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Mekhanik [1.2K]
2 years ago
10

3. How much work is done when you pull a 6 N wagon for 5 meters?

Physics
1 answer:
8090 [49]2 years ago
3 0

Answer:

<h2>30 J</h2>

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question

force = 6 N

distance = 5 m

We have

workdone = 6 × 5 = 30

We have the final answer as

<h3>30 J</h3>

Hope this helps you

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Most reactive non-metals in order?
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Carbon
</span> Nitrogen
Oxygen
Fluorine
Phosphorus <span>               
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Chlorine <span>               
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5 0
3 years ago
In physics, why would an equation like y=mx+b be plotted as a straight line rather than as a parabola, as is done in math?
notka56 [123]

Answer:

Explanation:

In Both Physics and Math

y=mx+b is plotted as straight line where

m=slope of line

b=intercept on Y-axis

whereas Equation of parabola is something like this

y^2=4ax

or

x^2=4ay

Math is a tool to solve Physics problems so equations are same in math and physics

3 0
3 years ago
A sample of a diatomic ideal gas has pressure P and volume V. When the gas is warmed, its pressure triples and its volume double
Andreyy89

Answer:

Amount of  Energy transferred =8.5PV

Explanation:

Given:

Initial volume=V

Initial pressure=P

Final volume=2V

Final pressure=3P

Now w know that the Energy transferred in constant pressure pressure is given by

E_1=nc_pdT\\\\E_1=n\times\dfrac{7R}{2}dT\\E_1=3.5(nRdT)\\E_1=3.5(V(2P-P))\\E_1=3.5PV

Now the Energy transferred in constant volume process is given by

E_2=nc_vdT\\\\E_2=n\times\dfrac{5R}{2}dT\\E_1=2.5(nRdT)\\E_1=2.5(V(3V-V))\\E_1=5PV

The total Energy transferred is given by

E_{total}=E_1+E_2\\E{total}=3.5PV+5PV\\=8.5PV

3 0
3 years ago
Read 2 more answers
Plz help on this ASAP !!
makkiz [27]

It would most likely be answer B considering the other ones don’t make sense in this question

4 0
3 years ago
Objects with masses of 120 kg and a 420 kg are separated by 0.380 m. (a) Find the net gravitational force exerted by these objec
SOVA2 [1]

Answer:

F_{net} = 6.879\times 10^{- 7}\ N

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

Now,

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

F = \frac{GMm}{\frac({d}{2}^{2})}

F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

F = \frac{GMm}{\frac({d}{2}^{2})}

F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

F_{net} = F + F'

F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N

6 0
3 years ago
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