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evablogger [386]
3 years ago
12

Two carts on an air track have the same mass and speed and are traveling toward each other. If they collide and stick together,

find the total momentum and total kinetic energy of the system.
Physics
1 answer:
lora16 [44]3 years ago
7 0

Answer:

p = 0, K_{A} = m\cdot v^{2} (before collision), K_{B} = 0 (after collision).

Explanation:

The total momentum is obtained using the Principle of Momentum Conservation:

m\cdot v_{A} - m \cdot v_{A} = 2\cdot v_{B}

It is trivial to find that final speed and total momentum of the system are zero:

p = 0

The total kinetic energy of the system becomes zero due to the inellastic collision and the same masses and speeds. Total kinetic energies before and after collision are, respectively:

K_{A} = m\cdot v^{2}

K_{B} = 0

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A carton is given a push across a horizontal, frictionless surface. The carton has a mass m, the push gives it an initial speed
sammy [17]

Answer and Explanation:

Data provided in the question

Carbon mass = m

Initial speed = v_i

Coefficient = μk

Based on the above information, the expressions are as follows

a. By using the energy considerations the expression for the carton moving distance is

As we know that

Fd = \frac{1}{2} m (v_i^2- v_f^2)

where,

v_f = 0

F = u_kmg

(\mu_kg) d = \frac{1}{2} m v_i^2

d = \frac{\frac{1}{2}v_i^2}{\mu_kg}

d = \frac{v_i^2}{2 \mu_kg}

b. The initial speed of the carton if the factor of 3 risen, so the expression is

v_i^1 = 3v_i

d^i = \frac{(3v_i^2)}{2\mu_kg}

= \frac{9v_i^2}{2\mu_kg}

d^i = 9(d)

8 0
3 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
yKpoI14uk [10]

Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

where ρ is density and P is pressure

Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

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