What is the law of gravity?
1 answer:
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Answer:
The height of the other ball go after the collision is 2.304 m.
Explanation:
Given that,
Mass of ball = 6.00 kg
Height = 12.0 m
Mass of bar =5.00 kg
Length = 4.00 m
Suppose we need to calculate how high will the other ball go after the collision
We need to calculate the velocity of ball
Using formula of velocity
We need to calculate the angular momentum
Using formula of angular momentum
Put the value into the formula
We need to calculate the angular momentum
Using formula of angular momentum
Put the value into the formula
After collision the ball leaves with velocity
We need to calculate the velocity after collision
Using formula of the velocity
We need to calculate the height
Using formula of height
Put the value into the formula
Hence, The height of the other ball go after the collision is 2.304 m.
Answer:
kinematic viscosity in SUS is = 671.64 SUS
Explanation:
given data
kinetic viscosity = 145 mm^2/s
we know
1 mm = 0.1 cm
so kinetic viscosity in cm is
other unit of kinetic viscosity is centistokes
so 1.45 cm^2/s will be 145 cst
if the temperature is 260°f , then cst value should be multiplied by 4.632. therefore kinematic viscosity in SUS is = 4.362 *145 = 671.64 SUS
Answer: 5.21 s
Explanation:
Given
Helicopter ascends vertically with
Height of helicopter
When the package leaves the helicopter, it will have the same vertical velocity
Using equation of motion
So, package will take 5.21 s to reach the ground
(a) the initial kinetic energy of the projectile is equal to:
The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to
b) If the projectile is fired with an angle of
, its velocity has 2 components, one in the x-direction and one in the y-direction:
This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:
the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory:
The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to
b) If the projectile is fired with an angle of
This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:
the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory: