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katrin2010 [14]
3 years ago
15

if you push a box a distance of 2000 meters with a force of 1 newton, how many calories have you used

Physics
1 answer:
kap26 [50]3 years ago
4 0
Note that
1 J = 0.239 cal

By definition,
Work = Force x Distance

Therefore work done is
W = (1 N)*(2000 m) = 2000 J

In calories,
W = (2000 J)*(0.239 cal/J) = 478 cal

Answer: 478 calories

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3 years ago
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A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum
Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , m_1= 0.473 kg

mass of puck 2 , m_2= 0.819 kg

initial speed of puck 1 , u_1=2.76\frac{m}{s}

initial speed of puck 2 , u_2=0.00\frac{m}{s}

Final speed of puck 1 and puck 2 be v_1\, and\, v_2  respectively

Apply conservation of linear momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

=>0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2

=>1.594=0.5775\times v_1+ v_2 -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

u_2-u_1=v_1-v_2

=>0-2.76=v_1-v_2 ------(B)

From equation (A) and (B)

v_1=-0.739\frac{m}{s}

and v_2=2.02\frac{m}{s}

Thus the final speed of puck 1 is 0.739 m/s towards west  and puck 2 is 2.02 m/s towards east .

       

3 0
3 years ago
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may vary depending on the organization.

4 0
3 years ago
A charge q1= 3nC and a charge q2 = 4nC are located 2m apart. Where on the line passing through these charges is the total electr
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Answer:

Explanation:

Electric field due to a charge Q at a point d distance away is given by the expression

E = k Q / d , k is a constant equal to 9 x 10⁹

Field due to charge = 3 X 10⁻⁹ C

E = E = \frac{9\times 10^9\times3\times10^{-9}}{d^2}

Field due to charge = 4 X 10⁻⁹ C

E = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}

These two fields will be equal and opposite to make net field zero

\frac{9\times 10^9\times3\times10^{-9}}{d^2} = [tex]\frac{9\times 10^9\times4\times10^{-9}}{(2-d)^2}[/tex]

\frac{3}{d^2} =\frac{4}{(2-d)^2}

\frac{2-d}{d} =\frac{2}{1.732}

d = 0.928

5 0
3 years ago
A solid of density 8000 kgm.. weighs 0.8 kgf in air. When it is completely
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Explanation:

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