Answer:
<em>J=36221 Kg.m/s</em>
Explanation:
<u>Impulse-Momentum Theorem</u>
These two magnitudes are related in the following way. Suppose an object is moving at a certain speed
and changes it to
. The impulse is numerically equivalent to the change of linear momentum. Let's recall the momentum is given by

The initial and final momentums are, respectively

The change of momentum is

It is numerically equal to the Impulse J


We are given

The impulse the car experiences during that time is

J=-36221 Kg.m/s
The magnitude of J is
J=36221 Kg.m/s
Answer:
a)W= - 720 J
b)ΔU= 330 J
Explanation:
Given that
P = 0.8 atm
We know that 1 atm = 100 KPa
P = 80 KPa
V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)
V₂ = 3 L = 0.003 m³
Q= - 390 J ( heat is leaving from the system )
We know that work done by gas given as
W = P (V₂ -V₁ )
W= 80 x ( 0.003 - 0.012 ) KJ
W= - 0.72 KJ
W= - 720 J ( Negative sign indicates work done on the gas)
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in the internal energy
Now by putting the values
- 390 = - 720 + ΔU
ΔU= 720 - 390 J
ΔU= 330 J
Answer: First, we determine the circumference of the Mars by the equation below.
C = 2πr
Substituting the known values,
C = 2(π)(3,397 km) = 6794π km
To determine the tangential speed, we divide the circumference calculated above by the time it takes for Mars to complete one rotation and that is,
tangential speed = 6794π km / 24.6 hours = 867.64 km/h
C. The bowling ball and the bicycle
p = mv
P of bike = 12x5 = 60
P of rock = 2x20 = 40
P of ball = 5 x 10 = 60
Answer:
The vector form is as shown in the attachment
Explanation:
The figure as shown in the diagram, indicates that the car is moving along the road at a constant speed. Centripetal acceleration comes into play for an object moving in a circular motion at uniform speed. The centripetal acceleration is the acceleration experienced by an object while in uniform circular motion.
Mathematically from centripetal acceleration; a = v2/r
The equation shows that there is an inverse relationship between the acceleration and the radius of curvature as such the radius of curvature at the point A will be more than the radius of curvature at the point C, this shows that the centripetal acceleration at point C will be more than the centripetal acceleration at point A.
The attachment shows the figure and the representation in vectorial form.