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2 years ago

A particle travels in a circle of radius 14 m at a constant speed of 21 m/s. What is the magnitude of the acceleration (in m/s2)

Physics

1 answer:

Salsk061 [2.6K]2 years ago

4 0

The magnitude of the acceleration is equal to 31.5 $m/s^2$

Given the following data:

Radius = 14 meters.

Speed = 21 m/s

To determine the magnitude of the acceleration:

In Science, the acceleration of a particle in a circle or circular track is known as centripetal acceleration.

<h3>How to calculate centripetal acceleration.</h3>

Mathematically, centripetal acceleration is given by this formula:

$a = \frac{V^2}{r}$

Substituting the given parameters into the formula, we have;

$a = \frac{21^2}{14}\\\\a=\frac{441}{14}$

Centripetal acceleration = 31.5 $m/s^2$

Read more on centripetal acceleration here: brainly.com/question/25780931

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A person is running at 2 m/s and 20 s later is running at 22 m/s. What is

AysviL [449]

I think it’s 1 m/s^2

8 0

3 years ago

A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

3 years ago

What is the frequency of radiation emitted by a photon of light if the energy released during its transition to ground state is

Tpy6a [65]

Energy (E) = Planck's constant (h) * frequency (nu)
$3.611*10^{-15}J = 6.626 * 10^{-34} * frequency$
therefore, frequency ~ $5.45*10^{18}$

8 0

3 years ago

Read 2 more answers

(b) A ball is thrown from a point 1.50 m above the ground. The initial velocity is 19.5 m/s at

Nataly_w [17]

The answers are:

(i) 6.35m

(ii) 20.2 m/s

It seems like you already have the answer, but let me show you how to get it:

You have two givens:

Vi = 19.5m/s

Θ = 30°

dy = 1.50m (This is not the maximum height, just to be clear)

When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components.

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = 9.75 m/s

Vix = CosΘVi = (Cos30°)(19.5m/s) = 16.89 m/s

Okay, so we have that down now. The next step is to decide which kinematics equation you will use. Because you have no time, you need to use the kinematic equation that is not time dependent.

(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

Where:

Vf = final velocity

Vi = inital velocity

a = 9.8m/s²

d = displacement

We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

$dy = \dfrac{Vfy^{2}-Viy^{2}}{2a}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}$

$dy = 4.85m$

So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT ABOVE THE GROUND.

4.85m + 1.50m = 6.35m

(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$

$Vfy^{2} = 0m/s^{2}+2(9.8m/s^{2}(6.35m)$

$\sqrt{Vfy^{2}} = \sqrt{124.46m^{2}/s^{2}}$

$Vfy= 11.16m/s$

OKAY! We are at our last step. Now to get the resultant velocity, we apply the Pythagorean theorem.

$Vf^{2} = Vfx^{2} + Vfy^{2}$

$\sqrt{Vf^{2}} = \sqrt{(16.89m/s)^{2}+(11.16m/s)^{2}}$

$Vf =20.2m/s$

The ball was falling at 20.2m/s before it hit the ground.

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3 years ago

Which of the following statements about jurisprudence is TRUE?

valentina_108 [34]

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3 years ago