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3 years ago

A large body collides with a stationary small body from the rear end. How do the bodies behave if the collision is inelastic?.

1 answer:

6 0

If the collision is inelastic, there is every possibility that the large body will drag the small stationary body along with it in the direction of the collision. Some amount of heat, light and sound energy will also be produced due to the kinetic energy of the large body. I hope the answer helps you.

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Before a rifle is fired, the linear momentum of the bullet-rifle system is zero.

marusya05 [52]

Hi there!

II. Linear momentum of the system is zero.

This is an example of a RECOIL collision. With the Law of Conservation of Momentum, momentum remains constant before and after the collision.

Thus, the total momentum would also be equivalent to zero after the collision.

3 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$

$C= 5.32x10^{-6}$ F

C= 5.32 uF because u is the symbol for micro that is equal to $10^{-6}$

8 0

3 years ago

A 1.5 kg ball is dropped from a height of 2.Gm. Assuming energy is

Vikki [24]

Answer:

Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.

Explanation:

6 0

2 years ago

Two 60 cm parallel disks are separated by 40 cm and are aligned directly on top of each other. Both disks are black surfaces wit

Crazy boy [7]

Answer:

775.48 W

Explanation:

given,

diameter of disk = 0.6 cm

length of the disk = 0.4 m

T₁ = 450 K T₂ = 450 K T₃ = 300 K

$\dfrac{d}{r_1}=\dfrac{0.4}{0.3}$ = 1.33

now,

the value of view factor (F₁₂)corresponding to 1.33

F₁₂ = 0.265

F₁₃ = 1 - 0.265 = 0.735

now,

net rate of radiation heat transfer from the disk to the environment:

$=\dot{Q_{1-3}+Q_{2-3}} = 2 \dot{Q_{1-3}}$

= 2 F₁₃ A₁ σ (T₁⁴ - T₃⁴)

= 2 x 0.735 x π x (0.3)² x (5.67 x 10⁻⁸ W/m²) (450⁴ - 300⁴)

= 775.48 W

Net radiation heat transfer from the disks to the environment = 775.48 W

3 0

4 years ago

In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of t

a_sh-v [17]

Answer:

a) 35.94 ms⁻²

b) 65.85 m

Explanation:

Take down the data:

ρ = 1000kg/m3

a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:

Ptot = Pgas + Pwater

However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:

Ptot = Pgas

= 6.46 × 10⁵ Pa

The change in pressure is given by the continuity equation:

ΔP = 1/2ρv²

where v is the velocity of the water as it exits the tank.

Calculating:

6.46 × 10⁵ =1/2 ×1000×v²

solving for v, we get v = 35.94 ms⁻²

b) The Bernoulli's equation will be applicable here.

The water is coming out with the same pressure, therefore, the equation will be:

ΔP = ρgh

6.46 × 10⁵ = 1000 x 9.81 x h

h = 65.85 meters

7 0

3 years ago