Answer:
The time where the avergae speed equals the instaneous speed is T/2
Explanation:
The velocity of the car is:
v(t) = v0 + at
Where v0 is the initial speed and a is the constant acceleration.
Let's find the average speed. This is given integrating the velocity from 0 to T and dividing by T:
v_ave = v0+a(T/2)
We can esaily note that when <u><em>t=T/2</em></u><u><em> </em></u>
v(T/2)=v_ave
Now we want to know where the car should be, the osition of the car is:
Where x_A is the position of point A. Therefore, the car will be at:
<u><em>x(T/2) = x_A + v_0 (T/2) + (1/8)aT^2</em></u>
Use the concept of beat frequency to find the applicable final freqeuncy for 20Hz beat frequency.
Beat can be defined as 'the interference pattern between two sounds of slightly different frequencies0
The expression for beat frequency is given as
Where,
Final frequency
Initial frequency
The beat frequency for us is 25Hz and the initial frequency is 240Hz, then
Being an absolute value, two values are possible, both in addition and subtraction:
The two possible values are
Answer:
A) B = 24 ft
B) H = 24.08 ft
C) M.A = 12.04
D) P = 13.7 lb
Explanation:
A)
Minimum allowable length of base of ramp can be found as follows:
Slope = H/B
where,
Slope = 1/12
H = Height of Ramp = 2 ft
B = Length of Base of Ramp = ?
Therefore,
1/12 = 2 ft/B
B = 2 ft * 12
<u>B = 24 ft</u>
B)
The length of the slope of ramp can be found by using pythagora's theorem:
L = √H² + B²
where,
H = Perpendicular = height = 2 ft
B = Base = Length of Base of Ramp = 24 ft
L = Hypotenuse = Length of Slope of Ramp = ?
Therefore,
H = √[(2 ft)² + (24 ft)²]
<u>H = 24.08 ft</u>
D)
The mechanical advantage of an inclined plane is given by the following formula:
M.A = L/H
M.A = 24.08 ft/2 ft
<u>M.A = 12.04</u>
D)
Another general formula for Mechanical Advantage is:
M.A = W/P
where,
W = Ideal Load = 165 lb
P = Ideal Effort Force = ?
Therefore,
12.04 = 165 lb/P
P = 165 lb/12.04
<u>P = 13.7 lb</u>
Answer:
(a) 17.37 rad/s^2
(b) 12479
Explanation:
t = 95 s, r = 6 cm = 0.06 m, v = 99 m/s, w0 = 0
w = v / r = 99 / 0.06 = 1650 rad/s
(a) Use first equation of motion for rotational motion
w = w0 + α t
1650 = 0 + α x 95
α = 17.37 rad/s^2
(b) Let θ be the angular displacement
Use third equation of motion for rotational motion
w^2 = w0^2 + 2 α θ
1650^2 = 0 + 2 x 17.37 x θ
θ = 78367.87 rad
number of revolutions, n = θ / 2 π
n = 78367.87 / ( 2 x 3.14)
n = 12478.9 ≈ 12479
Answer:
Maximum Tension=224N
Minimum tension= 64N
Explanation:
Given
mass =8 kg
constant speed = 6m/s .
g=10m/s^2
Maximum Tension= [(mv^2/ r) + (mg)]
Minimum tension= [(mv^2/ r) - (mg)]
Then substitute the values,
Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N
Minimum tension= [8 × 6^2)/2 -(8×9.8)]
=64N
Hence, Minimum tension and maximum Tension are =64N and 2224N respectively
