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3 years ago

How much kinetic energy does a 55 kg object have that is traveling 20m/s?

Physics

2 answers:

shutvik [7]3 years ago

4 0

Answer:

11000 J

Explanation:

Apply the formula: K = 1/2.m.v²

K = 1/2.55.20²

K = 1/2.55.400

K = 55.200

K = 11000 Joules

Lunna [17]3 years ago

4 0

Answer:

Explanation:

K.E = 1/2 × mv2

K.E = 0.5 ×55 (20)2

K.E = 0.5 × 55 × 400

K.E = 11000 J.

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Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car w

Len [333]

Answer:

Explanation:

Given Data

Car A speed v=22.0 m/s

Car B speed v=29.0 m/s

Car A distance S=300 m behind Car B

Car A acceleration a=2.40 m/s²

To find

Time required For Car A to take over Car B

Solution

We can represent Car A Coordinate by using equation of simple motion

$X_{A} =vt+1/2at^{2}\\X_{A} =22t+(1/2)(2.40)t^{2}$

And Coordinates of car B equals

$X_{B}=300+29t\\$

Car A is overtake car B when:

$X_{A}=X_{B}\\ 22t+(1/2)(2.4)t^{2}=300+29t\\1.2t^{2}-7t-300=0\\ time=19.0s$

Option (C) 19.0s is correct one

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4 years ago

Earth is slightly closer to the sun in january than in july. How does the area swept out by earth’s orbit around the sun during

Anna007 [38]

This is because both areas are the same.

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3 years ago

A volleyball player’s hand applies a 39 N force while in contact with a volleyball for 2 seconds. What is the impulse on the bal

nikklg [1K]

Given:
F = 39 N, the force applied
t = 2 s, the time interval in which the force is applied.

By definition, the impulse is
$I = \int_{0}^{2} F(t) dt = \int _{0}^{2} 39dt = 39*2=78 \, N-s$

Answer: 78 N-s

3 0

3 years ago

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Which of the following processes could NOT be used to separate dissolved particles from the liquid in a solution? *

DIA [1.3K]

Answer:

Vaporation

Explanation:

In the vaporization or boiling, the passage of particles from the liquid state to the gaseous state occurs completely

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4 years ago

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A bin is given a push across a horizontal surface. The bin has a mass m, the push gives it an initial speed of 1.60 m/s, and the

emmasim [6.3K]

Answer:

The bin moves 0.87 m before it stops.

Explanation:

If we analyze the situation and apply the law of conservation of energy to this case, we get:

Energy Dissipated through Friction = Change in Kinetic Energy of Bin (Loss)

F d = (0.5)(m)(Vi² - Vf²)

where,

F = Frictional Force = μR

but, R = Normal Reaction = Weight of Bin = mg

Therefore, F = μmg

Hence, the equation becomes:

μmg d = (0.5)(m)(Vi² - Vf²)

μg d = (0.5)(Vi² - Vf²)

d = (0.5)(Vi² - Vf²)/μg

where,

Vf = Final Velocity = 0 m/s (Since, bin finally stops)

Vi = Initial Velocity = 1.6 m/s

μ = coefficient of kinetic friction = 0.15

g = 9.8 m/s²

d = distance moved by bin before coming to stop = ?

Therefore,

d = (0.5)[(1.6 m/s)² - (0 m/s)²]/(0.15)(9.8 m/s²)

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3 years ago