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Vesna [10]
3 years ago
5

What is the torque about the center of the sun due to the gravitational force of attraction of the sun on the planet?

Physics
1 answer:
Anon25 [30]3 years ago
3 0
Torque, moment, or moment of force is the tendency of a force to rotate an object around an axis,fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist to an object. Mathematically, torque is defined as the cross product of the vector by which the force's application point is offset relative to the fixed suspension point (distance vector) and the force vector, which tends to produce rotational motion. So torque about the center of the sun due to the gravitational force of attraction of the sun on the planet = (Gxm1xm2 / r^2). r sin(theta) = Gxm1xm2 /r^2). r sin0° = 0
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White dwarfs are too small to see with telescopes true or false
trasher [3.6K]

Answer:

False

Explanation:

A white dwarf star that is easy to locate and see with small telescopes.

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3 years ago
You and your friends find a rope that hangs down 19m from a high tree branch right at the edge of a river. You find that you can
tangare [24]

Answer:

Explanation:

Given

length of rope L=19\ m

velocity while running v=2\ m/s

when the person jumps off the bank and hang on the rope then we can treat the person as pendulum with Time period T which is given by

T=2\pi \sqrt{\frac{L}{g}}

T=2\pi \sqrt{\frac{19}{9.8}}

T=2\pi \times 1.392

T=8.74\ m/s

Greatest Possible distance will be covered when person reaches the other extreme end  of assumed pendulum (velocity=zero)

therefore he must hang on for 0.5 T time

time=0.5\times 8.74=4.37\ s

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3 years ago
What does the atomic mass number represent
UNO [17]

Answer:

the number of neutrons and protons in an atom

6 0
3 years ago
When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the
dlinn [17]

To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first

F_1 = F \frac{{343}}{(343-V)}

Where F is the actual frequency and v is the velocity of the ambulance

Now the source is moving away from the observer.

F_2 = F\frac{343}{(343+V)}

We are also so told the perceived frequency decreases by 11.9%

F_2 = F_1 - 9.27\% \text{ of } F_1

F_2 = F_1-0.0927F_1

F_2 = 0.9073F_1

Equating,

F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})

\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}

0.9073(343+V) = 343-V

(0.9073)(343)+(0.9073)V = 343-V

V+0.9073V = 343-(0.9073)(343)

Solving for V,

V = 16.67 m/s

5 0
3 years ago
A train is moving with a velocity of 20 m/s. when the brakes are applied the acceleration is reduced to -0.6 m/s².calculate the
Serhud [2]

Answer:

x = 333.33 [m]

Explanation:

To solve this problem we must use the following kinematics equation.

v_{f}^{2} = v_{i}^{2}-(2*a*x)

where:

Vf = final velocity = 0

Vi = initial velocity = 20 [m/s]

a = desacceleration = 0.6 [m/s^2]

x = distance [m]

Note: the final speed is zero as the body finishes its movement.

Now replacing:

0 = (20)^2 - (2*0.6*x)

1.2*x = 400

x = 333.33 [m]

3 0
3 years ago
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