Ask question

Ask question

All categories

3 years ago

5

How much force would a 55kg person experience on a roller coaster if they travel through a loop with a radius of 55kg at a speed

of 14.1 m/s?

1 answer:

Inessa [10]3 years ago

5 0

Answer:

200 N

Explanation:

For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.

The radially inward acceleration, or the centripetal acceleration is given by :

a = v² / r

where v is the speed at which the body is moving and r is the radius of the circle

Given-

m = 55kg

v = 14.1 m/s

r= 55m

We know that F = ma

⇒ F = m ( v²/ r )

⇒ F = 55 x 14.1 x 14.1 / 55

⇒ F =14.1 x 14.1 = 200 N

∴ <em>The force acting is 200 N</em>.

You might be interested in

How could you increase the force advantage of a lever?Select one:a. Make the effort length longer.b. Make the effort length shor

11Alexandr11 [23.1K]

The ideal mechanical advantage of a lever (IMA) is given by:

$IMA=\frac{Le}{Lr}$

Where:

Le = Effort of the arm

Lr = Resistance arm.

Therefore, we can increase the force adventage by increasing the effort arm or reducing the load arm

Answer:

a. Make the effort length longer.

4 0

1 year ago

A wave with a frequency of 60 Hz is traveling along a string whose linear mass density is 230 g/m and whose tension is 65 N. If

matrenka [14]

To develop this problem we will use the concepts related to Speed in a string that is governed by Tension (T) and linear density (µ)

$V = \sqrt{\frac{T}{\mu}}$

Our values are given as:

$f = 60Hz\\\mu = 230 g/m = 0.230kg/m\\T = 65N\\P = 75w$

Replacing we have that the velocity is

$V = \sqrt{\frac{T}{\mu}}$

$V = \sqrt{\frac{65}{0.230}}$

$V = 16.81m/s$

From the theory of wave propagation the average power wave is given as

$P =\frac{1}{2} \mu \omega^2 A^2 V$

Where,

A = Amplitude

$\omega = 2\pi f \rightarrow$ Angular velocity

$A^2 = \frac{2P}{\mu \omega^2 V}$

$A^2 = \frac{2P}{\mu (2\pi f)^2 V}$

Replacing,

$A^2 = \sqrt{\frac{2(75)}{(0.230)(2\pi 60)^2(16.81)}}$

$A = 0.0165m$

Therefore the amplitude of the wave should be 0.0165m

8 0

3 years ago

A defibrillator passes 14.0 A of current through the torso of a person for 0.0300 s. How much charge moves in coulombs?

krek1111 [17]

<h2>Answer:</h2>

4.2 C

<h2>Explanation:</h2>

The charge (Q) moving is the product of the current(I) flowing through the torso of the person and the time taken (t) for the flow.

i.e

Q = I x t

Where;

I = current = 14.0A

t = time taken = 0.0300s

Substituting the values of I and t into the equation above gives

Q = 14.0 x 0.0300

Q = 4.2 C

Therefore quantity of charge moving is 4.2 C

4 0

3 years ago

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

$T = \frac{2 \pi}{w}$

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

7 0

3 years ago

A leopard with a mass of 55.00 kg climbs 12.0 m up a tree. What is it’s gain in GPE

PSYCHO15rus [73]

The answer is: 6,468 j

4 0

3 years ago