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2 years ago

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According to Newton's third law, two objects interacting under a force (such as gravity) both feel the same force. If the planet

s pull on the Sun as much as the Sun pulls on the planets, why are we able to approximate the Sun as a fixed position when studying the planetary orbits?

1 answer:

asambeis [7]2 years ago

8 0

Explanation:

That's because the Sun's acceleration is much smaller

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A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin

kap26 [50]

Answer:

a) $\omega \approx 219.911\,\frac{rad}{s}$ , b) $\alpha = 16.916\,\frac{rad}{s^{2}}$ , c) $a_{t} = 1.776\,\frac{m}{s^{2}}$ , d) $a_{n} = 5077.889\,\frac{m}{s^{2}}$ , e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation, f) Zero, g) $v = 23.091\,\frac{m}{s}$ .

Explanation:

a) The maximum angular velocity of the fan is:

$\omega = (35\,\frac{rev}{s} )\cdot (\frac{2\pi\,rad}{1\,rev} )$

$\omega \approx 219.911\,\frac{rad}{s}$

b) The angular acceleration of the fan is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{219.911\,\frac{rad}{s}-0\,\frac{rad}{s}}{13\,s}$

$\alpha = 16.916\,\frac{rad}{s^{2}}$

c) The magnitude of the tangential aceleration is:

$a_{t} = (16.916\,\frac{rad}{s^{2}} )\cdot (0.105\,m)$

$a_{t} = 1.776\,\frac{m}{s^{2}}$

d) The magnitude of the centripetal acceleration is:

$a_{n} = (219.911\,\frac{rad}{s} )^{2}\cdot (0.105\,m)$

$a_{n} = 5077.889\,\frac{m}{s^{2}}$

e) The direction of the centripetal acceleration experimented by the gum goes to the center of rotation.

f) When fan is at full speed, it rotates at constant rate and, hence, there is no angular acceleration. Besides, the tangential acceleration experimented by the gum is zero.

g) The linear speed of the gum is:

$v = (219.911\,\frac{rad}{s} )\cdot (0.105\,m)$

$v = 23.091\,\frac{m}{s}$

5 0

3 years ago

Two soccer players start from rest, 36 m apart. They run directly toward each other, both players accelerating. The first player

Cerrena [4.2K]

A) Both players are moving by uniformly accelerated motion, and we can write the position at time t of each of the two players as follows:
$x_1(t)= \frac{1}{2}a_1 t^2$
$x_2(t)=d- \frac{1}{2}a_2 t^2$
where
$a_1 = 0.58 m/s^2$ is the acceleration of the first player
$a_2=0.42 m/s^2$ is the acceleration of the second player
$d=36 m$ is the initial distance between the two players
and where I put a negative sign in front of the acceleration of the second player, since he's moving in the opposite direction of the first player.

The time t at which the two players collide is the time t at which $x_1 = x_2$ , therefore:
$\frac{1}{2}a_1 t^2 = d- \frac{1}{2}a_2 t^2$
from whic we find
$t= \sqrt{ \frac{2d}{a_1+a_2} }= \sqrt{ \frac{2 \cdot 36 m}{0.58 m/s^2+0.42 m/s^2} }=8.5 s$

b) We can use the equation of $x_1(t)$ to find how far the first player run in t=8.5 s:
$x_1(t)= \frac{1}{2}a_1 t^2= \frac{1}{2}(0.58 m/s^2)(8.5 s)^2=21.0 m$

8 0

3 years ago

What is the difference between velocity and speed /vector and scalar?

stepladder [879]

Answer:

Speed only defines the magnitude of how fast an object is moving from one point to another. This is a scalar quantity (Only Value)

Velocity defines both how fast an object is moving and also in what direction the object is moving. This is a vector quantity (Value + Direction)

7 0

3 years ago

I know the acceleration due to gravity (ie 9.8 m/s2) will have a negative sign when falling down, a positive one when going up.

rewona [7]

Answer:

Yes

Explanation:

Accerelation is measured by change in velocity. So naturally, if an object is slowing down, its velocity is decreasing so acceleration is negative. If it is speeding up velocity is increasing so positive acceleration.

(Velocity final - Velocity initial)/t

Note that this does not apply only to gravity, but to all linear accelerations

6 0

2 years ago

Find the instantaneous acceleration at t=ls for an object moving along a straight axis with velocity function:

Maru [420]

The answer is A.
Explanation:
We know that the average acceleration a for an interval of time Δt is expressed as:

a = Δv
Δt
where Δv is the change in velocity that occurs during Δt.
e formula for the instantaneous acceleration a is almost the same, except that we need to indicate that we're interested in knowing what the ratio of Δv to Δt approaches as Δt approaches zero.

We can indicate that by using the limit notation.

So, the formula for the instantaneous acceleration is:

a = lim Δv
Δt→0 Δt

8 0

3 years ago