The speed of sound in water is about 1.5 x 10³ m/s. What is the speed of sound in water in km/h?
1 answer:
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Assuming the the car is travelling to the right (→):
The information we know:
Using one of the equations of motion:
When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.
Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration
ω^2r
where
= mass of earth
G = gravitational constant = 6.6742 x 10-11<span> m</span>3<span> s</span>-2<span> kg</span><span>-1
</span>
= radius of earth
ω = angular velocity
<span>r = radius of orbit
To convert to angular velocity:
</span>Tangential velocity = rω
ω = 5000/r
Then,
r = 2557110.465 m
Therefore, the distance of the centers of the earth and the satellite is 2.6 x 10^6 m.
<span>
</span>
Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration
where
G = gravitational constant = 6.6742 x 10-11<span> m</span>3<span> s</span>-2<span> kg</span><span>-1
</span>
ω = angular velocity
<span>r = radius of orbit
To convert to angular velocity:
</span>Tangential velocity = rω
ω = 5000/r
Then,
r = 2557110.465 m
Therefore, the distance of the centers of the earth and the satellite is 2.6 x 10^6 m.
<span>
</span>
Answer:
29,7 m
Explanation:
We need to devide the problem in two parts:
A) Energy
B) MRUV
<u>Energy:</u>
Since no friction between pint (1) and (2), then the energy conservatets:
Energy = constant ----> Ek(cinética) + Ep(potencial) = constant
Ek1 + Ep1 = Ek2 + Ep2
Ek1 = 0 ; because V1 is zero (the ball is "dropped")
Ep1 = m*g*H1
Ep2= m*g*H2
Then:
Ek2 = m*g*(H1-H2)
By definition of cinetic energy:
m*(V2)²/2 = m*g*(H1-H2) --->
Replaced values: V2 = 14,0 m/s
<u>MRUV:</u>
The decomposition of the velocity (V2), gives a for the horizontal component:
V2x = V2*cos(α)
Then the traveled distance is:
X = V2*cos(α)*t.... but what time?
The time what takes the ball hit the ground.
Since: Y3 - Y2 = V2*t + (1/2)*(-g)*t²
In the vertical axis:
Y3 = 0 ; Y2 = H2 = 2 m
Reeplacing:
-2 = 14*t + (1/2)*(-9,81)*t²
solving the ecuation, the only positive solution is:
t = 2,99 sec ≈ 3 sec
Then, for the distance:
X = V2*cos(α)*t = (14 m/s)*(cos45°)*(3sec) ≈ 29,7 m
Answer:
total number of electron in 1 litter is 3.34 × electron
Explanation:
given data
mass per mole = 18 g/mol
no of electron = 10
to find out
how many electron in 1 liter of water
solution
we know molecules per gram mole is 6.02 × molecules
no of moles is 1
so
total number of electron in water is = no of electron ×molecules per gram mole × no of moles
total number of electron in water is = 10 × 6.02 × × 1
total number of electron in water is = 6.02× electron
and
we know
mass = density × volume ..........1
here we know density of water is 1000 kg/m
and volume = 1 litter = 1 × m³
mass of 1 litter = 1000 × 1 ×
mass = 1000 g
so
total number of electron in 1 litter = mass of 1 litter ×
total number of electron in 1 litter = 1000 ×
total number of electron in 1 litter is 3.34 × electron