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3 years ago

MATHPHYSSSSSSSS PLEASEEEEEE IM SORRY YOU PROBABLY HATE ME

Physics

1 answer:

inysia [295]3 years ago

3 0

Answer:

3.1 m/s

Explanation:

First, find the time it takes for the cat to land. Take down to be positive.

Given:

Δy = 0.61 m

v₀ = 0 m/s

a = 9.81 m/s²

Find: t

Δy = v₀ t + ½ at²

(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²

t = 0.353 s

Now find the horizontal velocity needed to travel 1.1 m in that time.

Given:

Δx = 1.1 m

a = 0 m/s²

t = 0.353 s

Find: v₀

Δx = v₀ t + ½ at²

(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²

v₀ = 3.1 m/s

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180 J of work is done when lifting a box up to a shelf that is 3 m high. What is the mass of the box?

ale4655 [162]

Explanation:

Work done= MGH

180=m×10×3

180/30=m

m=6kg

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2 years ago

James gently releases a ball at the top of a slope But does not push the ball. Space the ball rolls down the slope. Which force

NARA [144]

On an incline, the force causing the ball to move downwards would be gravity. Additionally, the component of gravity causing this ball to move downwards would be mgsintheta.

Hope this helps!

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3 years ago

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What is the radius of a communications satellite’s orbital path that is in a uniform circular orbit around Earth and has a perio

pantera1 [17]

If the period of a satellite is T=24 h = 86400 s that means it is in geostationary orbit around Earth. That means that the force of gravity Fg and the centripetal force Fcp are equal:

Fg=Fcp

m*g=m*(v²/R),

where m is mass, v is the velocity of the satelite and R is the height of the satellite and g=G*(M/r²), where G=6.67*10^-11 m³ kg⁻¹ s⁻², M is the mass of the Earth and r is the distance from the satellite.

Masses cancel out and we have:

G*(M/r²)=v²/R, R=r so:

G*(M/r)=v²

r=G*(M/v²), since v=ωr it means v²=ω²r² and we plug it in,

r=G*(M/ω²r²),

r³=G*(M/ω²), ω=2π/T, it means ω²=4π²/T² and we plug that in:

r³=G*(M/(4π²/T²)), and finally we take the third root to get r:

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3 years ago

Suppose a gliding 2-kg cart bumps into, and sticks to, a stationary 5-kg cart. If the speed of the gliding cart before the colli

Thepotemich [5.8K]

Answer:

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

Explanation:

Given:

Mass of Glidding Cart = m₁ = 2 kg

Mass of Stationary Cart = m₂ = 5 kg

Initial velocity of Glidding Cart = u₁ = 14 m/s

Initial velocity of Stationary Cart = u₂ = 0 m/s

To Find:

Final velocity of the coupled carts after the collision = $v_{f}=?$

Solution:

Law of Conservation of Momentum:

For a collision occurring between two objects in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

It is denoted by "p" and given by

Momentum = p = mass × velocity

Hence by law of Conservation of Momentum we hame

Momentum before collision = Momentum after collision

Here after collision both are stuck together so both will have same final velocity,

$m_{1}\times u_{1}+m_{2}\times u_{2}=(m_{1}+m_{2})\times v_{f}$

Substituting the values we get

$2\times 14 + 5\times 0 =(2+5)\times v_{f}$

$v_{f}=\dfrac{28}{7}=4\ m/s$

Therefore,

Final velocity of the coupled carts after the collision is

$v_{f}=4\ m/s$

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3 years ago

True or false conceptual physics 2 questions from chapter 25.

Arlecino [84]

Answer:

Explanation:

1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.

2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.

3) TRUE, W = q.VBA.

4) FALSE; eV is a unit for work (or) energy.

5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.

6) FALSE; in the direction of electric field the potential decreases.

7) FALSE; equipotential surface is perpendicular to the electric field lines.

8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.

9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.

10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.

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3 years ago