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3 years ago

In which situation is no work being done?

2 answers:

8 0

I would say that work is being done in all the situations because technically in physics work is done whenever force is applied through a distance.

neonofarm [45]3 years ago

4 0

A. because the box is being carried at a constant velocity, therefor there is no net force necessary to keep it it motion. It changes the direction in motion, but it does no work on the box

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An astronaut, while revolving in a

sergey [27]

Answer:

Why the hell would he throw a spoon out like if I were in space i would little letter in . bottle and see when someone finds it (link my ph. number obviously)

Explanation:

so yea lol

8 0

3 years ago

Solution A has a speciﬁc heat of 2.0 J/g◦C. Solution B has a speciﬁc heat of 3.8 J/g◦C. If equal masses of both solutions start

fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

We have a formula for such condition,

$Q=m.c.\Delta T$ .....................................(1)

where:

$\Delta T$ = temperature difference

Q= heat energy

m= mass of the body

c= specific heat of the body

Proving mathematically:

According to the given conditions

we have equal masses of two solutions A & B, i.e. $m_A=m_B$

equal heat is supplied to both the solutions, i.e. $Q_A=Q_B$

specific heat of solution A, $c_{A}=2.0 J.g^{-1} .\degree C^{-1}$

specific heat of solution B, $c_{B}=3.8 J.g^{-1} .\degree C^{-1}$

$\Delta T_A$ & $\Delta T_B$ are the change in temperatures of the respective solutions.

Now, putting the above values

$Q_A=Q_B$

$m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1$

Which proves that solution A attains a higher temperature than solution B.

7 0

3 years ago

Is the same side of the moon always dark? Explain please.

Reil [10]

Yeah we always see the same side of the moon because it orbits around earth with that side constantly facing earth, if that's not your question than just comment and i'll try to answer...

3 0

3 years ago

Read 2 more answers

If a vector 2ỉ +3j +8k is perpendicular to the vector 4j-4i + ak, then the value of a is

natita [175]

Explanation:

if two vectors are perpendicular to each other then thier dot product is equal to zero so (2i+3j+8k).(-4i+4j+ak)=0 (2)(-4) +(3)(4)+(8)(a) =0 -8+12+8a=0 4+8a=0 8a=-4

$a = \frac{ - 4}{8}$

$a = \frac{ - 1}{2}$

4 0

3 years ago

Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) resp

Sophie [7]

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: $x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ , where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

x-values:

$x_{cm}$ = $\frac{m_{1}x_{1} + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12}$ = $\frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41}$ = $\frac{81}{41}$ = 1.9756

y-values:

$y_{cm}$ = $\frac{m_{1}y_{1} + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}$ = $\frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12}$ = $\frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41}$ = $\frac{-90}{41}$ = -2.1951

center of mass:

(1.9756, -2.1951)

6 0

3 years ago