Answer:
933.804423995 m/s²
Explanation:
= Charge on particle 1 = 67 µC
= Charge on particle 2 = -25 µC
r = Distance between the particles = 47 cm
k = Coulomb constant = 
m = Mass of particle = 73 g
Electric force is given by
The magnitude of force is 68.1677229516 N
Acceleration is given by
The acceleration is 933.804423995 m/s²
Answer:
A planet's mass has no effect on its orbit around the Sun.
Explanation:
The kepler's third law tells us:
where 
is the orbit period and 
is the semi-major axis.
As we can see from the equation, the period depends only on the measure of the semi-major axis of the orbit, that is, how far a planet is from the sun.
The equation tells us that the closer a planet is to the sun, the faster it will go around it.
The mass does not appear in the equation to calculate the period.
This is why it is concluded from the third law of Kepler that<u> the period, or the orbit of a planet around the sun, does not depend on its mass.</u>
the answer i: A planet's mass has no effect on its orbit around the Sun.
Answer:
0.705 m/s²
Explanation:
a) The sprinter accelerates uniformly from rest and reaches a top speed of 35 km/h at the 67-m mark.
Using newton's law of motion:
v² = u² + 2as
v = final velocity = 35 km/h = 9.72 m/s, u = initial velocity = 0 km/h, s = distance = 67 m
9.72² = 0² + 2a(67)
134a = 94.484
a = 0.705 m/s²
b) The sprinter maintains this speed of 35 km/h for the next 88 meters. Therefore:
v = 35 km/h = 9.72 m/s, u = 35 km/h = 9.72 m/s, s = 88 m
v² = u² + 2as
9.72² = 9.72² + 2a(88)
176a = 9.72² - 9.72²
a = 0
c) During the last distance, the speed slows down from 35 km/h to 32 km/h.
u = 35 km/h = 9.72 m/s, v = 32 km/h = 8.89 m/s, s = 200 - (67 + 88) = 45 m
v² = u² + 2as
8.89² = 9.72² + 2a(45)
90a = 8.89² - 9.72²
90a = -15.4463
a = -0.1716 m/s²
The maximum acceleration is 0.705 m/s² which is from 0 to 67 m mark.
Answer:
Suppose the micrometeoroid weighed 1 g = .001 kg
Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)
Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground
v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s
KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules
One can see that 28000 Joules could be damaging amount of energy
Answer:
final kinetic energy of the hammer is 10 kJ
Explanation:
As we know that there is no non conservative force on the system
So here we can use the theory of mechanical energy conservation
So we will have
here we know that
from above expression now
so final kinetic energy of the hammer is 10 kJ
