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2 years ago

If a cart of 10 kg mass has a force of 5 newtons exerted on it, what is its acceleration? m/s2

Physics

2 answers:

kherson [118]2 years ago

6 0

Answer:

.5 m/s2

Explanation:

Elza [17]2 years ago

4 0

Answer:

= 0.5 m/s²

Explanation:

According to Newton's second law of motion, the resultant force is directly proportion to the rate of change of linear momentum.

Therefore; F = ma , where F is the Force, m is the mass and a is the acceleration.

Thus; a = F/m

but; F = 5 N, and m = 10 kg

 a = 5 /10

= 0.5 m/s²

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2 years ago

You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

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-- The vertical component of the ball's velocity is 14 sin(51°) = 10.88 m/s

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-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
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As usual when we're discussing this stuff, we completely ignore air resistance.

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2 years ago

Read 2 more answers

3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided

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Answer:

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2 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$