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Rasek [7]
3 years ago
13

Consider this situation: A rope is used to lift an actor upward off the stage. Of the forces listed, identify which act upon the

actor.
Normal


Gravity


Applied


Friction


Tension


Air Resistance
Physics
2 answers:
zhenek [66]3 years ago
6 0
Gravity, and air resistance act upon the actor.
sveticcg [70]3 years ago
4 0

Answer:

Gravity

Tension

Air Resistance

Explanation:

Since actor is lifted upwards with the help of rope which is connected to the actor

so here forces that will act upon it

1) Gravity

Gravity is due to gravitational force of earth which will act towards center of earth or vertically downwards

2) Tension

This is a type of electromagnetic force which will act along the string and it will help the actor to balance against his weight

3) Air Resistance

this is due to friction of air which will act opposite to the motion of actor in air

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calculate the frequency and time period of sound wave of 35 m wave length propagating at a speed of 3500m/s​
zheka24 [161]

Answer:

Frequency = 100 Hz

Time period = 0.01 sec

Explanation:

f= 3500/35 = 100

T = 1/f = .01

5 0
3 years ago
The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0
nirvana33 [79]

(a) 7.25\cdot 10^{-8}m

Wien's displacement law is summarized by the equation

\lambda = \frac{b}{T}

where

\lambda is the peak wavelength

b=2.898\cdot 10^{-3} m \cdot K is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays  

X-rays  1 nm - 1 pm

ultraviolet  380 nm - 1 nm

visible light  750 nm - 380 nm

infrared  25 \mu m - 750 nm

microwaves  1 mm - 25 \mu m

radio waves  > 1 mm

The peak wavelength of this star is

\lambda=7.25\cdot 10^{-8}m=72.5 nm

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect  much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0
3 years ago
Suppose that in a lightning flash the potential difference between a cloud and the ground is 0.96×109 V and the quantity of char
Dvinal [7]

(a) 2.98\cdot 10^{10} J

The change in energy of the transferred charge is given by:

\Delta U = q \Delta V

where

q is the charge transferred

\Delta V is the potential difference between the ground and the clouds

Here we have

q=31 C

\Delta V = 0.96\cdot 10^9 V

So the change in energy is

\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

K=\frac{1}{2}mv^2

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

K=2.98\cdot 10^{10} J

Therefore by re-arranging the equation, we find the final speed of the car:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s

5 0
4 years ago
A force of 7.50N is applied to a spring whose spring constant is 0.289 N/cm. Find its change in length.​
kotykmax [81]

Hook's law states that

F=k\Delta x

where F is the applied force, k is the spring constant, and \Delta x is the change in length.

Plugging your values, we have

7.5 = 0.289\Delta x \implies \Delta x = \dfrac{7.5}{0.289}\approx 25.95

4 0
3 years ago
Please help me with this question ​
mariarad [96]

Answer:

Please help me with this question Please help me with this questionPlease help me with this question Please help me with this question Please help me with this question Please help me with this question Please help me with this question Please help me with this questionPlease help me with this question

5 0
2 years ago
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