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zalisa [80]
3 years ago
5

10 points!HELP!!!! PLEASE, URGENT (PHYSICS)

Physics
1 answer:
Mariulka [41]3 years ago
4 0
90% of the input energy becomes useful. That's 180 J. The other 10% is wasted. That's 20 J.
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PLEASEEEE HELP :( it would mean a lot
Elis [28]

Answer:

Each object exerts a force on the other, and the two forces are equal and in opposite directions

Explanation:

Newton's 3rd Law Of Motion:

For every action, there is an equal and opposite reaction.

5 0
3 years ago
Read 2 more answers
A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

5 0
3 years ago
The next four questions refer to the situation below.
Anna11 [10]

Answer:

 t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in},      t_{out} = \frac{D}{v_s +v_r}

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

         v_{sg 1} = v_{sr} + v_{rg}

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

           v_{sg1} = D / t_{out}

           D = v_{sg1}  t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

        v_{sg 2} =  v_{sr}  - v_{rg}

         v_{sg 2} = D / t_{in}

         D = v_{sg 2}  t_{in}

with the distance is the same we can equalize

           v_{sg1} t_{out} = v_{sg2} t_{in}

          t_{out} =  t_{in}

           t_{out} = \frac{v_s - v_r}{v_s+v_r} t_{in}

This must be the answer since the return time is known. If you want to delete this time

            t_{in}= D / v_{sg2}

we substitute

            t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

            t_{out} = \frac{D}{v_s +v_r}

7 0
2 years ago
Which sentence is correct?
Alexeev081 [22]

Answer:

pretty sure its B if it isnt im so so sorry

Explanation:

7 0
3 years ago
The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece
Brilliant_brown [7]

Answer:

10.01 cm

Explanation:

Given that,

The time delay between transmission and the arrival of the reflected wave of a signal using ultrasound traveling through a piece of fat tissue was 0.13 ms.

The average propagation speed for sound in body tissue is 1540 m/s.

We need to find the depth when the reflection occur. We know that, the distance is double when transmitting and arriving. So,

v=\dfrac{2d}{t}\\\\d=\dfrac{vt}{2}\\\\d=\dfrac{1540\times 0.13\times 10^{-3}}{2}\\\\d= $$0.1001\ m

or

d = 10.01 cm

So, the reflection will occur at 10.01 cm.

8 0
2 years ago
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