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maria [59]
3 years ago
10

two objects were lifted by machine one object had a massive 2 kg it was lifted at a speed of 2 m/s the other had a massive 4 kg

and was lifted at a rate of three meters per second which object has more kinetic energy
Physics
1 answer:
rosijanka [135]3 years ago
8 0
Kinetic energy =(1/2) (mass) (speed²)

First object:  KE = (1/2) (2 kg) (2m/s)² =  4 joules during the lift.

Second object:  KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.

The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy. 
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Of all the planets in our solar system, Jupiter has the greatest gravitational strength. If a 1.5 kg pair of running shoes would
Andre45 [30]

Answer:

gₓ = 23.1 m/s²

Explanation:

The weight of an object is on the surface of earth is given by the following formula:

W = mg

where,

W = Weight of the object on surface of earth

m = mass of object

g = acceleration due to gravity on the surface of earth = strength of gravity on the surface of earth

Similarly, the weight of the object on Jupiter will be given as:

W_{x} = mg_{x}

where,

Wₓ = Weight of the object on surface of Jupiter = 34.665 N

m = mass of object = 1.5 kg

gₓ = acceleration due to gravity on the surface of Jupiter = strength of gravity on the surface of Jupiter = ?

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34.65 N = (1.5 kg)g_{x}

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<u>gₓ = 23.1 m/s²</u>

7 0
3 years ago
Based on the law of conservation of energy, how can we reasonably improve a machine’s ability to do work? Move the machine to a
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7 0
3 years ago
Read 2 more answers
2. Suppose that a parallel-plate capacitor has circular plates with radius R = 30 mm and a plate separation of d = 5.0 mm. Suppo
seropon [69]

Answer:

The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

Explanation:

Given that,

Radius of plate = 30 mm

Separation = 5.0 mm

Frequency = 60 Hz

Suppose the maximum potential difference is 100 V and r= 130 mm.

We need to calculate the angular frequency

Using formula of angular frequency

\omega=2\pi f

Put the value into the formula

\omega=2\times\pi\times60

\omega=376.9\ rad/s

When r>R, the magnetic field is inversely proportional to the r.

We need to calculate the maximum value of the induced magnetic field that occurs at r = R

Using formula of magnetic filed

B_{max}=\dfrac{\mu_{0}\epsilon_{0}R^2\timesV_{max}\times\omega}{2rd}

Where, R = radius of plate

d = plate separation

V = voltage

Put the value into the formula

B_{max}=\dfrac{4\pi\times10^{-7}\times8.85\times10^{-12}\times(30\times10^{-3})^2\times100\times376.9}{2\times130\times10^{-3}\times5.0\times10^{-3}}

B_{max}=2.901\times10^{-13}\ T

Hence, The maximum value of the induced magnetic field is 2.901\times10^{-13}\ T.

7 0
3 years ago
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