration a = 9.8sin28
Acceleration a = 4.6 m/s^2
Using the third equation of motion
V^2 = U^2 - 2as
Substitute the a and the U into the equation. Where V = 0
0 = 8^2 - 2 × 4.6 × S
9.2S = 64
S = 64/9.2
S = 6.956 m
S = 7.0 m
Answer:
B
Explanation:
That's the answer. Hope it helped!
Answer:
960 m
Explanation:
Given that,
- Speed = 120 m/s
- Time taken = 4 minutes
We have to find the distance covered.
Firstly, let's convert time in seconds.
→ 1 minute = 60 seconds
→ 4 minutes = (4 × 60) seconds
→ 4 minutes = 240 seconds
Now, we know that,
→ Distance = Speed × Time
→ Distance = (4 × 240) m
→ Distance = 960 m
Therefore, distance covered is 960 m.
Answer:
0.12959085 J
Explanation:
k = Coulomb constant = 
q = Charge = 1.55 μC
d = Distance between charge = 0.5 m
Electric potential energy is given by
In this system with three charges which are equidistant from each other
The potential energy of the system is 0.12959085 J
