Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
Think of the formula force=mass x acceleration. even though they have the same acceleration, a train has more mass. is that helpful?
Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
Answer:
Fall at equal acceleration with similar displacements.
Explanation:
-
Objects under free fall with no air resistance, are falling under the sole influence of gravity. So, under that conditions, objects with different masses will fall with the same rate of acceleration
Answer:
Explanation:
We know,
..............(1)
where,
η = Efficiency of the engine
T₁ = Initial Temperature
T₂ = Final Temperature
Q₁ = Heat available initially
Q₂ = Heat after reaching the temperature T₂
Given:
η =0.280
T₁ = 3.50×10² °C = 350°C = 350+273 = 623K
Q₁ = 3.78 × 10³ J
Substituting the values in the equation (1) we get
or
or
⇒ 
Now,
The entropy change (
) is given as:
or
substituting the values in the above equation we get
