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3 years ago

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EX 6-1 A ball is twirled on a 0.870 - m-long string with a constant speed of 3.36 m / s . Calculate the acceleration of the ball

. Be sure to specify the direction of the acceleration.

1 answer:

Elina [12.6K]3 years ago

4 0

Answer:

$a=12.97\ m/s^2$

Explanation:

Given that,

The length of a string, l = 0.87 m

Speed of the ball, v = 3.36 m/s

We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2$

So, the acceleration of the ball is $12.97\ m/s^2$ .

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Determine the angle of an incline that would yield a constant velocity, given the coefficient of kinetic friction is 0.10.

azamat

Answer:

$\theta=5.71^{o}$

Explanation:

In order to solve this problem, we mus start by drawing a free body diagram of the given situation (See attached picture).

From the free body diagram we can now do a sum of forces in the x and y direction. Let's start with the y-direction:

$\sum F_{y}=0$

$-W_{y}+N=0$

$N=W_{y}$

so:

$N=mgcos(\theta)$

now we can go ahead and do a sum of forces in the x-direction:

$\sum F_{x}=0$

the sum of forces in x is 0 because it's moving at a constant speed.

$-f+W_{x}=0$

$-\mu_{k}N+mg sen(\theta)=0$

$-\mu_{k}mg cos(\theta)+mg sen(\theta)=0$

so now we solve for theta. We can start by factoring mg so we get:

$mg(-\mu_{k} cos(\theta)+sen(\theta))=0$

we can divide both sides into mg so we get:

$-\mu_{k} cos(\theta)+sen(\theta)=0$

this tells us that the problem is independent of the mass of the object.

$\mu_{k} cos(\theta)=sen(\theta)$

we now divide both sides of the equation into $cos(\theta)$ so we get:

$\mu_{k}=\frac{sen(\theta)}{cos(\theta)}$

$\mu_{k}=tan(\theta)$

so we now take the inverse function of tan to get:

$\theta=tan^{-1}(\mu_{k})$

so now we can find our angle:

$\theta=tan^{-1}(0.10)$

so

$\theta=5.71^{o}$

8 0

3 years ago

To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a

irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

= 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

W = F x S joules

Substituting the values in the above equation

W = 1.5 x 10⁶ x 355 · sin θ

= 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J

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3 years ago

if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object

Paul [167]

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3 years ago

Find the heat energy is required to change 2Kg of ice at 0 C to water at 20 C ( specific latent heat of fusion of water = 336000

katrin2010 [14]

We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.

We will find that we must give 840,000 Joules.

First, we must change of phase from ice to water.

We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.

So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:

2*336,000 J = 672,000 J

Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.

Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.

We know that:

specific heat of capacity of water = 4200 J/kg°C

This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.

Then to increase 1°C of 2kg of water we need twice that amount:

2*4,200 J = 8,400 J

And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:

20*8,400 J = 168,000 J

Then the total amount of energy that we must give is:

E = 672,000 J + 168,000 J = 840,000 J

If you want to learn more, you can read:

brainly.com/question/12474790

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3 years ago

A diffraction grating produces a first-order bright fringe that is 0.18 m away from the central bright fringe on a flat screen.

Georgia [21]

Answer:

The wavelength of the light is 562.5 nm

Solution:

As per the question:

Order, n = 1

Slit separation, d = $2.5\times 10^{- 6}\ m$

Distance from the bright fringe, y = 0.18 m

Distance between the screen and the grating, D = 0.8 m

Now,

We know from the eqn for diffraction:

$n\lambda = dsin\theta$

n = 1

$\lambda = dsin\theta$ (1)

Also,

For very small angle, $\theta$ :

$sin\theta$ ≈ $tan\theta = \frac{y}{D} = \frac{0.18}{0.8} = 0.225$

Using the above value in eqn (1):

$\lambda = 2.5\times 10^{- 6}\times 0.225 = 5.625\times 10^{- 7}\ m = 562.5\ nm$

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3 years ago