Answer:
A-500 N
Explanation:
The computation of the tension in the chain is shown below
As we know that
F = ma
where
F denotes force
m denotes mass = 7
And, a denotes acceleration
Now for the acceleration we have to do the following calculations
The speed (v) of the hammer is
v = Angular speed × radius
where,
Angular seed = 2 × π ÷ Time Period
So, v = 2 × π × r ÷ P
v = 2 × 3.14 × 1.8 ÷ 1
= 11.304 m/s
Now
a = v^2 ÷ r
= 70.98912 m/s^2
Now the tension is
T = F = m × a
= 7 × 70.98912
= 496.92384 N
= 500 N
-- As far as we know, the forces on the wheelbarrow are balanced.
-- That tells us that the net force on the wheelbarrow is zero, just
as if there were no forces acting on it at all.
-- That tells us that the wheelbarrow's acceleration is zero ... its
speed and direction of motion are not changing.
-- That tells us that the wheelbarrow is moving in a straight line
at a constant speed. It's very possible that relative to us, the speed
may be zero, but we can't tell that from the given information.
Answer:
875 N
Explanation:
From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because
Force = mass * acceleration, while
Acceleration = velocity / time, basically
Force = mass * velocity / time.
We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.
Going by the formula I stated, we can then say that
Force = 250 * 3.5 / 1
Force = 875 N
This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N
Answer:
v’= 9.74 m / s
Explanation:
The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.
Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer
f₁ ’= f₀ (v + v₀)/v
Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest
f₂’= f₁’ v/(v - vs)
Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’
v’= vo = vs
Let's replace
f₂’= f₀ (v + v’)/v v/(v -v ’)
f₂’= f₀ (v + v’) / (v -v ’)
(v –v’ ) f₂’ / f₀ = v + v ’
v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)
v’ (1 + 1.059) = 340 (1.059 - 1)
v’= 20.06 / 2.059
v’= 9.74 m / s
