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zhuklara [117]
3 years ago
9

HELP ME

Physics
1 answer:
Mazyrski [523]3 years ago
3 0

Answer:

I THINK IT'S <em>D.</em><em>.</em><em>.</em><em>.</em>

<em>HOPE </em><em>SO</em>

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A ball that is attached to a string travels in a horizontal, circular path, as shown in Figure 1. At time t0 , the ball has a sp
Levart [38]

Answer:

F₁ = 4 F₀

Explanation:

The force applied on the string by the ball attached to it, while in circular motion will be equal to the centripetal force. Therefore, at time t₀, the force on ball F₀ is given as:

F₀ = mv₀²/r   --------------- equation (1)

where,

F₀ = Force on string at t₀

m = mass of ball

v₀ = speed of ball at t₀

r = radius of circular path

Now, at time t₁:

v₁ = 2v₀

F₁ = mv₁²/r

F₁ = m(2v₀)²/r

F₁ = 4 mv₀²/r

using equation (1):

<u>F₁ = 4 F₀</u>

5 0
3 years ago
Which of the following changes would increase the focal length of a lens? a) decrease rodius of curvature of surfoce c) increase
anastassius [24]

Answer:

Increase the radius of curvature of surface.

Explanation:

Radius of curvature is the measurement that is two times the focal length, for a given lens. It lies on either side of the lens.

Focal length is the distance which is half of the radius of curvature. Radius of curvature is a measure of the radius of the circle . Focal length is the distance between the center of curvature of the lens and the point where all the rays are brought to a focus for a distant object.

8 0
3 years ago
1. Draw magnetic field (B field) line diagrams for the following. Make sure the diagrams are large enough so the direction and s
larisa [96]

I attached the requested diagram.

<em>In the case of the magnetic field in a bar</em> by convention, the direction of the field is taken out of the north pole and towards the south pole of the magnet. These types of images are commonly made of some ferrous material.

<em>In the case of the horseshoe </em>magnet, the highly concentrated magnetic field is distinguished between its legs. In the figure it is shown in a contribution from North to South, again by agreement, however outside the two poles, the magnetic field falls rapidly. A horseshoe magnet is basically a bent bar magnet.

4 0
3 years ago
two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
3 years ago
Consider an ideal gas at 27.0 degrees Celsius and 1.00 atmosphere pressure. Imagine the molecules to be uniformly spaced, with e
My name is Ann [436]

To solve the exercise it is necessary to keep in mind the concepts about the ideal gas equation and the volume in the cube.

However, for this case the Boyle equation will not be used, but the one that corresponds to the Boltzmann equation for ideal gas, in this way it is understood that

PV =NkT

Where,

N = Number of molecules

k = Boltzmann constant

V = Volume

T = Temperature

P = Pressure

Our values are given as,

N = 1

k = 1.38*10^{-23}J/K

T = 27\°C = 27\°C + 273 = 300K

P = 1atm = 101325Pa

Rearrange the equation to find V we have,

V = \frac{NkT}{P}

V = \frac{1(1.38*10^{-23})(300K)}{101325Pa}

V = 4.0858*10^{-26}m^3

We know that length of a cube is given by

V = L^3

Therefore the Length would be given as,

L = V^{1/3}

L = (4.0858*10^{-26})^{1/3}

L = 3.445*10^{-9}m

Therefore each length of the cube is 3.44nm

7 0
3 years ago
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