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lys-0071 [83]
3 years ago
8

Calculate the linear momentum (in kg · m/s) of a pickup truck that has a mass of 1035 kg and is travelling eastward at 31.0 m/s.

Physics
1 answer:
Musya8 [376]3 years ago
8 0

Answer:

32085 kg.m/s is the correct answer..

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The more twists per foot in a pair of wires, the more resistant the pair will be to ____.
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The more twist per foot in a pair of wires, the more resistant the pair will be to cross talk. A cross talk in network planning and design is a disturbance produced by electromagnetic interference beside a circuit or a cable pair. A telecommunication signal interrupts a signal in an adjacent circuit and can source the signals to turn out to be confused and cross over each other.  
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3 years ago
HELP I NEED THE ANSWER NOW
Setler79 [48]

Answer:

Speed of light

Explanation:

The famous Einstein's equation is:

E=mc^2

where

E is the energy

m is the mass

c=3\cdot 10^8 m/s is the speed of light

In this equation, Einstein summarized the following fact: mass can be converted into energy, and the amount of energy released in such a process is given by the equation.

An example of application of this equation is the nuclear fusion process. In a nuclear fusion, two lighter nuclei combine into a heavier nucleus. However, the mass of the heavier nucleus is slightly less than the sum of the masses of the two original nuclei: some of the mass of the original nuclei has been converted into energy, accorging to the previous equation.

4 0
3 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
How much work does gravity do on the compressor during this displacement?
gayaneshka [121]

Explanation and Examples

let the mass of the compressor be

mass (m):

height in x axis is (h1)

height in y axis be (h2):

Height difference: h2-h1

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Put the values that you require and get the answer.

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2 years ago
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