The switch on some accumulaters is usually used to

n a presentation about measuring mass, one of your classmates states, "Two objects of the same size will always have the same ma

Neporo4naja [7]

Answer:
The statement is not correct.

Reason:
If two objects have the same mass, their respective densities determine their volumes.
By definition,
density = mass/volume.
or volume = mass/density.

Therefore if two objects have the same mass, the mode dense object (with higher density) will occupy less volume than the other object.

4 0

3 years ago

Read 2 more answers

It is 5.0 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 km/h

Ostrovityanka [42]

Answer:

Walking burns up more energy,1740000J

Explanation:

Given that the displacement is 5.0km, and running at 10km/h and uses, walking at 3km/hr and uses 290watts:

Energy consumption for running is calculated as:

$700watts=700j/s,d=5000m,v=\frac{25}{9}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{25}{9}}\times 700j/s\\\\=1260000J$

Energy consumption for walking is calculated as:

$290watts=290j/s,d=5000m,v=\frac{5}{6}m/s\\\\\therefore E_c=\frac{d}{v}\times Energy \ usage\\\\=\frac{5000}{\frac{5}{6}}\times 290j/s\\\\=1740000J$

Walking is a slower process hence the need for more energy over longer periods raltive to running the same distance.

Hence walking burns more energy; 1,740,000J. It burns more because you walk for a greater period of time.

5 0

3 years ago

50. A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 335 kV. The secondar

Sedaia [141]

Answer:

Explanation:

a ) The transformer steps up the voltage from 12000 V to 335000 V . Voltage in primary is 12000 V and in the secondary it is 335000 V in old transformer

If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils

the formula is

n₂ / n₁ = voltage in secondry / voltage in primary

n₂ / n₁ = 335000 / 12000

ratio of turns in old transformer

= 27.9

ratio of turns in new transformer

n₃ / n₁ = 750 / 12 ( n₃ is no of turns in the secondary of new transformer )

= 62.5

T he ratio of turns in the new secondary compared with the old secondary

n₃ / n₂ = 62.5 / 27.9

= 2.24

b ) Current in secondary / current in primary

= turns in primary / turns in secondary

current output ratio of old

= Current in secondary / current in primary

= n₁ / n₂

= 12 / 335

= .03582

current output ratio of new

= Current in secondary / current in primary

= n₁ / n₃

= 12 / 750

= .016

The ratio of new current output to old output (at 335 kV) for the same power

= .016 / .03582

= .4466

c ) power loss in new

= (current in secondary )² x resistance of secondary

=( .016 x current in primary )² x R

= 2.56 X 10⁻⁴ X ( current in primary )² x R

power loss in old

= (current in secondary )² x resistance of secondary

=( .03582 x current in primary )² x R

= 12.83 X 10⁻⁴ X ( current in primary )² x R

ratio of new line power loss to old

= 2.56 / 12.83

= .199

7 0

3 years ago

You have been hiking and come back to the camp with really cold hands.you put your hands over the wonderful fire your father has

bekas [8.4K]

The answer would be radiation because heat waves are being transferred

5 0

3 years ago

The temperature of a plastic cube is monitored while the cube is pushed 3.4 m across a floor at constant speed by a horizontal f

Nesterboy [21]

Answer:

$\Delta E_{floor} = 51 J$

Explanation:

The work (W) done on the cube to be pushed across the floor is equal to the total thermal energy (ΔE) of the system:

$W = \Delta E_{T} = \Delta E_{cube} + \Delta E_{floor}$ (1)

Also, the work done on the cube by the horizontal force is giving by:

$W = F \cdot d$ (2)

<em>where F: force applied to the cube , d: displacement of the cube </em>

<em>By equaling the equations (1) and (2)</em>, we can find the thermal energy of the floor:

$\Delta E_{cube} + \Delta E_{floor} = F \cdot d$

$\Delta E_{floor} = F \cdot d - \Delta E_{cube}$

$\Delta E_{floor} = 20 N \cdot 3.4 m - 17 J$

$\Delta E_{floor} = 51 J$

So, the increase in the thermal energy of the floor is 51 J.

Have a nice day!

7 0

3 years ago