To solve this problem it is necessary to apply the concept related to wavelength, specifically when the wavelength is observed from a source that is in motion to the observer.
By definition the wavelength is given defined by,
Where
= Observed wavelength
= Wavelength of the source
c = Speed of light in vacuum
u = Relative velocity of the source to the observer
According to our data we have that the wavelength emitted from the galaxy is 1875nm which is equal to the wavelength from the source, while the wavelength from the observer is 
Therefore replacing in the previous equation we have,
Solving for u,
Therefore the speed of the gas relative to earth is 0.02635 times the speed of light.
Answer:
The final angular velocity is 20rad/s
Explanation:
We are given;
mass, m = 12 kg
radius, r = 0.25 m
Work done;W = 75 J
Moment of inertia of cylinder, I = (1/2) mr²
Thus,
I = (1/2) x 12 x 0.25² = 0.375 kg.m²
Now, from work energy theorem,
Work done = Change in kinetic energy
So, W = KE_f - KE_i
Now, Initial Kinetic Energy (KE_i) = 0
Final Kinetic Energy; KE_f = (1/2)Iω²
So, KE_f = (1/2) x 0.375 x ω²
KE_f = 0.1875 ω²
Now, W = 75 J
Thus,
From, W = KE_f - KE_i, we have;
75 = 0.1875 ω² - 0
75 = 0.1875 ω²
ω² = 75/0.1875
ω² = 400
ω = √400
ω = 20 rad/s
Answer:
100 m/s
Explanation:
Mass the mass of Bond's boat is m₁. His enemy's boat is twice the mass of Bond's i.e. m₂ = 2 m₁
Initial speed of Bond's boat is 0 as it won't start and remains stationary in the water. The initial speed of enemy's boat is 50 m/s. After the collision, enemy boat is completely stationary. Let v₁ is speed of bond's boat.
It is the concept of the conservation of momentum. It remains conserved. So,
Putting all the values, we get :
So, Bond's boat is moving with a speed of 100 m/s after the collision.
Answer:
Initial speed = 20 m/s
Final speed = 35 m/s
Time to speed up = 5 seconds
Explanation:
Directly from the information given:
Initial speed = 20 m/s
Final speed = 35 m/s
Time to speed up = 5 seconds
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
