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3 years ago

14

The energy wasted in using a machine is 600j. if the machine is 70% efficient. calculate the volume of water pumb by the machine

through a height of 15m.

1 answer:

WINSTONCH [101]3 years ago

3 0

Answer:

yhgigy6ftu5cg l8vbbnnbbgtccccvhklhaywje nc 62bbnzmakbdbvfvdbf93bdldmffmfkqhdv

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A weight of 6 kg increases the speed of its force from 2 m / s to 4 m / s. By how many joules does the kinetic energy of the bod

Leviafan [203]

Answer:

36 J

Explanation:

Let's start off with the Kinetic Energy formula: 1/2m $v^{2}$

So to find the <em>change</em> in Kinetic Energy, you would first have to find the Kinetic Energy in the beginning and in the end.

*Note: Mass: you incorrectly wrote weight = 6 kg- because weight is a force you have to write 6 Newtons. If you are talking about the <em>mass</em>, you would write 6 kg. If you are talking about weight, you would write 6 Newtons. The difference will make significant changes to the answer, so I will give you the answer for both.

If 6 kg is the MASS:

Beginning: find the kinetic energy.

Plug in all parts of the formula: 1/2mv^2 = $\frac{1}{2}$ (6)(2 squared)= $\frac{1}{2}$ (6)(4)= 12 J

End: find the kinetic energy.

Plug in all parts of the formula: 1/2mv^2= $\frac{1}{2}$ (6)(4 squared)= $\frac{1}{2}$ (6)(16)= 48 J

Answer:

36 J

If 6 kg is the WEIGHT:

We know that the weight formula is mg, or mass times acceleration due to gravity (which is always 9.8 m/s^2). Plug in the numbers:

Weight = mass x acceleration due to gravity

6 = mass x 9.8

6 = 9.8m

*Divide both sides*

mass = 0.6 kg

Now, we can use the mass to find the kinetic energy.

Beginning: find the Kinetic Energy

Plug in all parts of the formula: 1/2mv^2 = $\frac{1}{2}$ (0.6)(2 squared)= $\frac{1}{2}$ (0.6)(4)= 1.2 J

End: find the Kinetic Energy

Plug in all parts of the formula: 1/2mv^2= $\frac{1}{2}$ (0.6)(4 squared)= $\frac{1}{2}$ (0.6)(16)= 4.8 J

Answer:

3.6 J

(this answer is not very feasible, so 36 J is the way to go. But just remember, don't mix up weight and mass again- as you can see, they lead to different answers!)

4 0

3 years ago

What is the formular use in solving range of a projectile.

Afina-wow [57]

The range =vi².sin2Ф/g

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3 years ago

a battery of 9v is connected in series with resistors of 0.2ohm , 0.3ohm, 0.5ohm, and 5 ohm respectively. how much current would

Nonamiya [84]

Answer:

1.5 A

Explanation:

Applying

V = IR'....................... Equation 1

make I the subject of the equation

I = V/R'.................. Equation 2

Where V = Voltage, I = current, R' = Total resistance.

From the question,

In a series connection,

R' = 0.2+0.3+0.5+5 = 6 ohm.

Given: V = 9V

Substitute into equation 2

I = 9/6

I = 1.5 A.

Note: Since all the resistors are connected in series, thesame current flows through them

Therefore the current flowing through the 5 ohm resistor = 1.5 A

5 0

3 years ago

Can someone help meeeeee... show how to solve it plzzzzzzzz

liubo4ka [24]

<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant.

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

In this case we have a gravitation force $F_{1}=16units$ , given by the formula written at the beginning. Let’s rename the distance $r$ as $d$ :

$F_{1}=G\frac{m_{1}m_{2}}{d^2}$ (1)

And we are asked to find the gravitation force $F_{2}$ with a given distance of $\frac{d}{2}$ :

$F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}$

$F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}$ (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate $G m_{1}m_{2}$ in both equations:

From (1):

$Gm_{1}m_{2}=F_{1}{d}^{2}$ (3)

From (2):

$Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}$ (4)

If (3)=(4):

$F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}$ (5)

Now we have to find $F_{2}$ :

$F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}$

$F_{2}=4F_{1}$ (6)

If $F_{1}=16 units$ :

$F_{2}=(4)(16 units)$

$F_{2}=64 units$ >>>>This is the new force of attraction

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3 years ago

How much net force is required to accelerate a 0.5 kg toy car, initially at rest to a velocity of 2.4 m/s in 6s?

Jet001 [13]

Force = 0.20N .F = m ×a .& a = v/t then the f = m×v/t

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4 years ago