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Nostrana [21]
3 years ago
14

Why does a ball accelerate as it rolls down a hill?

Physics
2 answers:
Anika [276]3 years ago
6 0
The ball accelerates because of gravity.
Allushta [10]3 years ago
5 0
Because there is a 'net force' on it.  That means that all the different
forces on it do NOT all cancel out.  Whenever there is a net force on
an object ... not zero force ... the object accelerates.  (That means that
either its speed or its direction changes.)  We've known this for about
300 years now.
The main force on the ball is the force of gravity.
You might be interested in
An fm radio station broadcasts at 98. 6 mhz. What is the wavelength of the radiowaves?.
victus00 [196]

The wavelength of the radio waves is 3.04 cm.

<h3>Calculation:</h3>

λf = c

λ = c/f

where,

λ = wavelength

c = speed of light

f = frequency

Given,

f = 98.6 MHz = 98.6 × 10⁶

c = 3 × 10⁸

To find,

λ =?

Put the values in the formula,

λ = c/f

λ = 3 × 10⁸/98.6 × 10⁶

  = 0.0304 × 10²  m

  = 3.04 cm

Therefore, the wavelength of the radio waves is 3.04 cm.

Learn more about the calculation of wavelength here:

brainly.com/question/8422432

#SPJ4

3 0
2 years ago
The position of a ball is given by x(t) (2.3 m/s) .t + (5.3m/s3) t3. What is the position 348 m at time t 4 s? You are correct Y
Georgia [21]

Answer:

(a) 348.4 m

(b) 256.7 m/s

(c)  127.2 m/s^2

Explanation:

x = 2.3 t + 5.3 t^{3}

(a) at t = 4 s

x = 2.3 x 4 + 5.3 x 4 x 4 x 4

x = 348.4 m

(b) The derivative of displacement function gives the value of instantaneous velocity.

So, v = dx / dt = 2.3 + 5.3 x 3 x t^2

v = 2.3 + 15.9 t^2

Put t = 4 s

So, v = 2.3 + 15.9 x 4 x 4

v = 256.7 m/s

(c) The derivative of velocity function with respect to time gives the value of instantaneous acceleration.

So, a = dv / dt = 5.3 x 3 x 2 x t

a = 31.8 t

Put t = 4 s

a = 31.8 x 4 = 127.2 m/s^2

4 0
4 years ago
Calculate the voltage difference in a circuit with a resistance of 25 Ω if the current in the circuit is 0.5 A.
weeeeeb [17]

Answer:

check the attachments for the solution

8 0
2 years ago
The force of gravity on the moon is only 16.6% the force of gravity on Earth. For each item listed in the table below, fill in w
Dmitry [639]

==> Take an Earth weight. Don't forget the unit.

==> Multiply it by 0.166 .

==> The product is its Moon weight. Don't forget the unit.

Example:

If I weigh 1,000 lbs on Earth.

Multiply 1,000 lbs by 0.166 .

I weigh 166 lbs on the Moon

4 0
3 years ago
A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with t
sergij07 [2.7K]

Answer: 135.29\º

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

<u>*Note the normal is perpendicular to the plane, with a 90\º angle with the surface</u>

And this can be visualized in the figure, where the angle {\theta}_{1} with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which  the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.  

According to Snell’s Law:  

n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})     (1)  

Where:  

n_{1}=1 is the first medium refractive index  (the air)

n_{2}=1.309 is the second medium refractive index  (the ice)

\theta_{1}=25.5\º is the angle of the incident ray  

\theta_{2} is the angle of the refracted ray  

Now, firstly we have to find \theta_{2} and then, by geometry, find \alpha and \beta which sum the <u>angle between the reflected and the refracted light</u>.

Finding \theta_{2} from (1):

(1)sin(25.5\º)=(1.309)sin(\theta_{2})    

sin(\theta_{2})=0.328

\theta_{2}=arcsin(0.328)

\theta_{2}=19.201\º   (2)

Remembering that the Normal makes  a 90\º angle with the surface, we can say:

90\º=\theta_{1}+\beta    (3)

Finding \beta:

\beta=90\º-25.5\º=64.5\º    (4)

Doing the same with  \theta_{2} and  \alpha}:

90\º=\theta_{2}+\alpha    (5)

Finding \alpha:

\alpha=90\º-19.201\º=70.79\º    (6)

Adding both angles (4) and (6):

\alpha+\beta=70.79\º+64.5\º    (7)

\alpha+\beta=135.29\º>>>This is the angle between the reflected and the refracted light.

5 0
3 years ago
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