Answer:
Time taken = 10400 s
Explanation:
Given:
Initial speed of the train, 
Final speed of the train, 
Displacement of the train, 
Using Newton's equation of motion,

Now, using Newton's equation of motion for displacement,

Now, plug in the value of
in the above equation. This gives,

Now, plug in 234000 m for
, 25 m/s for
and 20 m/s for
. Solve for
.

Therefore, the time taken by the train is 10400 s.
Ans: Beat Frequency = 1.97HzExplanation:
The fundamental frequency on a vibrating string is

<span> -- (A)</span>
<span>here, T=Tension in the string=56.7N,
L=Length of the string=0.66m,
m= mass = 8.3x10^-4kg/m * 0.66m = 5.48x10^-4kg </span>
Plug in the values in Equation (A)
<span>so </span>

<span> = 197.97Hz </span>
<span>the beat frequency is the difference between these two frequencies, therefore:
Beat frequency = 197.97 - 196.0 = 1.97Hz
-i</span>
Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
Answer:
2 kg
Explanation:
Acceleration = 5 m/s^2
Force = 10 N
Force = mass * acceleration
mass = force / acceleration
mass = 10 / 5
mass = 2 kg
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