Answer: Change in ball's momentum is 1.5 kg-m/s.
Explanation: It is given that,
Mass of the ball, m = 0.15 kg
Speed before the impact, u = 6.5 m/s
Speed after the impact, v = -3.5 m/s (as it will rebound)
We need to find the change in the magnitude of the ball's momentum. It is given by :
So, the change in the ball's momentum is 1.5 kg-m/s. Hence, this is the required solution.
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Answer:
d = 84 m
Explanation:
As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object
here we know that
initial speed 
final speed 
time taken by the car to stop

now the distance moved by the car before it stop is given as

now we have


Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
Answer:
The equilibrium temperature is
21.97°c
Explanation:
This problem bothers on the heat capacity of materials
Given data
specific heat capacities
copper is Cc =390 J/kg⋅C∘,
aluminun Ca = 900 J/kg⋅C∘,
water Cw = 4186 J/kg⋅C∘.
Mass of substances
Copper Mc = 235g
Aluminum Ma = 135g
Water Mw = 825g
Temperatures
Copper θc = 255°c
Water and aluminum calorimeter θ1= 16°c
Equilibrium temperature θf =?
Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water
McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)
Substituting our data into the expression we have
235*390(255-θf)=
(135*900+825*4186)(θf-16)
91650(255-θf)=(3574950)(θf-16)
23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6
Collecting like terms and rearranging
23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf
8.2*10^6=3.66*10^6θf
θf=80.5*10^6/3.6*10^6
θf =21.97°c
Answer:
A). A virtual image cannot be formed on a screen.
Explanation:
A virtual image can not be formed on a screen.
For image:
1.A virtual image can be viewed by the unaided eye.
2. A real image must be erect or maybe inverted.
3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.
4.A virtual image can be photographed.
So the option A is correct.