HELP WILL MARK BRAINLIEST

Suppose a 65.5 kg gymnast climbs a rope. What is the tension in the rope if she climbs at a constant speed

natali 33 [55]

The tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Given:

Mass of gymnast, m = 65.5 kg

The speed 'v' of gymnast is constant

Solution:

Consider the free-body diagram of the system as shown below.

Balancing forces along the vertical axis we get:

ΣFy = 0

Thus, we get:

F = ma - (1)

where, m is mass of gymnast

a is acceleration of gymnast (a = 0m/s², as the speed is constant)

Also,

F = T - mg -(2)

where, T is tension in the rope

g is acceleration due to gravity

Equating (1) & (2), we get:

ma = T - mg

Re-arranging the equation, we get:

T = m(a+g)

Applying values in above equation we get:

T = (65.5 kg)(0 m/s²+9.8 m/s²)

T = 641.9 N

Therefore, the tension in the rope when the gymnast climbs it at constant speed is 641.9 N.

Learn more about tension here:

<u>brainly.com/question/14294135</u>

#SPJ4

7 0

2 years ago

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

5 0

3 years ago

Identify which method of thermal energy transfer would be fastest through a vacuum, which would be fastest through a gas, and wh

Blizzard [7]

Answer:

vacuum-radiation

gas-convection

solid-conduction

Explanation:

6 0

3 years ago

Most often frequency is measured by the number of waves that pass a point in one blank

AlladinOne [14]

I’m pretty sure it’s hertz
but if it’s wrong i apologize

8 0

3 years ago

Which is true of electricity generated both from coal and from nuclear reactions?

Fynjy0 [20]

Neither source is a renewable one.

6 0

3 years ago