Answer:
200N
Explanation:
mass(m) = 10 kg
acceleration(a) = 20 m/s^2
Force = mass * acceleration
= 10*20
= 200 N
Force = 200N
C. A mechanical wave generally travels faster in gases than liquids.
Wow ! This one could have some twists and turns in it.
Fasten your seat belt. It's going to be a boompy ride.
-- The buoyant force is precisely the missing <em>30N</em> .
-- In order to calculate the density of the frewium sample, we need to know
its mass and its volume. Then, density = mass/volume .
-- From the weight of the sample in air, we can closely calculate its mass.
Weight = (mass) x (gravity)
185N = (mass) x (9.81 m/s²)
Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u>
-- For its volume, we need to calculate the volume of the displaced water.
The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³. So the volume of the
displaced water (in cm³) is the same as the number of grams in it.
The weight of the displaced water is 30N, and weight = (mass) (gravity).
30N = (mass of the displaced water) x (9.81 m/s²)
Mass = (30N) / (9.81 m/s²) = 3.058 kilograms
Volume of displaced water = <u>3,058 cm³</u>
Finally, density of the frewium sample = (mass)/(volume)
Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)
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I'm thinking that this must be the hard way to do it,
because I noticed that
(weight in air) / (buoyant force) = 185N / 30N = <u>6.1666...</u>
So apparently . . .
(density of a sample) / (density of water) =
(weight of the sample in air) / (buoyant force in water) .
I never knew that, but it's a good factoid to keep in my tool-box.
Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = 
/ vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
