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vladimir2022 [97]

3 years ago

10

A car with a mass of 1.5 × 103 kilograms is traveling west at a velocity of 22 meters/second. It hits a stationary car with a ma

ss of 9.0 × 102 kilograms. If the collision is inelastic, what is the final direction and approximate velocity of the two cars?
A. 14 meters/second to the west
B. 14 meters/second to the east
C. 22 meters/second to the east
D. 22 meters/second to the west

2 answers:

stepan [7]3 years ago

5 0

Answer:

A. 14 meters/second to the west

Explanation:

The collision is inelastic, which means that:

- only the total momentum is conserved (not the kinetic energy)

- the two cars stick to each other and continue their motion together after the collision

Therefore, we can write the law of conservation of momentum as follows:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2 )v$

where:

$m_1 = 1.5 \cdot 10^3 kg$ is the mass of the first car

$u_1 = +22 m/s$ is the initial velocity of the first car (let's take as positive the west direction)

$m_2 = 9.0 \cdot 10^2 kg$ is the mass of the second car

$u_2 = 0$ is the initial velocity of the second car

$v$ is the final velocity of the two cars

Re-arranging the equation, we get:

$v=\frac{m_1 u_1}{m_1 +m_2}=\frac{(1.5\cdot 10^3 kg)(+22 m/s)}{1.5\cdot 10^3 kg+9.0\cdot 10^2 kg}=+13.8 m/s$

So, approximately 14 m/s, and the direction is still west since the sign is positive.

Licemer1 [7]3 years ago

3 0

A. 14 meters/second to the west

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Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is $\frac{1}{6}$

So, the equation is

$\frac{1}{6} = 1 - \frac{T_2}{T_1}$

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is $\frac{1}{3}$

Therefore the next equation is

$\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}$

Now solve both the equations

solve equations (1) and (2)

$2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)$

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Now put the T_2 value in any of the above equation

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7 0

3 years ago

a charge of 2 * 10^-9C is placed at the origin, and another charge of 4 * 10^-9C is placed at x = 1.5m. find the point between t

damaskus [11]

Answer:

x₁ = 0.62 m

Explanation:

In this exercise the force is electric, given by Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

This force is a vector, since the three charges are in a line we can reduce the vector sum to a scalar sum.

For the sense of force let us use that charges of the same sign repel and charges of the opposite sign attract.

∑ F = F₁₂ - F₂₃

They ask us to find the point where the summaries of the force is zero.

F₁₂ - F₂₃ = 0

F₁₂ = F₂₃

let's fix a reference system located in the first charge (more to the left), the distance between the two charges is d = 1.5 m and x is the distance to the location of the second sphere

k q₁q₂ / x² = k q₂q₃ / (d-x) ²

q₁ (d-x) ² = q₃ x²

let's solve

d² - 2 x d + x² = $\frac{q_3}{q_1}$ x²

x² (1 - $\frac{q_3}{q_1}$ ) - 2x d + d² = 0

we substitute the values

x² (1- 4/2) - 2 1.5 x + 1.5² = 0

x² (-1) - 3.0 x + 2.25 = 0

x² + 3 x - 2.25 = 0

let's solve the quadratic equation

x = [-3 ± $\sqrt{ 3^2 + 4 \ 2.25}$ ] / 2

x = [-3 ± 4.24] / 2

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x₂ = 3.62 m

since it indicates that the charge q₂ e places between the spheres, the correct solution is

x₁ = 0.62 m

7 0

3 years ago

mass of the planet is 12 times that of earth and its radius is thrice that of earth , then find the escape velocity on that plan

Over [174]

Answer:

The escape velocity on the planet is approximately 178.976 km/s

Explanation:

The escape velocity for Earth is therefore given as follows

The formula for escape velocity, $v_e$ , for the planet is $v_e = \sqrt{\dfrac{2 \cdot G \cdot m}{r} }$

Where;

$v_e$ = The escape velocity on the planet

G = The universal gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m = The mass of the planet = 12 × The mass of Earth, $M_E$

r = The radius of the planet = 3 × The radius of Earth, $R_E$

The escape velocity for Earth, $v_e_E$ , is therefore given as follows;

$v_e_E = \sqrt{\dfrac{2 \cdot G \cdot M_E}{R_E} }$

$\therefore v_e = \sqrt{\dfrac{2 \times G \times 12 \times M}{3 \times R} } = \sqrt{\dfrac{2 \times G \times 4 \times M}{R} } = 16 \times \sqrt{\dfrac{2 \times G \times M}{R} } = 16 \times v_e_E$

$v_e$ = 16 × $v_e_E$

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Andrej [43]

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4 years ago

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Maksim231197 [3]

For this, we will use the formula:

<span>a = [V₀⁻V₁]÷t

</span>Where:

a - Acceleration
<span>V₀ - Final Velocity
</span>V₁ - Initial Velocity
<span>t - Time

</span>Given:
a = -2.0 m/s²
V₁ = 15.0 m/s
V₀ = 0 m/s ; because the car stopped

Find:
t = ?

Solution:

a = [V₀⁻V₁]÷t

First, we should derive the formula to the element that we are looking for (which is time).

t = [V₀⁻V₁]÷a
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It took the car 7.5 seconds to stop

3 0

3 years ago