Answer:
21.2 degrees
Explanation:
Let gravitational acceleration constant g = 9.81 m/s2 directed downward. We can calculate the g component that is parallel to the 25 degree incline:
This parallel component would produce an acceleration of the block. But since the net acceleration is only 0.7 m/s2, there's a friction acceleration that hinders g parallel. This acceleration can be calculated by
The force of friction would be
Friction is a product of normal force and its coefficient
For the crate to slide down at constant speed, the net acceleration must be 0. In other words, the parallel g must be the same as acceleration caused by friction. Let this incline angle be α
Answer:
the required frequency of waves is 2.066 Hz
Explanation:
Given the data in the question;
μ = 1.50 kg/m
T = 6000 N
Amplitude A = 0.500 m
P = 2.00 kW = 2000 W
we know that, the average power transmit through the rope can be expressed as;
p = 
vμω²A²
p = √(T/μ)μω²A²
so we solve for ω
ω² = 2P / √(T/μ)μA²
we substitute
ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²
ω² = 4000 / 23.71708
ω² = 168.65
(2πf)² = ω²
so
(2πf)² = 168.65
4π²f² = 168.65
f² = 168.65 / 4π²
f² = 4.27195
f = √4.27195
f = 2.066 Hz
Therefore, the required frequency of waves is 2.066 Hz
Answer:
A. Mrŵ² = ųMg
Ŵ = (ųg/r)^½
B.
Ŵ =[ (g /r)* tan á]^½
Explanation:
T.v.= centrepetal force = mrŵ²
Where m = mass of block,
r = radius
Ŵ = angular momentum
On a horizontal axial banking frictional force supplies the Pentecostal force is numerically equal.
So there for
Mrŵ² = ųMg
Ŵ = (ųg/r)^½
g = Gravitational pull
ų = coefficient of friction.
B. The net external force equals the horizontal centerepital force if the angle à is ideal for the speed and radius then friction becomes negligible
So therefore
N *(sin á) = mrŵ² .....equ 1
Since the car does not slide the net vertical forces must be equal and opposite so therefore
N*(cos á) = mg.....equ 2
Where N is the reaction force of the car on the surface.
Equ 2 becomes N = mg/cos á
Substituting N into equation 1
mg*(sin á /cos á) =mrŵ²
Tan á = rŵ²/g
Ŵ =[ (g /r)* tan á]^½
6 cubic feet I’m pretty sure that’s the answer
Answer:
The rate at which bus 1 is going is 55 mph
The rate at which bus 1 is going is 35 mph
Explanation:
As per the question:
Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h
Thus
Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.
R' = R - 20
The distance is given as the product of rate and time:
d = Rt (1)
Now, the total distance given is 270 miles:
d + d' = 270
Now, using eqn (1):
Rt + R't = 270
3(R + R - 20) = 270
6R = 270 + 60
R = 55 mph
R' = R - 20 = 55 - 20 = 35 mph
