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bearhunter [10]
2 years ago
11

If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac

tor of 4? (enter an integer, as in it will be rotating ________ times faster). (3 pts)
Physics
1 answer:
maks197457 [2]2 years ago
5 0
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ =  (r₁/ r₂)² · ω<span>₁
     
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ =  (4)² · ω<span>₁
     = 16 </span>· ω<span>₁

Hence, the protostar will be rotating 16 </span><span>times faster.</span>
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Calculate the distance the marble travels during the first 3.0 seconds. [Show all work, including the equation and substitution
Alenkinab [10]

D = V0t + 0.5at^2

Where d is the distance

V0 is the initial velocity

A is the acceleration

T is time

From the graph a = 4/3 m/s2

D = 0(3) + 0.5( 4/3 m/s2) ( 3 s)^2

D = 6 m

3 0
2 years ago
A circular saw blade with radius 0.175 m starts from rest and turns in a vertical plane with a constant angular acceleration of
ANEK [815]

Answer:

The distance the piece travel in horizontally axis is

L=3.55m

Explanation:

a=2 \frac{rev}{s^{2}} \\h=0.820m\\r = 0.125 m&#10;\\d=150rev

d= 155 rev = 155(2\pi ) = 310\pi rad

a= 2.0 \frac{rev}{s^{2} } = 2.0(2\pi )  = 4.0\pi \frac{rev}{s^{2} }

d=d_{i}+vo*t+\frac{1}{2}*a*t^{2} \\ di=0\\vo=0\\d=\frac{1}{2}*a*t^{2}\\t=\sqrt{\frac{2*d}{a}}\\t=\sqrt{\frac{2*310 rad}{4\frac{rad}{s^{2}}}} \\t=12.449

w=a*t\\w=4\frac{rad}{s^{2}}*12.449s\\ w=49.79 \frac{rad}{s}

Now the angular velocity is the blade speed so:

V=w*r\\V=49.79 \frac{rad}{s}*0.175m\\V=8.7 \frac{m}{s}

assuming no air friction effects affect blade piece:

time for blade piece to fall to floor

t=\sqrt{\frac{2*h}{g}}\\t=\sqrt{\frac{2*0.820m}{9.8\frac{m}{s^{2} } }}\\t=0.409s

Now is the same time the piece travel horizontally

L=t*V\\L=0.409s*8.7\frac{m}{s}\\L=3.55m

blade piece travels  HORIZONTALLY = (24.5)(0.397) = 9.73 m  ANS

6 0
2 years ago
Little Tammy lines up to tackle Jackson to (unsuccessfully) prove the law of conservation of momentum. Tammy’s mass is 34.0 kg a
Naily [24]

Answer:

So Tammy must move with speed 4.76 m/s in opposite direction of Jackson

Explanation:

As per law of conservation of momentum we know that there is no external force on it

So here we can say that initial momentum of the system must be equal to the final momentum of the system

now we have

m_1v_1 + m_2v_2 = 0

final they both comes to rest so here we can say that final momentum must be zero

now we have

34 v + 54 (3 m/s) = 0

v = -4.76 m/s

8 0
2 years ago
Which changes would result in a decrease in the gravitational force between two objects? Check all that apply.
Rufina [12.5K]

Answer: 1. decreasing the mass of both objects

2. decreasing the mass of one of the objects

3. increasing the distance between the objects

Explanation: Hope that helped! (:

4 0
2 years ago
A 2530-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force so that
gavmur [86]

Answer:

A = 1.4 m/s²

B = -0.10493 m/s³

a = 1.29507 m/s²

T = 28095.8271 N

T = 1.13198 W

Explanation:

t = Time taken

g = Acceleration due to gravity = 9.81 m/s²

The equation

v(t)=At+Bt^2

Differentiating with respect to time

\frac{dv}{dt}=\frac{d(At+Bt^2)}{dt}\\\Rightarrow 1.4=A+2Bt

At t = 0

1.4=A

Hence, A = 1.4 m/s²

B=\frac{v-At}{t^2}\\\Rightarrow B=\frac{2.18-1.4\times 1.8}{1.8^2}\\\Rightarrow B=-0.10493\ m/s^3

B = -0.10493 m/s³

At t = 5 seconds

a=1.4+2\times -0.010493\times 5=1.29507\ m/s^2

a = 1.29507 m/s²

T=m(a+g)\\\Rightarrow T=2530(1.29507+9.81)\\\Rightarrow T=28095.8271\ N

T = 28095.8271 N

Weight of rocket

W=2530\times 9.81=24819.9\ N

\frac{T}{W}=\frac{28095.8271}{24819.9}\\\Rightarrow \frac{T}{W}=1.13198\\\Rightarrow T=1.13198W

T = 1.13198 W

3 0
2 years ago
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