Answer:
First frequency f = 153.846 Hz
Second period T = 1.625 ms
Step by stepplanation:
T = 6.5ms= 6.5*(10^-3)
But T = 1/f
f = 1/T
f = 1(6.5*(10^-3))
f = 153.846 Hz
Now the frequency f is increased by a factor of 3
= 3*153.846
= 461.538
New frequency = 461.538+153.846
= 615.384 Hz
New period T = 1/615.384
Period T = 1.625*10^-3
T= 1.625 ms
Explanation:
(a)
Time period, T1 = 6.5 ms
= 6.5 × 10^-3 s
Frequency is the reciprocal of time period.
f1 = 1 / T1 = 1 / (6.5 × 10^-3) = 153.85 Hz
Now the frequency becomes 3 times
f2 = 3 × f1 = 3 × 153.85 = 461.55 Hz
The new period is
T2 = 1 / f2 = 1 / 461.55
= 0.0022 second = 2.22 ms
P.E. = mgh
PE = 40 × 10 × 10
PE = 4000 Joule
16.3 ft/s
Let d=distance
and
x = length of shadow.
Therfore,
x=(d + x)
= 6/15
So,
15x = 6x + 6d
9x = 6d.
x = (2/3)d.
As we know that:
dx=dt
= (2/3) (d/dt)
Also,
Given:
d(d)=dt
= 7 ft/s
Thus,
d(d + x)=dt
= (7/3)d (d/dt)
Substitute, d= 7
d(d + x) = 49/3 ft/s.
Hence,
d(d + x) = 16.3 ft/s.
If the wagon travels 18.75 m, then the work done on the wagon is
(18.75 m) x (the steady force applied to the wagon all the way, in Newtons) .
The unit is Joules .
4m/s/s
a=f/m
m=5kg
f=20N
20/5=4
(N=kg-m/s/s)
