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Hoochie [10]
3 years ago
9

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?

Physics
2 answers:
Alex_Xolod [135]3 years ago
4 0
A.900 watts That would be your correct answer 
Bess [88]3 years ago
3 0

Answer : Power = 900 watts

Explanation :

It is given that,

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. The power delivered is given by the following relation :

P=VI

or

P=I^2R

or

P=\dfrac{V^2}{R}

In this case, we use relation (1)

P=12\ V\times 75\ A

P=900\ W

The correct option is (A) " 900 Watts ".

Hence, this is the required solution.

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What’s the answer to this
ladessa [460]
Choices  1,  2,  and 4 . . . . . Yes

Choices  3  and 5 . . . . . No
6 0
3 years ago
Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given poin
qaws [65]

Answer:

30 V

Explanation:

Given that:

The uniform electric field = 50 N/C

Voltage = 80 V

distance = 1.0 m

The potential difference of the electric field = Δ V

E_d = V₁ - V₂

50 × 1 = 80V - V₂

50 - 80 V = - V₂

-30 V = - V₂

V₂ = 30 V

5 0
3 years ago
What is the average speed of a car that travelled 400 miles in 6 hours?
Reil [10]

Answer:

About 66 miles per hour

Explanation:

Based on the information given we can assume the car traveled the same number of miles every hour meaning all we need to do is divide.

400/6 ≈ 66 miles per hour

8 0
3 years ago
Read 2 more answers
A projectile rolls off a cliff with a velocity of 40 m/s. The cliff is 60 meters high.
masya89 [10]

Answer:

1) t = 3.45 s, 2)  x = 138 m, 3) v_{y} = -33.81 m /s, 4) v = 52.37 m / s ,

5) θ = -40.2º

Explanation:

This is a projectile exercise, as they indicate that the projectile rolls down the cliff, it goes with a horizontal speed when leaving the cliff, therefore the speed is v₀ₓ = 40 m / s.

1) Let's calculate the time that Taardaen reaches the bottom, we place the reference system at the bottom of the cliff

      y = y₀ + v_{oy} t - ½ g t²

When leaving the cliff the speed is horizontal  v_{oy}= 0 and at the bottom of the cliff y = 0

      0 = y₀ - ½ g t2

      t = √ 2y₀ / g

      t = √ (2 60 / 9.8)

      t = 3.45 s

2) The horizontal distance traveled

     x = v₀ₓ t

     x = 40 3.45

     x = 138 m

3) The vertical velocity at the point of impact

     v_{y} = I go - g t

     v_{y} = 0 - 9.8 3.45

     v_{y} = -33.81 m /s

the negative sign indicates that the speed is down

4) the resulting velocity at this point

   v = √ (vₓ² + v_{y}²)

   v = √ (40² + 33.8²)

   v = 52.37 m / s

5) angle of impact

    tan θ = v_{y} / vx

    θ = tan⁻¹ v_{y} / vx

    θ = tan⁻¹ (-33.81 / 40)

    θ = -40.2º

6) sin (-40.2) = -0.6455

7) tan (-40.2) = -0.845

8) when the projectile falls down the cliff, the horizontal speed remains constant and the vertical speed increases, therefore the resulting speed has a direction given by the angle that is measured clockwise from the x axis

6 0
3 years ago
Describe the differences between the change in momentum when a car is stopped with the brakes and when a car is stopped in a col
tresset_1 [31]

Explanation:

When a car is breaking, the brakes apply pressure/force onto the wheels(car) which allows it to slow down.

When the car collides with an object, it is exerting a force upon that object to which it applies an equal and opposite force on the car.

I think this is what you are asking for.

Hope This Helps :)

7 0
3 years ago
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