Answer:
This is due to impulse
Explanation:
Impulse equal to mΔv and FΔt
You can set these equal as mΔv = FΔt
When a boxer punches a tissue, it is like punching a cushion or a pillow. The time that the hit takes is much grater than if they were to hit something solid. In addition, the change in velocity of the boxer's arm would be much greater when they hit a punching bag. In this equation, the greater the time, the less force that is needed.
Answer:
The height of the image will be "1.16 mm".
Explanation:
The given values are:
Object distance, u = 25 cm
Focal distance, f = 1.8 cm
On applying the lens formula, we get
⇒ 
On putting estimate values, we get
⇒ 
⇒ 
⇒ 
As a result, the image would be established mostly on right side and would be true even though v is positive.
By magnification,
and 
⇒ 
⇒ 
⇒ 
Answer:
Explanation:
Given
Each student exert a force of 
Let mass of car be m
there are 18 students who lifts the car
Total force by 18 students 
therefore weight of car 
mass of car 
(b)
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
Answer:
a)1815Joules b) 185Joules
Explanation:
Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;
F = ke where;
F is the applied force
k is the elastic constant
e is the extension of the material
From the formula, k = F/e
F1/e1 = F2/e2
If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;
k = 60/0.5
k = 120N/m
a) To get the work done in stretching the spring 5.5m from its position,
Work done by the spring = 1/2ke²
Given k = 120N/m, e = 5.5m
Work done = 1/2×120×5.5²
Work done = 60× 5.5²
Work done = 1815Joules
b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;
Work done = 1/2ke²
Work done =1/2× 120×1.5²
Works done = 60×1.5²
Work done = 135Joules
