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2 years ago

8

Outside the space shuttle, you and a friend pull on two ropes to dock a satellite whose mass is 700 kg. The satellite is initial

ly at position <3.3, -1.6, 2.0> m and has a speed of 6 m/s. You exert a force <-400, 500, 250> N. When the satellite reaches the position <2.3, 2.6, -11.8> m, its speed is 6.18 m/s. How much work did your friend do?

1 answer:

MaRussiya [10]2 years ago

3 0

Answer:

your friend did a work of 1717.34 J

Explanation:

The sum of the work made by your friend and you is equal the change in the kinetic energy, so:

$W_a+W_b=\frac{1}{2}mv_f^{2}-\frac{1}{2}mv_i^{2}$

Where $W_a$ is the work that you did, $W_b$ is the work that your friend did, m is the mass of the satellite and $v_f$ and $v_i$ are the final and initial speeds

Additionally the work that you did is equal to:

$W_a=F.s$

Where F and s are vectors. So, the vector s that said the displacement is calculated as:

s = <2.3, 2.6, -11.8> - <3.3, -1.6, 2.0>

s = <-1 , 4.2 , -13.8>

Therefore, $W_a$ is:

$W_a=.$

$W_a=(-400*-1)+(500*4.2)+(250*-13.8)\\W_a=400+2100-3450\\W_a=-950$

Then, replacing the values on initial equation and solving for $W_b$ , we get:

$-950+W_b=\frac{1}{2}(700)(6.18)^{2}-\frac{1}{2}(700)(6)^{2}\\-950+W_b=767.24\\W_b=767.24+950\\W_b=1717.34J$

Finally, your friend did a work of 1717.34 J

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An 80 g, 40 cm long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15 g ball of clay trav

fiasKO [112]

Answer:

θ = 12.95º

Explanation:

For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height

Let's start by finding the speed of the bar plus clay ball system, using amount of momentum

The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)

Initial before the crash

p₀ = m v₀

Final after the crash before starting the movement

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v

v = v₀ m / (m + M)

v = 2.0 0.015 / (0.015 +0.080)

v = 0.316 m / s

With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy

Lower

Em₀ = K = ½ (m + M) v²

Higher

$E_{mf}$ = U = (m + M) g y

Em₀ = $E_{mf}$

½ (m + M) v² = (m + M) g y

y = ½ v² / g

y = ½ 0.316² / 9.8

y = 0.00509 m

Let's look for the angle the height from the pivot point is

L = 0.40 / 2 = 0.20 cm

The distance that went up is

y = L - L cos θ

cos θ = (L-y) / L

θ = cos⁻¹ (L-y) / L

θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)

θ = 12.95º

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2 years ago

The winning time for a 500.0 mile circular race track was 3 hours and

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Explanation:

We have,

Distance traveled in a circular track is 500 miles

The winning time was 3 hours and 13 minutes. It means time is 3.217 hours.

The driver's average speed is given by total distance divided by total time taken. Its formula can be written as :

$v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph$

At the end of the race, the driver reaches the point form where he has started. It means the displacement of the driver is equal to 0. Hence, driver's average velocity is equal to 0.

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3 years ago

Which of the following is an example or an orginasm mantaining homeostasis

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a cat grooming itself

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2 years ago

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

$\theta=274rev$

Explanation:

From the question we are told that:

Angular velocity $\omega=510rpm$

Mass $m=40.kg$

Diameter d $75=>0.75m$

Off Time $t=40.0s$

Oscillation at Power off $N=210$

Generally the equation for Angular displacement is mathematically given by

$\theta_{\infty}=\frac{w+w_0}{t}t$

$w=\frac{2*\theta_{\infty}}{t}-w_0$

$w=\frac{28210}{40*(\frac{1}{60})}-510$

$w=120rpm$

Generally the equation for Time to come to rest is mathematically given by

$t=(\frac{\omega_0}{\omega_0-\omega})t$

$t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})$

$t=0.87min$

Therefore Angular displacement is

$\theta =(\frac{120+510}{2})0.87$

$\theta=274rev$

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2 years ago

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

GREYUIT [131]

Answer:

$\omega = 22.67 rad/s$

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

$W = \frac{1}{2}I\omega^2$

now we know that work done is product of force and displacement

so here we have

$W = F.d$

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now for moment of inertia of the disc we will have

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(7 kg)(0.21^2)$

$I = 0.154 kg m^2$

now from above equation we will have

$39.6 = \frac{1}{2}(0.154)\omega^2$

$\omega = 22.67 rad/s$

5 0

2 years ago

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