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2 years ago

A 10,000J battery is depleted in 2h. What power consumption is this? *

Physics

1 answer:

Tomtit [17]2 years ago

8 0

Answer:

P = 1.4 W

Explanation:

Given that,

The work done or the energy of the battery, E = 10,000 J

Time, t = 2 h

We need to find the power consumption. Let it is P. Power is the rate of doing work. So,

$P=\dfrac{W}{t}\\\\P=\dfrac{10,000}{2\times 3600}\\\\P=1.38\ W$

P = 1.4 W

So, the power of the battery is 1.4 W.

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Thanks + BRAINLIST <br><br> Please need correct answer asappp

nirvana33 [79]

Answer:

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2 years ago

A boy is pulling a cart by a force of 100N. The frictional force experienced by the cart is 20N. The force causing the motion of

Serga [27]

Answer:

3 is right i guss look : 100N-20N=80N

4 0

2 years ago

A mass of 4kg suspended by a light string 2m long and at rest is projected horizontally with a velocity of 1.5 m/s. find the ang

Dafna11 [192]

Answer:

19.5°

Explanation:

The energy of the mass must be conserved. The energy is given by:

1) $E=\frac{1}{2}mv^2+mgh$

where m is the mass, v is the velocity and h is the hight of the mass.

Let the height at the lowest point of the be h=0, the energy of the mass will be:

2) $E=\frac{1}{2}mv^2$

The energy when the mass comes to a stop will be:

3) $E=mgh$

Setting equations 2 and 3 equal and solving for height h will give:

4) $h=\frac{v^2}{2g}$

The angle ∅ of the string with the vertical with the mass at the highest point will be given by:

5) $cos\phi=\frac{l-h}{l}$

where l is the lenght of the string.

Combining equations 4 and 5 and solving for ∅:

6) $\phi={cos}^{-1}(\frac{l-h}{l})={cos}^{-1}(1-\frac{h}{l})={cos}^{-1}(1-\frac{v^2}{2gl})$

8 0

3 years ago

It is found experimentally that the electric field in a certain region of Earth's atmosphere is directed vertically down. At an

Maru [420]

Answer:

$q=7.965*10^-^6C$

Explanation:

From the question we are told that

Altitude of $d_1 390m$

Magnitude $M_1=60.0 N/C$

Altitude of $d_2=240 m$

Magnitude is $M_2= 100 N/C$

Distance of cube $d_c=150 m$

Generally the flux $\phi$ is mathematical given as

$\phi=60(150)^2cos180+100(150)^2*cos0$

$\phi=-9*10^5$

Generally Quantity of charge q is mathematically given as

$q=\varepsilon _0 *\phi$

$q=8.85*10^-^1^2 *9*10^5$

$q=7.965*10^-^6C$

5 0

2 years ago

2 76 Consider a household that uses 14,000 kWh of elec- (tricity per year and 3400 L of fuel oil during a heating sea- son. The

NNADVOKAT [17]

Answer:

The total amount of CO₂ produced will be = 20680 kg/year

The reduction in the amount of CO₂ emissions by that household per year = 3102 kg/year

Explanation:

Given:

Power used by household = 14000 kWh

Fuel oil used = 3400 L

CO₂ produced of fuel oil = 3.2 kg/L

CO₂ produced of electricity = 0.70 kg/kWh

Now, the total amount of CO₂ produced will be = (14000 kWh × 0.70 kg/kWh) + (3400 L × 3.2 kg/L)

⇒ The total amount of CO₂ produced will be = 9800 + 10880 = 20680 kg/year

Now,

if the usage of electricity and fuel oil is reduced by 15%, the reduction in the amount of the CO₂ emission will be = 0.15 × 20680 kg/year = 3102 kg/year

6 0

3 years ago