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2 years ago

A baby carriage is sitting at the top of a hill that is 35 m high. The

Physics

2 answers:

DerKrebs [107]2 years ago

7 0

Answer:

$\boxed {\boxed {\sf E_p= 515.025 \ J}}$

Explanation:

Potential energy is found using this formula:

$E_p= mgh$

The mass of the baby in the carriage is 1.5 kilograms and the height is 35 meters. Assuming this is on Earth, the gravitational acceleration is 9.81 meters per square second.

$m=1.5 \ kg \\g= 9.81 \ m/s^2\\h= 35 \ m$

Substitute the values into the formula.

$E_p= (1.5 \ kg )( 9.81 \ m/s^2)(35 \ m )$

Multiply the first numbers together.

$E_p= 14.715 \ kg*m/s^2 (35 \ m )$

Multiply again.

$E_p= 515.025 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 515.025 kg*m²/s² is equal to 515.025 J

$E_p= 515.025 \ J$

The potential energy is 515.025 Joules.

Lemur [1.5K]2 years ago

5 0

P=M(mass)G(Gravity)H(Height)

Gravity=9.8

M=1.5 G=9.8 H=35

so multiply all

=514.5 potential energy

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When the distance between two masses is doubled, the gravitational attraction between

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The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

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where

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m1, m2 are the masses of the two objects

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In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

$r'=2r$

Therefore, the new force will be

$F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}$

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3 years ago

A 790kg car moving at 7 m/s takes a turn around a circle with a radius of 20 m. Determined the net force (in Newton’s) acting up

prisoha [69]

1935.5 N is the "net force" acting on a car.

<u>Explanation</u>:

Given that,

Mass of the car is 790 kg.

Velocity of the car is 7 m/s. (v)

It turned around with 20 m. (r)

We know that, Net force = m × a

$\text { Here, acceleration of the car is radial acceleration } a_{\mathrm{rad}}=\frac{v^{2}}{r}$

$\mathrm{a}_{\mathrm{rad}}=\frac{7^{2}}{20}$

$\mathrm{a}_{\mathrm{rad}}=\frac{49}{20}$

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3 years ago

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3 years ago

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2 years ago

If two objects A and B have the same kinetic energy but A has three times the momentum of B, what is the ratio of their inertias

Thepotemich [5.8K]

Answer:

$\frac{inertia_B}{inertia_A}=9$

Explanation:

First of all, let's remind that:

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- The momentum of an object is given by $p=mv$

- The inertia of an object is proportional to its mass, so we can write $I=km$ , where k just indicates a constant of proportionality

In this problem, we have:

- $K_A = K_B$ (the two objects have same kinetic energy)

- $p_A = 3 p_B$ (A has three times the momentum of B)

Re-writing both equation we have:

$\frac{1}{2}m_A v_A^2 = \frac{1}{2}m_B v_B^2\\m_A v_A = 3 m_B v_B$

If we divide first equation by second one we get

$v_A = 3 v_B$

And if we substitute it into the first equation we get

$m_A (3 v_B)^2 = m_B v_B^2\\9 m_A v_B^2 = m_B v_B^2\\m_B = 9 m_A$

So, B has 9 times more mass than A, and so B has 9 times more inertia than A, and their ratio is:

$\frac{I_B}{I_A}=\frac{km_B}{km_A}=\frac{9m_A}{m_A}=9$

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3 years ago