Answer:
a) 4.31 m/s²
b) 215.5 m
Explanation:
a) According to Newton's first law of motion
The net force applied to particular mass produced acceleration, a, according to
F = ma
F = 140 N
m = 32.5 kg
a = ?
140 = 32.5 × a
a = 140/32.5 = 4.31 m/s²
b) Using the equations of motion, we can obtain the distance travelled by the object in t = 10 s
u = initial velocity of the probe = 0 m/s (since it was initially at rest)
a = 4.31 m/s²
t = 10 s
s = distance travelled = ?
s = ut + at²/2
s = 0 + (4.31×10²)/2 = 215.5 m
Answer:
I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .
Explanation:
15m/s - 0m/s divided by 5 s = 3m/s
I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry
Answer:
0.66c
Explanation:
Use length contraction equation:
L = L₀ √(1 − (v²/c²))
where L is the contracted length,
L₀ is the length at 0 velocity,
v is the velocity,
and c is the speed of light.
900 = 1200 √(1 − (v²/c²))
3/4 = √(1 − (v²/c²))
9/16 = 1 − (v²/c²)
v²/c² = 7/16
v = ¼√7 c
v ≈ 0.66 c
The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.
If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²). That's (2E)/4 = 0.5E .
