Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :

So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
Answer:
(b) 10 Wb
Explanation:
Given;
angle of inclination of magnetic field, θ = 30°
initial area of the plane, A₁ = 1 m²
initial magnetic flux through the plane, Φ₁ = 5.0 Wb
Magnetic flux is given as;
Φ = BACosθ
where;
B is the strength of magnetic field
A is the area of the plane
θ is the angle of inclination
Φ₁ = BA₁Cosθ
5 = B(1 x cos30)
B = 5/(cos30)
B = 5.7735 T
Now calculate the magnetic flux through a 2.0 m² portion of the same plane
Φ₂ = BA₂Cosθ
Φ₂ = 5.7735 x 2 x cos30
Φ₂ = 10 Wb
Therefore, the magnetic flux through a 2.0 m² portion of the same plane is is 10 Wb.
Option "b"
Answer:
126.99115 g
Explanation:
50 g at 90 cm
Stick balances at 61.3 cm
x = Distance of the third 0.6 kg mass
Meter stick hanging at 50 cm
Torque about the support point is given by (torque is conserved)

The mass of the meter stick is 126.99115 g
Answer:
Answer:
1.1 x 10^9 ohm metre
Explanation:
diameter = 1.5 mm
length, l = 5 cm
Potential difference, V = 9 V
current, i = 230 micro Ampere = 230 x 10^-6 A
radius, r = diameter / 2 = 1.5 / 2 = 0.75 x 10^-3 m
Let the resistivity is ρ.
Area of crossection
A = πr² = 3.14 x 0.75 x 0.75 x 10^-6 = 1.766 x 10^-6 m^2
Use Ohm's law to find the value of resistance
V = i x R
9 = 230 x 10^-6 x R
R = 39130.4 ohm
Use the formula for the resistance



ρ = 1.1 x 10^9 ohm metre
Explanation:
Answer: a) 95.07m b) 81.88 m
Explanation:
a)
For finding the distance when vehicle is going downhill we have the formula as:
Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)
Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31
Reaction time= 0.28
So putting values we get
Stop sight distance= 0.28*72.4 *1 + 
Stop sight distance= 95.07 m
b)
For finding the distance when vehicle is going uphill we have the formula as:
Stop sight distance= Velocity*Reaction time + Velocity² / 2*g*(f constant- Grade value)
Now by AASHTO, we have for v= 45 mph= 72.4 kph, f= 0.31
Reaction time= 0.28
So putting values we get
Stop sight distance= 0.28*72.4 *1 + 
Stop sight distance= 81.88 m