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Alona [7]
3 years ago
9

The battery for a certain cell phone is rated at 3.70 V. According to the manufacturer it can produce math]3.15 \times 10^{4} J[

/math] of electrical energy, enough for 5.25 h of operation, before needing to be recharged. Find the average current that this cell phone draws when turned on.
Physics
1 answer:
AysviL [449]3 years ago
5 0

Answer:

The average current that this cell phone draws when turned on is 0.451 A.

Explanation:

Given;

voltage of the phone, V = 3.7 V

electrical energy of the phone battery, E = 3.15 x 10⁴ J

duration of battery energy, t = 5.25 h

The power the cell phone draws when turned on, is the rate of energy consumption, and this is calculated as follows;

P = \frac{E}{t}

where;

P is power in watts

E is energy in Joules

t is time in seconds

P = \frac{3.15*10^4}{5.25*3600s} = 1.667 \ W

The average current that this cell phone draws when turned on:

P = IV

I = \frac{P}{V} =\frac{1.667}{3.7} = 0.451 \ A

Therefore, the average current that this cell phone draws when turned on is 0.451 A.

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A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
3 years ago
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The acceleration due to gravity vector is always in the ____ direction.
finlep [7]
It goes in the downward direction
4 0
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A sailboat sits tied up at the dock.
shutvik [7]

Answer:B

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5 0
3 years ago
A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have
denpristay [2]
1) The total mechanical energy of the rock is:
E=U+K
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
E_i=U=mgh
where m=4 kg is the mass, g=9.81 m/s^2 is the gravitational acceleration and h=5 m is the height.
Putting the numbers in, we find the potential energy
U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
E_f=K= \frac{1}{2}mv^2
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
K=U=196.2 J

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
W=196.2 J-0 J=196.2 J
6 0
3 years ago
Two sticky spheres are suspended from light ropes of length LL that are attached to the ceiling at a common point. Sphere AA has
a_sh-v [17]

Answer:

  h’ = 1/9 h

Explanation:

This exercise must be solved in parts:

* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy

starting point. Higher

         Em₀ = U = m g h

final point. Lower, just before the crash

         Em_f = K = ½ m v_{b}^2

energy is conserved

         Em₀ = Em_f

         m g h = ½ m v²

         v_b = \sqrt{2gh}

* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

         p₀ = 2m 0 + m v_b

final instant. Right after the crash

         p_f = (2m + m) v

       

the moment is preserved

         p₀ = p_f

         m v_b = 3m v

         v = v_b / 3

         

          v = ⅓ \sqrt{2gh}

* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together

Starting point. Lower

          Em₀ = K = ½ 3m v²

Final point. Higher

          Em_f = U = (3m) g h'

          Em₀ = Em_f

          ½ 3m v² = 3m g h’

           

we substitute

         h’=  \frac{v^2}{2g}

         h’ =  \frac{1}{3^2} \  \frac{ 2gh}{2g}

         h’ = 1/9 h

6 0
3 years ago
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