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kati45 [8]
1 year ago
7

In this lab you will use a cart and a track to explore Newton's second law of motion. You will vary two different variables and

determine how each one affects how quickly the cart moves along the track. In the space below, write a scientific question that you can answer by doing this experiment.

Physics
2 answers:
Veseljchak [2.6K]1 year ago
7 0

Answer:

Sample Response: How do force and mass affect the acceleration of an object?

Explanation:

Savatey [412]1 year ago
3 0

<u>Answer:</u>

<em>Newtons II law: </em>

<em>     </em>It  is defined as<em> "the net force acting on the object is a product of mass and acceleration of the body"</em> . Also it defines that the <em>"acceleration of an object is dependent on net force and mass of the body".</em>

Let us assume that,a string is attached to the cart, which passes over a pulley along the track. At another end of the string a weight is attached which hangs over the pulley. The hanging weight provides tension in the spring, and it helps in accelerating the cart. We assume that the string is massless and no friction between pulley and the string.

Whenever the hanging weight moves downwards, the cart will accelerate to right side.

<em>For the hanging weight/mass</em>

When hanging weight of mass is m₁ and accelerate due to gravitational force g.

           Therefore we can write F = m₁ .g

and the tension acts in upward direction T (negetive)

        Now, Fnet = m₁ .g - T

                          = m₁.a

So From Newtons II law<em> F =  m.a</em>

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A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the
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Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

\frac{dA}{dt} = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B \frac{dA}{dt}

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 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
1 year ago
Sketch a position-time graph for a bear starting
Dmitrij [34]

Explanation:

hopefully that makes sense. the position doesn't change over the 5 seconds, meaning it's stopped but time still continues. then when the slope is negative this shows the bear's position becoming negative (backing up, changing direction).

3 0
2 years ago
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the e
Lapatulllka [165]

Answer:

0.2cm towards the retina.

Explanation:

the focal length of the frog eye is

(1/f) = (1/10) + (1/0.8)

f = 0.74cm

Since the distance of the object is 15cm Hence

(1/0.74) = (1/15) + (1/V)

V = 0.78cm

Therefore the distance the retina is to move is

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3 0
1 year ago
A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's int
Fittoniya [83]

Answer:

The answer is A. There is no electric field on the interior of the conducting sphere.

Explanation:

A solid conducting sphere in a uniform electric field will exert force on the charges in the sphere to redistribute themselves in such a way that both the charges and the field inside the sphere would vanish.

3 0
2 years ago
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