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Radda [10]
2 years ago
6

Stars originate as large bodies of slowly rotating gas.Because of gravity, these clumps of gas slowly decrease in size.The angul

ar velocity of a star increases as it shrinks because of__.
A.) law of universal gravitation
B.) conservation of linear momentum
C.) conservation of energy
D.) conservation of mass
E.) conservation of angular momentum
Physics
1 answer:
Tamiku [17]2 years ago
4 0

Answer:

E.) conservation of angular momentum

Explanation:

The angular momentum is defined as:

L=r x mv

where r is the radius of the star, m is the mass and v the angular velocity.

and angular momentum is an amount that is conserved, so the angular momentum before the star is compressed must be equal to the angular momentum after the star was compressed:

r_{1} x mv_{1}=r_{2} x mv_{2}

the second radius is smaller than the first radius, since the star shrinked, the second angular velocity must be greater that the first.

In other words, the angular velicity increases as the star shrinks because of the conservation of angular momentum.

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An archer fires an arrow, which produces a muffled "thwok" as it hits a target. If the archer hears the "thwok" exactly 1 s afte
aniked [119]

Answer:

35,79 meters

Explanation:

So, we got an archer, and we got a target. Lets call the distance between this two d.

Now, the archer fires the arrow, that, in a time t_{arrow} travels the distance d with a speed v_{arrow} of 40 m/s and hits the target. We can see that the equation will be:

v_{arrow} * t_{arrow} = d\\ \\40 \frac{m}{s} * t_{arrow} = d

Immediately after this, the arrow produces a muffled sound, which will travel the distance d at  340 m/s in a time t_{sound}. Obtaining :

v_{sound} * t_{sound} = d\\ \\340 \frac{m}{s} * t_{sound} = d.

Finally, the sound reaches the archer, exactly 1 second after he fired the bow, so:

t_{arrow} + t _{sound} = 1 s.

This equation allows us to write:

t _{sound} = 1 s - t_{arrow}.

Plugging this  relationship in the distance equation for the sound:

340 \frac{m}{s} * t_{sound} = d \\ \\ 340 \frac{m}{s} * (1 s- t_{arrow}) = d.

Now, we can replace d from the first equation, and obtain:

40 \frac{m}{s} * t_{arrow} = d \\ 40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * (1 s- t_{arrow}).

Now, we can just work a little bit:

40 \frac{m}{s} * t_{arrow} = 340 \frac{m}{s} * 1 s - 340 \frac{m}{s} * t_{arrow} \\ \\ 40 \frac{m}{s} * t_{arrow} + 340 \frac{m}{s} * t_{arrow} = 340 m \\ \\ 380 \frac{m}{s} * t_{arrow} = 340 m \\ \\ t_{arrow} = \frac{340 m}{380 \frac{m}{s}} \\ \\ t_{arrow} = 0.8947 s.

Now, we can just plug this value into the first equation:

40 \frac{m}{s} * t_{arrow} = d

40 \frac{m}{s} * 340/380 s = 35,79 s = d

6 0
2 years ago
Seven seconds after a brilliant flash of lightning, thunder shakes the house. approximately how far was the lightning strike fro
tangare [24]
Very roughly 7,700 feet ... about 1.5 miles.
8 0
2 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
Oksana_A [137]
<h2>The K.E of the charge is 1.02 x 10⁻¹⁷ J</h2>

Explanation:

When the charge of 2e is placed in between the plates .

The force applied on this charge by plates is = q E

here q is the magnitude of charge = 2 e = 2 x 1.6 x 10⁻¹⁹ C

and E is the magnitude of electric field intensity

The work done = Force x displacement

Thus W = q E x S

here S is displacement

Therefore W = 2 x 1.6 x 10⁻¹⁹ x 4 x 8

= 1.02 x 10⁻¹⁷ J

This work will be converted into the kinetic energy of charge .

Thus K.E = 1.02 x 10⁻¹⁷ J

3 0
2 years ago
A high school physics instructor catches one of his students chewing gum in class. He decides to discipline the student by askin
KengaRu [80]

a) 219.8 rad/s

b) 20.0 rad/s^2

c) 2.9 m/s^2

d) 7005 m/s^2

e) Towards the axis of rotation

f) 0 m/s^2

g) 31.9 m/s

Explanation:

a)

The angular velocity of an object in rotation is the rate of change of its angular position, so

\omega=\frac{\theta}{t}

where

\theta is the angular displacement

t is the time elapsed

In this problem, we are told that the maximum angular velocity is

\omega_{max}=35 rev/s

The angle covered during 1 revolution is

\theta=2\pi rad

Therefore, the maximum angular velocity is:

\omega_{max}=35 \cdot 2\pi = 219.8 rad/s

b)

The angular acceleration of an object in rotation is the rate of change of the angular velocity:

\alpha = \frac{\Delta \omega}{t}

where

\Delta \omega is the change in angular velocity

t is the time elapsed

Here we have:

\omega_0 = 0 is the initial angular velocity

\omega_{max}=219.8 rad/s is the final angular velocity

t = 11 s is the time elapsed

Therefore, the angular acceleration is:

\alpha = \frac{219.8-0}{11}=20.0 rad/s^2

c)

For an object in rotation, the acceleration has two components:

- A radial acceleration, called centripetal acceleration, towards the centre of the circle

- A tangential acceleration, tangential to the circle

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

Here we have

\alpha =20.0 rad/s^2

d = 29 cm is the diameter, so the radius is

r = d/2 = 14.5 cm = 0.145 m

So the tangential acceleration is

a_t=(20.0)(0.145)=2.9 m/s^2

d)

The magnitude of the radial (centripetal) acceleration is given by

a_c = \omega^2 r

where

\omega is the angular velocity

r is the radius of the circle

Here we have:

\omega_{max}=219.8 rad/s is the angular velocity when the fan is at full speed

r = 0.145 m is the distance of the gum from the centre of the circle

Therefore, the radial acceleration is

a_c=(219.8)^2(0.145)=7005 m/s^2

e)

The direction of the centripetal acceleration in a rotational motion is always towards the centre of the axis of rotation.

Therefore also in this case, the direction of the centripetal acceleration is towards the axis of rotation of the fan.

f)

The magnitude of the tangential acceleration of the fan at any moment is given by

The tangential acceleration is given by

a_t = \alpha r

where

\alpha is the angular acceleration

r is the radius of the circle

When the fan is rotating at full speed, we have:

\alpha=0, since the fan is no longer accelerating, because the angular velocity is no longer changing

r = 0.145 m

Therefore, the tangential acceleration when the fan is at full speed is

a_t=(0)(0.145)=0 m/s^2

g)

The linear speed of an object in rotational motion is related to the angular velocity by the formula:

v=\omega r

where

v is the linear speed

\omega is the angular velocity

r is the radius

When the fan is rotating at maximum angular velocity, we have:

\omega=219.8 rad/s

r = 0.145 m

Therefore, the linear speed of the gum as it is un-stucked from the fan will be:

v=(219.8)(0.145)=31.9 m/s

7 0
2 years ago
A 3 kg rubber block is resting on wet concrete. The coefficient of static friction is 0.3. What is the minimum force that must b
mrs_skeptik [129]

Answer:

You would have to find the friction force of the rubber block which would be found with the equation of Normal force (mass*gravity) times cooeficient of friction which would give 8.82 N for the amount of friction and because you need more force than 8.82 N (assuming gravity is 9.8)

8 0
1 year ago
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