A baseball has a mass of 0.145 kg. A professional pitcher throws a baseball 67 mi/h, which is 30.0 m/s. What is the magnitude of
2 answers:
The concept of momentum tells us that it is equivalent to the product between the mass and the velocity of the object, that is to say that in general it can be written as
Where,
m = mass
v = Velocity
Our values are given as,
Replacing we have that,
Therefore the magniude of the momentum of the pitched baseball is
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Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v =
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c
Answer:
(A) 0.54 kg.m^{2}
(B) 0.0156 N
Explanation:
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here is the complete question:
A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horizontal circle about the other end of the rod at 5100 rev/min. (a) Calculate the rotational inertia of the system about the axis of rotation. (b) There is an air drag of 2.60 x 10^{-2} N on the ball, directed opposite its motion. What torque must be applied to the system to keep it rotating at constant speed?.
solution
mass of the ball (m) = 1.5 kg
length of the rod (L) = 0.6 m
angular velocity (ω) = 4900 rpm
air drag (F) = 2.60 x 10^{-2} N = 0.026 N
(take note that values from the original question are used, with the exception of the air drag which was not in the original question)
(A) because the rod is mass less, the rotational inertia of the system is the rotational inertia of the rod about the other end, hence rotational inertia = where m = mass of ball and L = length of rod
= = 0.54 kg.m^{2}
(B) The torque that must be applied to keep the ball in motion at constant speed = FLsin90
= 0.026 x 0.6 x sin 90 = 0.0156 N
Answer:
<h2>Energy needed 1680kJ</h2>Explanation:
The quantity of heat required to raise the temperature of water to 100 degrees is expressed as
Given data
mass of water = 5kg
initial temperature T1= 20 °C
final temperature T2= 100 °C
Specific heat capacity of water= 4 200 J/Kg °C