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Nuetrik [128]
2 years ago
14

Alculate the potential difference if 20J of energy are transferred by 8C of charge.

Physics
1 answer:
sveta [45]2 years ago
6 0

Answer:

V = 2.5 J/C

Explanation:

<u><em>Given:</em></u>

Energy = E = 20 J

Charge = Q = 8 C

<u><em>Required:</em></u>

Potential Difference = V = ?

<u><em>Formula:</em></u>

V = \frac{E}{Q}

<u><em>Solution:</em></u>

V = 20/8

V = 2.5 J/C

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A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is
tiny-mole [99]

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

8 0
2 years ago
Read 2 more answers
13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
2 years ago
You are piloting a small airplane in which you want to reach a destination that is 750 km due north of your starting location. O
alexira [117]

Answer:

v_wind = 101.46 km / h   ,  θ = 61.8

Explanation:

This is a velocity composition exercise.

Let's do the problem in parts. Let's start by knowing the speed of the plane without air.

           v = d / t

           v = 750 / 3.14

           v = 238.85 km / h

This is the speed of the plane relative to the Earth and it does not change.

In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and v_{y}

Let's use trigonometry to find the components of the plane's speed

          cos θ = v_N / v

          sin θ  = v_W / v

          v_N = v cos θ

          v_W = v sin θ

           

let's calculate

          V _N = 238.85 cos 22 = 221.46 km / h

           v_W = -238.85 sin 22 = -89.47

the negative sign is because the plane is going west and the positive sign is the east direction.

As it indicates that the destination of the avine is towards the north, the x component of the wind must be

              vₓ - v_W = 0

              vₓ = v-w

              vₓ = 89.47 km / h

in the direction to the East.

Now let's analyze the component of the wind in the Nort-South direction,

Indicate the travel time, let's calculate the speed that the component must have the speed of the plane

             v_total = d / t

             v_total = 750 / 4.32

             v_total = 173.61 km / h

This is the final speed of the plane, which can be written

              v_total = v_n - vy

               vy = v_n - v_total

               vy = 221.46 - 173.61

               vy = 47.85 km

this component is directed towards the south

Let's use the Pythagorean Theorem, to find the magnitude

             v_wind² = vₓ² + vy²

             v_wind = √ (89.47² + 47.85²)

             v_wind = 101.46 km / h

the address will then be found using trigonometry

             θ = Vy / vx

             θ = tan⁻¹ (vy / vx)

             θ = tan⁻¹1 (47.85 / 89.47)

             θ = 28.14

Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value

                  90- θtea = 90- 28.2

                  θ = 61.8

East of South

5 0
2 years ago
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
2 years ago
How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue t
hodyreva [135]

Answer:

The heat transferred  from water to skin  is 6913.5 J.

Explanation:

Given that,

Weight of water = 25.0 g

Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.

We need to calculate the heat transferred  from water to skin

Using formula for stream

Q=mc\Delta T

Put the value into the formula

Q=25\times10^{-3}\times4190\times(373-307)

Q=6913.5\ J

Hence, The heat transferred  from water to skin  is 6913.5 J.

3 0
2 years ago
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