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cricket20 [7]
2 years ago
8

A cell phone charger operates at 5 volts and can deliver a maximum current of 1.0 Amp. If your cell phone has an energy storage

capacity of 38.9~38.9 kiloJoules, what is the shortest possible time (in hours) it would take to fully charge your empty phone battery using this charger.
Physics
1 answer:
madam [21]2 years ago
7 0

Answer:

<em>it</em><em> </em><em>takes</em><em> </em><em>2</em><em>.</em><em>1</em><em>6</em><em> </em><em>hrs</em><em> </em><em>to</em><em> </em><em>fully</em><em> </em><em>charge</em><em> </em><em>the</em><em> </em><em>phone</em>

Explanation:

V = 5v

I = 1A

E = 38.9 Kj = 38900 j

t = ?

E = IVt

t = E/IV

t = 38900 ÷ (1 × 5)

t = 38900 ÷ 5

t = 7780 seconds

t = 7780/3600............( converting time in seconds to hours)

t = 2.16 hrs

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A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between
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Answer:

90x20=1800

Explanation:

just multiply 10 & 9 and then mutiply 90x20 or 20x90

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2 years ago
A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin
tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b)  Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

       sin 40 = voy / v₀

       v_{oy} = v₀ sin 40

       v_{oy} = 50.0 sin40

       v_{oy} = 32.14 m / s

       cos 40 = v₀ₓ / V₀

       v₀ₓ = v₀ cos 40

       v₀ₓ = 50.0 cos 40

       v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and v_{y} = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

     p₀ₓ = m v₀ₓ

Y Axis y

    p_{oy} = 0

After the break

X axis

   p_{fx} = m₂ v₂

Axis y

     p_{fy} = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis  

     0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

     v₃ = - m₁ v₁ / m₃

     v₃ = - (-10) 1 / 0.3

     v₃ = 33.3 m / s

X axis

     M v₀ₓ = m₂ v₂

     v₂ = v₀ₓ M / m₂

     v₂ = 38.3  2 / 0.7

     v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

                                         0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

       v_{fy}² = v_{oy}² - 2 gy

       0 =  v_{oy}²-2gy

       y =  v_{oy}² / 2g

       y = 32.14² / 2 9.8

       y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

      v_{f}² =  v_{y}² - 2 g y₂

      0 =  v_{y}² - 2 g y₂

     y₂ =  v_{y}² / 2g

     y₂ = 33.3²/2 9.8

     y₂ = 56.58 m

Total body height is

      Y = y + y₂

      Y = 52.7 + 56.58

     Y = 109.3 m

8 0
2 years ago
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Answer:

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If an area has all the wolves that it can support the wolf population has reached is what
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The wolf population in that area has reached its carrying capacity.
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After a model rocket reached its maximum height, it then took 5.0 seconds to return to the launch site. what is the approximate
kifflom [539]
Air resistance is ignored.
g = 9.8 m/s².
At maximum height, the vertical velocity is zero.

Let h =  the maximum height reached.
Let u =  the vertical launch velocity.

Because ot takes 5.0 seconds to reach maximum height, therefore
(u m/s) - (9.8 m/s²)*(5 s) = 0 
u = 49 m/s

The maximum height reached is
h = (49 m/s)*(5 s) - (1/2)*(9.8 m/s²)*(5 s)²
   = 122.5 m

Answer: 122.5 m
3 0
3 years ago
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