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Luden [163]
3 years ago
9

A trapeze artist on a rope is momentarily held to one side by a partner on a platform The rope makes an angle of 26.0 with the v

ertical. Insuch a condition of static equilibrium, what is the magnitude of the horizontal force being applied by the partner? The weight of the artist is 7.61 10 N
Physics
1 answer:
masya89 [10]3 years ago
7 0

Answer:

F_t=371.16N

Explanation:

The weight 7.61x10^2N

The angle 26.0

To determine the magnitude of horizontal force applied

tan(\alpha)=\frac{F_t}{F}

F_t=F*tan(\alpha)

F_t=7.61x10^2N*tan(26.0)

F_t=371.16N

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An ideal gas is maintained at a constant pressure of 70,000 Pa during an isobaric process and its volume decreases by 0.2m^3. Wh
Svetach [21]

Answer:

c. -14,000

Explanation:

Workdone by gas is the product of the pressure and the volume where there is a change of volume.

If v1 is the initial volume of the gas and v2 is the final volume of gas, the work done

= p(v2 - v1)

where p is the pressure

and p = 70,000 Pa

Given that volume decrease by 0.2m^3, v2 - v1 = -0.2

Workdone = 70000 ( -0.2)

Workdone = -14,000 J

Option c. -14,000

5 0
4 years ago
As an adult, what is the best way to remain healthy?
Zigmanuir [339]
1. Eat a balanced diet and regular exercise. This is the best way because eating right and exercising are is more beneficial than just exercising.
4 0
3 years ago
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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.6 multiply 1010 m (inside t
Doss [256]

We will apply the concepts related to energy conservation to develop this problem. In this way we will consider the distances and the given speed to calculate the final speed on the path from the sun. Assuming that the values exposed when saying 'multiply' is scientific notation we have the following,

d_1 = 4.6*10^{10}m

v_i = 9.3*10^4m/s \rightarrow \text{Initial velocity comet}

d_2 = 6*10^{12}m

The difference of the initial and final energy will be equivalent to the work done in the system, therefore

E_f = E_i +W

K_f +U_f = K_i +U_i + 0

\frac{1}{2} mv_f^2+\frac{-GMm}{d_2} = \frac{1}{2} mv_i^2+\frac{-GMm}{d_1}

Here,

m = Mass

v_f = Final velocity

G = Gravitational Universal Constant

M = Mass of the Sun

m = Mass of the comet

v_i = Initial Velocity

Rearranging to find the final velocity,

v_f = \sqrt{v_i^2+2GM(\frac{1}{d_2}-\frac{1}{d_1})}

Replacing with our values we have finally,

v_f = \sqrt{(9.3*10^4)+2(6.7*10^{-11})(1.98*10^{30})(\frac{1}{6*10^{12}}-\frac{1}{4.6*10^{10}})}

v_f = 75653.9m/s

Therefore the speed is 75653m/s

8 0
4 years ago
What is the youngest to oldest mesozoic era, precambrian time, cenozoic era, paleozoic era
Lynna [10]
Oldest to youngest *Precambrian Era*Paleozoic *Mesozoic*Cenozoic
6 0
4 years ago
You are in a spacecraft moving at a constant velocity. The front thruster rocket fires incorrectly, causing the craft to slow do
Alchen [17]

Answer:

It continue to move forward at a constant velocity which will be slower than before the front thruster was fired.

Explanation:

Before the front thruster was fired, the spacecraft was already moving at a particular velocity.

After the malfunction, the front thruster is fired and then the force exerted by that front thruster slows the spacecraft down, as we are told.

By using the rear thruster to exert a force equal to that from the front thrusters, a force equal in magnitude to that of the front thrusters is added, cancelling out the effect of the front thrusters. Because the spacecraft is already moving at a slower speed at this point compared to the beginning, it continues to move at that speed.

8 0
3 years ago
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