Answer:
i hope it will be useful for you
Explanation:
F=5.6×10^-10N
R=93cm=0.93m
let take m1 and m2 =m²
according to newton's law of universal gravitation
F=m1m2/r²
F=m²/r²
now we have to find masses
F×r²=m²
5.6×10^10N×0.93m=m²
5.208×10^-9=m²
taking square root on b.s
√5.208×10^-9=√m²
so the two masses are m1=7.2×10^-5
and m2=7.2×10^-5
Answer:
<em> 508Hz</em>
Explanation:
A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.
When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production
frequency is the number of oscillation a wave makes in one seconds.
f1-f2=beats
therefore f1=512Hz
f2=?
beats=4Hz
512Hz-f2=4Hz
f2=512-4
f2=508Hz
the original frequency of the violin is 508Hz
Answer:
S = 16 m
Explanation:
Given that
The frequency of the water waves, f = 4 Hz
The wavelength of the water waves, λ = 2 m
The time the waves reached the shore, t = 2 s
The relation between the velocity, wavelength, and the frequency of the wave is given by the relation,
v = f λ m/s
Substituting the given values in the above equation,
v = 4 x 2
= 8 m/s
The velocity of the water waves is v = 8 m/s
The distance between the shore and boat is given by
s = v x t
= 8 x 2
= 16 m
Hence, the distance between the boat and the shore is, s = 16 m
Answer:
The acceleration is 14.28 km/h^2
Explanation:
Step one:
Given data
initial speed u= 0 km/h
final speed v= 140km/h
time t= 9.8 seconds
Required
The acceleration of the car
Step two:
From a= v-u/t
substitute
a= 140-0/9.8
a=140/9.8
a=14.28 km/h^2
a)
for the puck :
F = force applied in the direction of pull
N = normal force on the puck in upward direction by the surface of table
W = weight of the puck in down direction due to force of gravity
b)
along the vertical direction , normal force balance the weight of the puck , hence the net force is same as the force of pull F .
so F = ma where m = mass of puck , a = acceleration
Fnet = F
c)
since the net force acts in the direction of force of pull F , hence the puck accelerates in the same direction .
