Answer:
Explanation:
Range of projectile R = 20 m
formula of range
R = u² sin2θ / g
u is initial velocity , θ is angle of projectile
putting the values
20 = u² sin2x 40 / 9.8
u² = 199
u = 14.10 m /s
At the initial point
vertical component of u
= u sin40 = 14.1 x sin 40
= 9.06 m/s
Horizontal component
= u cos 30
At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .
Horizontal component of velocity
u cos 30
Vertical component
= - u sin 30
= - 9.06 m /s
So its horizontal component remains unchanged .
change in vertical component = 9.06 - ( - 9.06 )
= 18.12 m /s
change in momentum
mass x change in velocity
= .050 x 18.12
= .906 N.s
Impulse = change in momentum
= .906 N.s .
Answers:
a) 
b) 
c) 
Explanation:
We have the following data:
is the spring constant
is the amplitude of oscillation
is the velocity of the block when
Now let's begin with the answers:
<h3>a) Mass of the block</h3>
We can solve this by the conservation of energy principle:
(1)
Where:
is the initial potential energy
is the initial kinetic energy
is the final potential energy
is the final kinetic energy
Then:
(2)
Isolating
:
(3)
(4)
(5)
<h3>b) Period</h3>
The period
is given by:
(6)
Substituting (5) in (6):
(7)
(8)
<h3>c) Maximum acceleration</h3>
The maximum acceleration
is when the force is maximum
, as well :
(9)
Being 
Hence:
(10)
Finding
:
(11)
(12)
Finally:

Answer:
5 hours
Explanation:
Let the required time be x hours. The time will be the same for both cars.
The cars will cover different distances because they are travelling at different speeds.
<em>D=S×T
</em>
The distance travelled by the slower car = 50×x miles.
The distance travelled by the faster car = 58×x miles.
The two distances differ by 40 miles.
58x−50x=40
8x=40
x=5 hours
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
A second method:
The difference in the distances is 40 miles
The difference in the speeds is #8mph.
The time to make up the 40 miles=
=5 hours
We can solve the problem by using the first law of thermodynamics:

where
is the change in internal energy of the system
is the heat absorbed by the system
is the work done by the system on the surrounding
In this problem, the work done by the system is

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

with a negative sign as well because it is released by the system.
Therefore, by using the initial equation, we find
