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3 years ago

What type of energy is stored in a battery?

Physics

2 answers:

dimulka [17.4K]3 years ago

8 0

Answer:

Batteries use chemistry, in the form of chemical potential, to store energy, just like many other everyday energy sources. For example, logs store energy in their chemical bonds until burning converts the energy to heat.

Explanation:

Naddika [18.5K]3 years ago

4 0

Electrical energy I’m leaning about this in class right now and I’m really good at it I have a 100 but if I get it wrong I’m sorry but I think it’s electrical energy it has to be

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2 years ago

Braddy connected the loose wire to the battery and created an electromagnet. He picked up 45 thumb tacks with his electromagnet,

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Explanation:

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4 0

3 years ago

Read 2 more answers

A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl

Taya2010 [7]

Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

Given that,

Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

$F-f=ma$

Put the value into the formula

$340-f=60\times2.0$

$f=340-60\times2.0$

$f=220\ N$

We need to calculate the coefficient of friction

Using formula of friction force

$f= \mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{220}{60\times9.8}$

$\mu =0.37$

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

$F = ma$

$a=\dfrac{F}{m}$

$a=\dfrac{220}{100}$

$a=2.2\ m/s^2$

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0

3 years ago

A textbook of mass 2.05 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d

blsea [12.9K]

Answer:

a. 7.38 N b. 40.87 N c. 0.113 kg-m²

Explanation:

a. Let T be the tension in the cord. For the textbook, T = ma since no other force acts on it and it is an horizontal force, and m = mass = 2.05 kg and a = acceleration. We find the acceleration from s = ut + 1/2at² where u = initial speed = 0 (since it starts from rest), s = distance moved = 1.30 m and t = time = 0.850 s.

Substituting these values into s,

1.30 m = 0 × 0.850 + 1/2a × 0.850² = 0 + 0.36125a

1.30 = 0.36125a

a = 1.30/0.36125 = 3.6 m/s²

Substituting this into T, we have

T = ma = 2.05 kg × 3.6 m/s² = 7.38 N

b. Let T be the tension in the cord attached to the book. The book has the only vertical forces acting on it as the tension, T(acting upwards) and its weight mg (acting downwards). So the net force acting on it is

T - mg = ma

T = m(a + g)

substituting a = 3.6 m/s² and g = 9.8 m/s² and m = 3.05 kg

T = 3.05(3.6 + 9.8) = 3.05 × 13.4 = 40.87 N

c. Since the tangential acceleration of the pulley is also the acceleration of the masses, the a = rα where r = radius of pulley = 0.200 m/2 = 0.100 m and α = angular acceleration of the pulley.

α = a/r = 3.6 m/s² ÷ 0.100 m = 36 rad/s²

Now, the torque on the pulley τ = Tr = Iα where I = moment of inertia of pulley about its rotational axis and T = tension in cord attached to book and r = radius of pulley = 0.200 m/2 = 0.100 m

From the equation above, I = Tr/α

Substituting the variables we have

I = 40.87 N × 0.100 m ÷ 36 rad/s² = 0.113 kg-m²

4 0

3 years ago

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