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exis [7]
3 years ago
13

How is the mass and speed of a particle related to its kinetic energy?

Physics
2 answers:
jasenka [17]3 years ago
5 0
The higher mass of a particle means it’ll be harder to move, slowing it down and the faster the particle is moving the higher the kinetic energy because there is more movement and pressure within the object with the energy
Furkat [3]3 years ago
4 0
The formula for KE:

1/2 x mass x (speed)^2 = KE

The formula should help you be able to dissect.
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A car traveled 176 miles on 6.4 gallons of fuel the car averaged how many miles per gallon
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It would be 27.5 miles per gallon
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When the direction of an object changes, what happens to its velocity?
Vikentia [17]

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I think that

It changes?

8 0
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How much tension is in a rope if it pulls a 5-kg bucket filled with water with an upward acceleration of 1m/s^2
liraira [26]
F=ma
Tension - weight = mass x acceleration
T - 5(9.81) = 5 x 1
T = 5 + 5(9.81)
T = 54.05 N
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With a 900 V source, voltage is divided across 3 series resistors of 300 V, 280 V, and:
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6 0
3 years ago
A person of mass 55 kg swings on a rope length 4 m from rest (when the rope makes an angle of 30 degrees with the vertical) and
vovangra [49]

Answer:

θ = 19.66°

Explanation:

To determine the angle that the rope makes with the vertical for the two people, you first take into account the potential energy of the first person before he swings on the rope:

U=mgh

h: distance to the ground

g: gravitational acceleration = 9.8m/s^2

m: mass of the first person = 55 kg

In the image attache below you can notice that the height h is:

h=4-4cos(30\°)=0.53m

Then, the potential energy is:

U=(55kg)(9.8m/s^2)(0.53m)=285.67J

When the first person picks up the second person (when the rope is exactly vertical), all the potential energy becomes kinetic energy. Next, when both people reaches the maximum height h' the energy must be equal to the initial potential energy of the first person:

U'=(m_1+m_2)gh'=285.67\ J

From the previous equation you can get h':

h'=\frac{285.67J}{(55kg+70kg)(9.8m/s^2)}=0.2332m

Finally, you obtain the angle between the rope at the height h,' and the vertical, by calculating the following:

h'=4-4cos(\theta)\\\\\theta=cos^{-1}(\frac{4-h'}{4})=cos^{-1}(\frac{4-0.2332}{4})=19.66\°

hence, the angle between the rope and the vertical, when the two people are in the rope is 19.66°

8 0
3 years ago
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