The force constant of the spring is 317.8 N/m.
We need to calculate the reduced mass
Using formula of reduced mass
Where, = atomic mass of H
= atomic mass of I
Put the value into the formula
We need to calculate the force constant of the spring
Using formula of frequency
Hence, The force constant of the spring is 317.8 N/m.
Magnitude of δv = 80.78 m/s
Angle , θ =
Here east direction is zero degree, that is zero degree is along positive X direction, north is along + Y direction, West is along - X direction and South is along -Y direction.
In this case initial velocity = 30.0 m/s south = -30 j
Final velocity = 75.0 m/s west = - 75 i
We know that change in velocity = Initial velocity - Final velocity
So δv = - 75 i - (-30 j)
= - 75 i + 30 j m/s
Magnitude of δv =
Angle , θ =
The equation to find work is <em>W=F*D</em> or <em>Work=Force*Distance.</em>
All we have to do is insert the values we know.
The distance traveled was 200 feet and the force was 50 pounds.
The amount of work done is 10,000 Joules.
Hope this helps!
The acceleration of textbook is downward at the slowest rate.
(a) is correct option.
The object is accelerating downward at the slowest rate.
(a). a textbook falling onto the surface of the moon
(b). a flower pot that has been pushed out a window
(c). a pebble falling off a cliff through the air
(d). a baseball thrown from third base to first base
We know that,
Acceleration is the product of mass of object and gravity.
We need to find the accelerating downward at the slowest rate
According to given data,
Acceleration depends upon weight and gravity.
We have some example of free fall.
(b), (c) and (d) are the example of free fall.
In which the gravity is same.
But in option (a) textbook falling onto the surface of the moon
The gravity on the moon is less than the earth.
So, we can say that the acceleration of textbook is downward at the slowest rate.
Hence, The acceleration of textbook is downward at the slowest rate.
Using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses of the object
u1 and u2 are the velocities before collision
v is the final collision
m1 = 300g = 0.3kg
u1 = 6.0m/s
m2 = 10g = 0.01kg
u2 = 30m/s
The bird's speed immediately after swallowing v
Substitute the given values into the formula
0.3(6) + 0.01(30) = (0.3+0.01)v
1.8+0.3 = 0.31v
2.1 = 0.31v
v = 2.1/0.31
<em>v = 6.77m/s</em>
<em>Hence the bird's speed immediately after swallowing is 6.77m/s</em>