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ad-work [718]
3 years ago
11

A battery can provide a current of 1.80 A at 2.60 V for 6.00 hr. How much energy (in kJ) is produced?A battery can provide a cur

rent of 1.80 A at 2.60 V for 6.00 hr. How much energy (in kJ) is produced?
Physics
1 answer:
love history [14]3 years ago
4 0

Answer:

The energy which is produced by a battery is 101.1 kJ.

Explanation:

The expression for the energy in terms of voltage, current and time is as follows;

E=VIt

Here, V is the voltage, I is the current and t is the time.

It is given in the problem that a battery can provide a current of 1.80 A at 2.60 V for 6.00 hr.

Calculate the energy of the battery.

E=VIt

Convert time from hour int seconds.

t=6 hr

t=(6)(60)(60)

t=21600 s

Put I= 1.80 A, V= 2.60 V and t= 21600 s in the expression of energy.

E=(2.60)(1.80)(21600)

E= 101.1 kJ

Therefore, the energy which is produced by a battery is 101.1 kJ.

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An asteroid is on a collision course with Earth. An astronaut lands on the rock to bury explosive charges that will blow the ast
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Answer:

The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters

Explanation:

Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;

v² = u² - 2·g·h

Where;

v = The final velocity

u = The initial velocity

g = The acceleration due to gravity ≈ 9.8 m/s²

h = The height she jumps

At the maximum height, h_{max} = 0.500 m, she jumps, v = 0, therefore, we have;

0² = u² - 2·g·h_{max}

u² = 2 × 9.8 × 0.5 = 9.8

u = √9.8 ≈ 3.13

u = 3.13 m/s

Her initial jumping velocity ≈ 3.13 m/s

Escape velocity, v_e = \sqrt{\dfrac{2 \cdot G \cdot M}{r} }

Where;

M = The mass of the asteroid

G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

r = The radius of the asteroid

The average density of the Earth = 5515 kg/m³

The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³

The escape velocity, she has, v_e ≈ 3.13 m/s is therefore;

3.13 = \sqrt{\dfrac{2 \times 6.67408 \times 10^{-11} \times 5515 \times \frac{4}{3} \times \pi \times r^3}{r} } = r \times \sqrt{3.084 \times 10^{-6}}

r = \dfrac{3.13}{ \sqrt{3.084 \times 10^{-6}}} \approx 1782.45

Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.

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