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1 year ago

How do you find acceleration due to gravity with time and height given?

Physics

1 answer:

Feliz [49]1 year ago

3 0

Here, height is given which will be the distance for a freely falling object.

The velocity will be

$v=\text{ }\frac{h}{t}$

and the acceleration will be

$a=\frac{v}{t}$

In this way, the formula works.

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Suppose that you measure the length of a spaceship, at rest relative to you, to be 400 m. how long will you measure it to be if

Fantom [35]

Calculate the length of a spaceship as follows:

l = l₀√1 - v²/c²

=(400 m)√1 - (0.75c)2 c²

=264.575m.

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2 years ago

Two infinite wires 20 cm apart each carry a current of 3 A into the paper. d I I d/2 d/2 At a distance d 2 below their midpoint,

Rzqust [24]

<u>Answer:</u>

<h3>As electric current is carried in a cable, around it, a magnetic field is created. The lines of the magnetic fields form concentric circles around the wire. The direction of the magnetic field hinges on the direction of the current. It can be calculated by pointing the thumb of your right hand in the direction of the moment, using the "right hand law." The position of your curled fingers is in the magnetic field lines. The magnetic field magnitude depends on the sum of current, and the distance from the wire carrying the charge.</h3>

<u></u>

<u>Explanation:</u>

Determine the direction of vector B magnitude B: $B: B=\mu_{0} * 1 /(2 \pi r): r=d / 2 * \sqrt{2}$

$\cos \alpha=1 / 2 \Rightarrow \alpha=450$

Resultant magnitude strength: $B=2 B^{*} \cos \alpha=B=u_{0}^{*} 1 /(2 \pi r)=4.24^{*} 10^{-6} T=4.24 u T$ its direction is pointing to the left.

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3 0

3 years ago

A force of 100N was necessary to lift a rock. A total of 150J of work was done. How far was the rock lifted?

Gwar [14]

Answer:

A total of 150 joules of work was done

Explanation:

4 0

3 years ago

A 3.0-kg object moves to the right with a speed of 2.0 m/s. It collides in a perfectly elastic collision with a 6.0-kg object mo

Zinaida [17]

Answer:

The kinetic energy of the system after the collision is 9 J.

Explanation:

It is given that,

Mass of object 1, m₁ = 3 kg

Speed of object 1, v₁ = 2 m/s

Mass of object 2, m₂ = 6 kg

Speed of object 2, v₂ = -1 m/s (it is moving in left)

Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

$E=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_1^2$

$E=\dfrac{1}{2}\times 3\ kg\times (2\ m/s)^2+\dfrac{1}{2}\times 6\ kg\times (-1\ m/s)^2$

E = 9 J

So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.

3 0

3 years ago

PLEASE ANSWER, I NEED HELP

Scorpion4ik [409]

1) The gravitational force between Ellen and the moon is $1.56\cdot 10^{-3} N$

2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.

Explanation:

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

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

$m_1, m_2$ are the masses of the two objects

r is the separation between them

In this problem, we have:

$m_1 = 47 kg$ is the mass of Ellen

$m_2 = 7.35\cdot 10^{22} kg$ is the mass of the moon

$r=3.84\cdot 10^8 m$ is the distance between Ellen and the moon

Substituting, we find the gravitational force between Ellen and the moon:

$F=(6.67\cdot 10^{-11})\frac{(47)(7.35\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.56\cdot 10^{-3} N$

We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).

Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:

$a=\frac{F}{m}$

where

a is the acceleration

F is the force exerted on the vehicle

m is the mass of the vehicle

As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.

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6 0

3 years ago