<u>Answer:</u> The Young's modulus for the wire is 
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:
where,
Y = Young's Modulus
F = force exerted by the weight = 
m = mass of the ball = 10 kg
g = acceleration due to gravity = 
l = length of wire = 2.6 m
A = area of cross section = 
r = radius of the wire = 
(Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm = 
Putting values in above equation, we get:
Hence, the Young's modulus for the wire is
Answer:
554.27N
Explanation:
(a) The max frictional force exerted horizontally on the crate and the floor is,
Substitute the values,
μs=0.5
mass=113kg
g=9.81m/s
Ff=μsN
=μsmg
=(0.5 x 113 x 9.81)
Ff=554.27N
Answer:
Disagree
Explanation:
We cannot see infared lights.
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F × t = m( v - v₀ )
we substitute
F × 0.00265 = 1.75( 0 - 5.25 )
F × 0.00265 = 1.75( - 5.25 )
F × 0.00265 = -9.1875
F = -9.1875 / 0.00265
F = -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F = - F
we substitute
F = - ( -3466.98 N )
F = 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
Answer:
a car
A sled sliding across snow or ice.
a ball down a hill
mercury
Explanation:
