Answer:
2. 4.63 x 10 to the -11th power
Explanation:
m1 = 3.1, m2= 6.3, d = 5.3, G = 6.67 × 10^-11
m1 · m2 = 3.1 · 6.3 = 19.5
d² = 5.3² = 28.1
19.5/28.1 = 0.694
(6.67 · 10^-11) · 0.694 = 4.63 × 10^-11
Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
Answer:
the amplitude of the wave
the energy of the wave
the type of wave
the type of medium
Answer:
Q = 1.095 x 10^-9 C
Let the force experienced by the top piece of tape be F
F = kQ²/r²
r = distance between the two pieces tape = 1.00cm = 1.00 x 10^ -2 m
1/4(pi)*Eo = k = 8.99 x 10^9 Nm²/C²
The electric force of repulsion between the two charges and the weight of the top piece of tape are equal so
F = kQ²/r² = mg
Where m is the mass of the top piece of tape and g is the acceleration due to gravity
On re-arranging the equation above,
Q² = mgr²/k
Q² = ((11.0 x 10^-6) x 9.8 x (1.00x10^-2)²)/(8.99 x 10^9)
Q = 1.095x10^-9 C
Explanation:
The charge Q on both pieces of tape are equal and both act with a force of repulsion on each other.
The force of repulsion between both tapes pushes the top piece of tape upwards. The weight of the top piece of tape acts vertically downward. Since the top tape is in a position of equilibrium, the two forces acting on the top piece of tape must be equal to each other. This assumption is backed up by newton's first law of motion which states that the summation of all forces acting on a body at rest must be equal to zero. That is
Fe (electric force) - Fg (gravitational force) = 0
Fe = Fg
kQ²/r² = mg
On substituting the respective values for all variables except Q and rearranging the equation Q = 1.09 x 10^-9
Answer:
The the average magnitude of the force exerted on the ball by the floor 
N
Explanation:
Given:
Mass of ball 
kg
Initial speed 
Rebound speed 
Contact time interval 
sec
For finding the average magnitude of the force on the ball by the floor is given by,
Here 
N
Therefore, the the average magnitude of the force exerted on the ball by the floor N
