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3 years ago

15. If an 800.-kg sports car slows to 13.0 m/s to check out an accident scene and the

Physics

1 answer:

goldfiish [28.3K]3 years ago

5 0

The final combined velocity after the collision is 20.2 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum: in fact, in absence of external forces, the total momentum of the two car and of the truck must be conserved before and after the collision.

This means that we can write the following equation:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$

where:

$m_1 = 800 kg$ is the mass of the sport car

$u_1 = 13.0 m/s$ is the initial velocity of the car (taking its direction as positive direction)

$m_2 = 1200 kg$ is the mass of the truck

$u_2 = 25.0 m/s$ is the initial velocity of the truck

$v$ is the final combined velocity of the car and the truck, after the collision

Re-arranging the equation and substituting the values, we find the velocity after the collision:

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{(800)(13)+(1200)(25)}{800+1200}=20.2 m/s$

And the positive sign indicates their final direction is the same as the initial direction of the two vehicles.

Learn more about momentum here:

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In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

Bond [772]

Answer:

$B_0 = 1.69 \times 10^{-4}\ T$

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

$v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}$

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

$v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}$

v_0 = 6496355.63 m/s

$v_0 = \dfrac{E}{B_0}$

$B_0 = \dfrac{E}{v_0}$

$B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}$

$B_0 = 1.69 \times 10^{-4}\ T$

5 0

3 years ago

A stone that is thrown vertically upwards was having a velocity of 15m/s after reaches 2/3 of its maximum height. What is the ma

attashe74 [19]

<em>The correct answer is option</em><em> B.</em> The maximum height that can be reached by the stone is determined as 11.5 m.

<h3>Maximum height attained by the stone </h3>

The maximum height attained by the stone when it is a 2/3 of its total height is calculated as follows;

v² = u² - 2gh

where;

v is final velocity at maximum height, v = 0
u is initial velocity
g is acceleration due to gravity

0 = u² - 2gh

2gh = u²

h = u²/2g

h = (15²)/(2 x 9.8)

h = 11.48 m

h = 11.5 m

Thus, the maximum height that can be reached by the stone is determined as 11.5 m

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4 0

2 years ago

Will give brainliest to right answer!

viva [34]

The correct answer is D.

8 0

4 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

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3 years ago

What are the units of measure for distance?

olga2289 [7]

The most common unit is meters (m for short). It is the base unit for distance or displacement in the metric system. If you are dealing with larger distances, you might use kilometers (I'm for short) which is just 1000 meters. On the other hand, centimeter (cm) are used for small distances and are 1/100 of a meter. Another common unit is millimeters (mm) which is 1/1000 of a meter.

6 0

4 years ago