KE= .5*M*V^2
.5*.06*50^2
=75 joules
Answer:
The relative density of the second liquid is 7.
Explanation:
From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.
Let us assume that the volume of the object is 'V'
Thus for the liquid in which the block is completely submerged
The buoyant force should be equal to weight of liquid
Mathematically

Thus for the liquid in which the block is 1/7 submerged
The buoyant force should be equal to weight of liquid
Mathematically

Comparing equation 'i' and 'ii' we see that

Since the first liquid is water thus 
Thus the relative density of the second liquid is 7.
Answer:
<h2>9.3kN</h2>
Explanation:
Step one:
given data
mass of bullet= 0.02kg
speed v=700m/s
time taken =1.5ms= 0.0015 seconds
Step two:
we know that from the first law
F=ma-----1 first law of motion
also, we know that
a=v/t----2
put a=v/t in equation 1 we have
F=mv/t
Step three:
substitute our given data to find force
F=0.02*700/0.0015
F=14/0.0015
F=9333.33N
F=9.3kN
<u>The average force exerted is 9.3kN</u>
Answer:
864 mT
Explanation:
The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.
The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.
Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.
The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR
Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.
Substituting these into B' = μ₀i/πR, we have
B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)
B = 10.8/1.25 × 10⁻⁵ T
B = 8.64 × 10⁻⁵ T
B = 864 × 10⁻³ T
B = 864 mT
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