Answer:
Hey
Yes, this is true.
As some people have it wrong, waves in the water (ocean) are not waves of moving water, rather the wave is moving through the water. A wave is a disturbance of a medium not the meduim moving.
Answer:
Wn = 9.14 x 10¹⁷ N
Explanation:
First we need to find our mass. For this purpose we use the following formula:
W = mg
m = W/g
where,
W = Weight = 675 N
g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²
m = Mass = ?
Therefore,
m = (675 N)/(9.8 m/s²)
m = 68.88 kg
Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:
gn = (G)(Mn)/(Rn)²
where,
gn = acceleration due to gravity on surface of neutron star = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg
Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m
Therefore,
gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)
gn = 13.27 x 10¹⁵ m/s²
Now, my weight on neutron star will be:
Wn = m(gn)
Wn = (68.88)(13.27 x 10¹⁵ m/s²)
<u>Wn = 9.14 x 10¹⁷ N</u>
Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
Answer:
r = 0.0548 m
Explanation:
Given that,
Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.
We need to find the radius of their circular path. The formula for the radius of path is given by :
m is mass of Singly charged uranium-238 ion, 
q is charge
So,
So, the radius of their circular path is equal to 0.0548 m.
Answer:
Explanation:
velocity of first projectile after 3 s
v = u - gt
v = 49.4 - 9.8 x 3
= 20 m /s
Velocity of second projectile after 3 s after being dropped from rest
v = u + gt
= 0 + 9.8 x 3
= 29.4 m /s
They will be moving in opposite direction at the time of meeting , so their relative velocity
= 20 + 29.4 = 49.4 m /s
From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .
