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3 years ago

6

An electric wheel chair is designed to run on a single 12-V battery rated to provide 100 ampere-hours. How much energy is stored

in this battery?

1 answer:

Paha777 [63]3 years ago

5 0

Explanation:

It is given that,

Voltage of the battery, V = 12 V

Current, I = 100 ampere-hours

Energy stored is given by the product of power and time taken. So,

$E=P\times t$

P is the power, $P=V\times I$

$P=12\times 100$

P = 1200 watts

This power can be used for 1 hour or 3600 seconds

Energy, $E=1200\times 3600$

E = 4320000 J

So, the energy stored in this battery is 4320000 J. Hence, this is the required solution.

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One differnce between magnetic poles and elcyrical charges is that

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Answer:

In the electric field, the like charges repel each other, and the unlike charges attract each other, whereas in a magnetic field the like poles repel each other and the unlike poles attract each other.

Explanation:

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Adaptive radiation is demonstrated by ____

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X-rays are high energy electrons that can cause damage when exposed under extreme conditions. The best technology that can block it is using a lightweight type of metal foam. It can take in high energy collisions which also exhibits high forces. it does not only block x-rays but also, neutron radiation and gamma rays.

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3 years ago

the speed of travel of the moon around the earth, using the formula for the speed of a moving object in a circular path

Svetach [21]

Answer: 1018.26 m/s

Explanation:

Approaching the orbit of the Moon around the Earth to a circular orbit (or circular path), we can use the equation of the speed of an object with uniform circular motion:

$V=\sqrt{G\frac{M}{r}}$

Where:

$V$ is the speed of travel of the Moon around the Earth

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M=5.972(10)^{24} kg$ is the mass of the Earth

$r=384400(10)^{3} m$ is the distance from the center of the Earth to the center of the Moon

Solving:

$V=\sqrt{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{5.972(10)^{24} kg}{384400(10)^{3} m}}$

$V=1018.26 m/s$ This is the speed of travel of the Moon around the Earth

5 0

3 years ago

One of the frequencies used to transmit and receive cellular telephone signals in the United States is 985 MHz. What is the wave

tia_tia [17]

Answer:

approximately 304358

Explanation:

wavelength = speed / frequency

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2 years ago

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$

where

m is the mass

a is the acceleration

In this problem, we have

$\sum F = 205.9 N$ is the net force

m = 150 kg is the mass

Solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{205.9}{150}=1.37 m/s^2$

C)

Since the motion of the motorbike is a uniformly accelerated motion, we can use the following suvat equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

For this motorbike, we have:

u = 0 (it starts from rest)

$a=1.37 m/s^2$

s = 350 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(1.37)(9.8)}=5.2 m/s$

4)

For this part of the problem, we can use the following suvat equation:

$v=u+at$

where

v is the final velocity

u is the initial velocity

a is the acceleration

t is the elapsed time

Here we have:

v = 5.2 m/s

u = 0

$a=1.37 m/s^2$

Solving for t, we find

$t=\frac{v-u}{a}=\frac{5.2-0}{1.37}=3.80 s$

Learn more about Newton's laws of motion and accelerated motion:

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6 0

3 years ago