Answer:
Weight of the body in air = 20gf ( Wa)
Weight of the body in water = 18gf ( Ws)
Weight of the body in liquid =18.2gf ( Wl)
Now,
The loss of weight in the water = 20 - 18 or, 2 gf
So the relative density of the body is = Wa / ( wa - ws )
= 20 / 2
= 10
2) The weight loss in liquid = 20 - 18.2 or, 1.8 gf
So the relative density of the liquid is =( Wa - Wl ) / ( Wa - Ws )
= 1.8 / 2
= 0.9
Explanation:
Answer:
A = 10 m amplitude
m = 3 kg mass of object
Vm = 5 m/s
w A = Vm where w = omega
w = 2 * pi * f
2 * pi * f 10 = 5
f = 5 / (20 * pi) = .0796 / sec
Answer:
You have the answer in your comments. I will be copying it so your question doesn't get deleted.
The answers is $0.58
$11.48
the video tape
the new shirt
Answer:
106.7 N
Explanation:
We can solve the problem by using the impulse theorem, which states that the product between the average force applied and the duration of the collision is equal to the change in momentum of the object:
where
F is the average force
is the duration of the collision
m is the mass of the ball
v is the final velocity
u is the initial velocity
In this problem:
m = 0.200 kg
u = 20.0 m/s
v = -12.0 m/s
Solving for F,
And since we are interested in the magnitude only,
F = 106.7 N
Answer:
a. 299,792,458 m/s
Explanation:
Since the speed of light in a vacuum is invariant and has the value of 299,792,458 m/s, we would measure this value of 299,792,458 m/s for the speed of light from the star as it arrives on Earth.
