As Saba was wearing high heels they are long from the bottom so they sank however Sana was wearing snow boots which means they were flat and so she didn’t sink.
Answer:
The centripetal force acting on the skater is <u>48.32 N.</u>
Explanation:
Given:
Radius of circular track is, 
Tangential speed of the skater is, 
Mass of the skater is, 
We are asked to find the centripetal force acting on the skater.
We know that, when an object is under circular motion, the force acting on the object is directly proportional to the mass and square of tangential speed and inversely proportional to the radius of the circular path. This force is called centripetal force.
Centripetal force acting on the skater is given as:

Now, plug in the given values of the known quantities and solve for centripetal force,
. This gives,

Therefore, the centripetal force acting on the skater is 48.32 N.
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:
41.27m/s
Explanation:
According to law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the velocity after impact
Given
m1 = 0.2kg
u1 = 43.7m/s
m2 = 45.9g = 0.0459kg
u2 = 30.7m/s
Required
Velocity after impact v
Substitute the given parameters into the formula
0.2(43.7)+0.0459(30.7) = (0.2+0.0459)v
8.74+1.409 = 0.2459v
10.149 = 0.2459v
v = 10.149/0.2459
v = 41.27m/s
Hence the speed of the golf ball immediately after impact is 41.27m/s
Answer:0.147 N-m
Explanation:
Given
Diameter of Pulley 
radius 
mass of first object 
mass of second object 
Now both masses will exert a torque a on Pulley
Torque due to first Pulley 

Torque due to second mass on Pulley 

Total Torque by masses 

so we need to apply a torque of magnitude 0.147 N-m opposite to the direction of