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3 years ago

A Jeep Rubicon has a mass of 2,630 kg. With a momentum of 55,200 kg m/s. What was the speed, in m/s, of the vehicle?

Physics

1 answer:

Annette [7]3 years ago

4 0

Answer:

The speed of the vehicle was 21 m/s

Explanation:

Momentum can be defined as mass in motion

Momentum depends on the variables mass and velocity

The momentum of an object is equal to the mass of the object times the

velocity of the object

The given is:

→ A Jeep Rubicon has a mass of 2,630 kg

→ With a momentum of 55,200 kg m/s

We need to find the speed of the vehicle

→ The momentum = mass × speed

Substitute the values above in the rule

→ 55,200 = 2,630 × speed

Divide both sides by 2,630

→ speed = 20.99 ≅ 21 m/s

<em>The speed of the vehicle was 21 m/s</em>

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A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

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Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

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Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

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2 years ago

According to the concept of punctuated equilibrium, _____.

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<span>According to the concept of punctuated equilibrium, </span>new species evolve suddenly over relatively short periods of time (a few hundred to a thousand years), followed by longer periods in which little genetic change occurs. Hope this helps. Have a nice day.

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2 years ago

A car company wants to ensure its newest model can stop in less than 450 ft when traveling at 60 mph. If we assume constant dece

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Answer:

The value of acceleration that accomplishes this is 8.61 ft/s² .

Explanation:

Given;

maximum distance to be traveled by the car when the brake is applied, d = 450 ft

initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s

final velocity of the car when it stops, v = 0

Apply the following kinematic equation to solve for the deceleration of the car.

v² = u² + 2as

0 = 88.02² + (2 x 450)a

-900a = 7747.5204

a = -7747.5204 / 900

a = -8.61 ft/s²

|a| = 8.61 ft/s²

Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .

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2 years ago

Remember to include your data, equation, and work when solving this problem.

andrezito [222]

Answer:

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We can solve this problem by using Newton's proposed universal gravitation law.

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Where:

F = gravitational force between the moon and Ellen; units [Newtos] or [N]

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m1= Ellen's mass [kg]

m2= Moon's mass [kg]

r = distance from the moon to the earth [meters] or [m].

Data:

G = 6.67 * 10^-11 [N^2*m^2/(kg^2)]

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This force is very small compare with the force exerted by the earth to Ellen's body. That is the reason that her body does not float away.

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