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3 years ago

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

Physics

1 answer:

Tomtit [17]3 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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A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw

meriva

Answer:

As 28m/s = 28m/s

Explanation:

r = the radius of the curve

m = the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

$a = \frac{v^{2}}{r}$

therefore

$\mu mg = m \frac{v^{2}}{r} \\\\ \mu = \frac{v^{2}}{rg}$

$v = \sqrt{\mu rg}$

$\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s$

As 28m/s = 28m/s

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3 years ago

b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 2.50 m/s

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Answer:

The horizontal distance is 0.64 m.

Explanation:

Initial velocity, u =2.5m/s

The maximum horizontal distance is

$R = \frac{u^2}{2g}\\\\R = \frac{2.5\times 2.5}{9.8}\\\\R = 0.64 m$

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3 years ago

Which statement about speed and/or velocity is true?

maw [93]

Answer: The right Answer is Velocity has both speed and direction.

Explanation:

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2 years ago

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What is the velocity of a 2000 kg truck with a momentum of 48,000 kg•m/s?

Goshia [24]

Answer:

Explanation:

momentum= mass* velocity

velocity= momentum/ mass

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2 years ago

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 1.6×104 m/s when at a distance

xz_007 [3.2K]

Answer:

v₂ = 7.6 x 10⁴ m/s

Explanation:

given,

speed of comet(v₁) = 1.6 x 10⁴ m/s

distance (d₁)= 2.7 x 10¹¹ m

to find the speed when he is at distance of(d₂) 4.8 × 10¹⁰ m

v₂ = ?

speed of planet can be determine using conservation of energy

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\dfrac{1}{2}mv_1^2-\dfrac{GMm}{r_1} = \dfrac{1}{2}mv_2^2-\dfrac{GMm}{r_2}$

$\dfrac{1}{2}v_1^2-\dfrac{GM}{r_1} = \dfrac{1}{2}v_2^2-\dfrac{GM}{r_2}$

$v_2^2= v_1^2 + \dfrac{2GM}{r_2}-\dfrac{2GM}{r_1}$

$v_2= \sqrt{v_1^2 +2GM(\dfrac{1}{r_2}-\dfrac{1}{r_1})}$

$v_2= \sqrt{(1.6\times 10^4)^2 +2\times 6.67 \times 10^{-11}\times 1.99 \times 10^{30}(\dfrac{1}{4.8\times 10^{10}}-\dfrac{1}{2.7\times 10^{11}})}$

v₂ = 7.6 x 10⁴ m/s

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2 years ago