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kolezko [41]
4 years ago
13

What are earths two main motions called

Physics
1 answer:
andre [41]4 years ago
3 0
Rotation and revolution
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Which of the following is an example of sustainable use of natural resources? Question 21 options: 1) clear-cutting a forest 2)
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4 years ago
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What is the torque τa about axis a due to the force f⃗ ? express the torque about axis a at cartesian coordinates (0,0)?
pentagon [3]

Answer:

\tau_a = F a sin \theta

Explanation:

The torque of a force is given by:

\tau = F d sin \theta

where

F is the magnitude of the force

d is the distance between the point of application of the force and the centre of rotation of the system

\theta is the angle between the direction of the force and d

In this problem, we have:

F, the force

a, the distance of application of the force from the centre (0,0)

\theta, the angle between the direction of the force and a

Therefore, the torque is

\tau_a = F a sin \theta

5 0
3 years ago
You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While
ANTONII [103]

To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.

When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:

W = F_T + m\omega^2r_E

Where

\omega = \frac{2\pi}{T} \rightarrow Angular velocity is equal to the Period, at this case Earth's period

r_E = 6.371*10^6m \rightarrow Radius of the Earth

m = mass

F_T= Force of Tension

W = mg \rightarrow Newton's second law

Replacing and re-arrange to find the Tension we have,

F_T = W- \frac{W}{g} (\frac{2\pi}{T})^2r_E

F_T = W(1-(\frac{2\pi}{T})^2\frac{r_E}{g})

F_T = (505)(1-(\frac{2\pi}{24hours})^2\frac{6.371*10^6}{9.8})

F_T = (505)(1-(\frac{2\pi}{24hours(\frac{3600s}{1hour})})^2\frac{6.371*10^6}{9.8})

F_T = (505)(1-(\frac{2\pi}{86400})^2\frac{6.371*10^6}{9.8})

F_T = 503.26N

Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N

7 0
3 years ago
A soccer ball is kicked from the ground with an initial speed v at an upward angle θ. A player a distance d away in the directio
Nady [450]

Answer:

The speed of player is given by

V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}

Explanation:

The time of flight for a projectile motion is given by

T=\frac{2vsin\alpha }{g}    (i)

where t is the time of flight, v is the initial speed, and α is the angle.

Now the person must also reach the impact point of ball in the same time as above.

Now the total distance D the player needs to cover is basically R horizontal range of projectile minus the distance d, range R is given by,

R=\frac{2v^{2} sin2\alpha }{g}

Now the distance the player must cover is given by

D= R-d

D= \frac{2v^{2} sin2\alpha }{g}  - d

 

D=\frac{2v^{2}sin2\alpha-gd}{g}  (ii)

Now the average speed of player is given by

V=\frac{D}{T}   (iii)

Replacing the values of D and T from eq. (i) and (ii) in eq. (iii).

V=\frac{\frac{2v^{2} sin2\alpha-gd }{g}}{\frac{2vsin\alpha }{g} }

V=\frac{2v^{2} 2sin\alpha.cos\alpha-gd }{2vsin\alpha}

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3 years ago
What happens when a charged insulator is placed near an uncharged metallic object?
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