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Rus_ich [418]
2 years ago
8

Help please help!!!!!!!!

Physics
1 answer:
Sidana [21]2 years ago
6 0

1. Electron

2. Neutron

3. Nucleus

4. Atom

5. Molecule

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A bug flying horizontally at 0.65 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, th
irina [24]

The angular momentum is defined as,

L=I\omega

Acording to this text we know for conservation of angular momentum that

L_i=L_f

Where L_iis initial momentum

L_f is the final momentum

How there is a difference between the stick mass and the bug mass, we define that

Mass of the bug= m

Mass of the stick=10m

At the point 0 we have that,

L_i=mvl

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity

vector from the point of reference (O), and ve is the velocity

At the end with the collition we have

L_f=(I_b+I_s)\omega

Substituting

L_f=(ml^2+\frac{10ml^2}{3})\omega

L_f=\frac{13}{10}ml^2w

m(0.65)l=\frac{13}{10}ml^2 \omega

\omega=\frac{1}{2l}

Applying conservative energy equation we have

\frac{1}{2}(I_b+I_s)\omega^2=mgh+10mgh'

\frac{1}{2}(ml^2+\frac{10ml^2}{3})(\frac{1}{2l})^2=mg(l-lcos\theta)+\frac{10}{2}mg(l-lcos\theta)

Replacing the values and solving

l=\frac{13}{0.54g}

Substituting

l=\frac{13}{0.54(9.8)}

l=2.45cm

7 0
3 years ago
One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious spa
Ludmilka [50]

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

<u>ΔP.E = 6.48 x 10⁸ J</u>

7 0
3 years ago
Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a
brilliants [131]

Answer:

Speed will be 30810 rpm      

Explanation:

We have given diameter of the tire d = 24 inch

So radius r=\frac{d}{2}=\frac{24}{2}=12imch

We have given linear velocity v = 35 mph

We know that linear velocity is given by v=\omega r

35=\omega \times 12

\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min

As we know that 1 mile = 63360 inch and 1 hour = 60 min

3 0
3 years ago
Describe the relationship between radius and mass?
algol [13]

Answer:

mass of the earth is directly proportional to its gravity and the radius is inversely proportional to its gravity. ... But when you go inside the surface of the earth inspiteof decreasing radius the force of gravity decreases.

Explanation:

8 0
3 years ago
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The author writes that each part of the ecosystem is as important as another. Based
Masja [62]
I agreeeeeee do you?
3 0
3 years ago
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