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2 years ago

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Sue and Sean are having a tug-of-war by pulling on opposite ends of a 5.0-kg rope. Sue pulls with a 15-N force. What is Sean's f

orce if the rope accelerates toward Sue at 2.0 m/s2?

1 answer:

ikadub [295]2 years ago

4 0

Answer:

5N

Explanation:

Sean's force is 5N because from the equation of Force, calculating for Sue's acceleration F=MA, A=F/M, A= 15N/5kg = 3m/s²

If Sue's force was 15N and he yielded 3m/s², If sean's accelerate towards Sue at 2m/s², his net acceleration of pull is 1m/s², therefore his force F=Ma = 5kg X 1m/s² + 5N

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Un tubo de acero de 40000 kilómetros forma un anillo que se ajusta bien a la circunferencia de la tierra. Imagine que las person

Darina [25.2K]

Answer:

82.76m

Explanation:

In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.

You use the formula for the circumference of the steel ring:

$C=2\pi r$ (1)

C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)

you solve for r in the equation (1):

$r=\frac{C}{2\pi}=\frac{4*10^7m}{2\pi}=6,366,197.724m$

Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:

$L=Lo[1+\alpha \Delta T]$ (2)

L: final length of the tube = ?

Lo: initial length of the tube = 4*10^7m

ΔT = change in the temperature of the steel tube = 1°C

α: thermal coefficient expansion of steel = 13*10^-6 /°C

You replace the values of the parameters in the equation (2):

$L=(4*10^7m)(1+(13*10^{-6}/ \°C)(1\°C))=40,000,520m$

With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:

$r'=\frac{C}{2\pi}=\frac{40,000,520m}{2\pi}=6,366,280.484m$

Finally, you compare both r and r' radius:

r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m

Hence, the distance to the ring from the ground is 82.76m

4 0

2 years ago

Angular and Linear Quantities: A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an a

serious [3.7K]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

$\omega = 1.25 rad/s \rightarrow$ The angular speed

$\alpha = 0.745 rad/s2 \rightarrow$ The angular acceleration

$r = 4.65 m \rightarrow$ The distance

The relation between the linear velocity and angular velocity is

$v = r\omega$

Where,

r = Radius

$\omega =$ Angular velocity

At the same time we have that the centripetal acceleration is

$a_c = \frac{v^2}{r}$

$a_c = \frac{(r\omega)^2}{r}$

$a_c = \frac{r^2\omega^2}{r}$

$a_c = r \omega^2$

$a_c = (4.65 )(1.25 rad/s)^2$

$a_c = 7.265625 m/s^2$

Now the tangential acceleration is given as,

$a_t = \alpha r$

Here,

$\alpha =$ Angular acceleration

r = Radius

$\alpha = (0.745)(4.65)$

$\alpha = 3.46425 m/s^2$

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

$|a| = \sqrt{a_c^2+a_t^2}$

$|a| = \sqrt{(7.265625)^2+(3.46425)^2}$

$|a| = 8.049 m/s^2 \approx 8.05 m/s2$

Therefore the correct answer is C.

7 0

2 years ago

If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the induc

mel-nik [20]

Answer:

Inner radius = 2 mm

Explanation:

In a coaxial cable, series inductance per unit length is given by the formula;

L' = (µ/(2π))•ln(R/r)

Where R is outer radius and r is inner radius.

We are given;

L' = 50 nH/m = 50 × 10^(-9) H/m

R = 2.6mm = 2.6 × 10^(-3) m

Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m

Plugging in the relevant values, we have;

50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)

Rearranging, we have;

(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)

0.25 = ln((2.6 × 10^(-3))/r)

So,

e^(0.25) = (2.6 × 10^(-3))/r)

1.284 = (2.6 × 10^(-3))/r)

Cross multiply to give;

r = (2.6 × 10^(-3))/1.284)

r = 0.002 m or 2 mm

5 0

2 years ago

Arrange the objects in order from least to greatest potential energy. Assume that gravity is constant. (Use PE = m × g × h.)

yawa3891 [41]

Answer:The objects in order from least to greatest potential energy:

3<1<4<2

Explanation:

1. Potential energy of a 15-kilogram stone found at a height of 3 meters:

$mgh=15 kg\times 9.8 m/s^2\times 3 m=441 J$

2. Potential energy 10 kilograms of water stored at a height of 9 meters:

$mgh=10 kg\times 9.8 m/s^2\times 9 m=882 J$

3. Potential energy 1-kilogram ball located 20 meters in the air:

$mgh=1 kg\times 9.8 m/s^2\times 20 m=196 J$

4. Potential energy box of books weighing 25 kilograms placed on a shelf 2 meters high:

$mgh=25 kg\times 9.8 m/s^2\times 2 m=490 J$

The objects in order from least to greatest potential energy:

3<1<4<2

8 0

2 years ago

Read 2 more answers

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. how much work is required to stop it

jeyben [28]

One way of thinking of energy is to say the energy is the ability of the body to do work. If you have a body that is moving that body has kinetic energy, as said above this energy can be used to do work. The amount of work done is equal to the amount of energy that body has.
In our case cylinder has the following kinetic energy:
$E_k=m\frac{v^2}{2}\\
E_k=40\frac{6^2}{2}\\
E_k=40\cdot18\\
E_k=720$J$
And that is exaclty how much work you would have to put in to stop the cylinder.

3 0

2 years ago