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Umnica [9.8K]
2 years ago
7

What is the similarity between pushing and lifting? What is the difference?

Physics
1 answer:
MA_775_DIABLO [31]2 years ago
7 0

Answer:

the similarity is that hands are applied and force

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Four identical particles of mass 0.980 kg each are placed at the vertices of a 4.14 m x 4.14 m square and held there by four mas
Zielflug [23.3K]

Answer:

a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²

b) I = 50.39 kg m²

c) I = 16.8 kg m²

Explanation:

a) Given data:

m = 0.98 kg

a = 4.14 * 4.14

The moment of inertia is:

I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}

For 4 particles:

I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

b) Distance from top left mass = x = a/2

Distance from bottom left mass = x = a/2

Distance from top right mass = x = √5 (a/2)

The total moment of inertia is:

I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}

c)

I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}

8 0
2 years ago
Your new motorcycle weighs 2450 N.
Snowcat [4.5K]

Answer:

Mass can never be negative. Everything has mass. Just like how they ask you to find area under the graph in maths. If the area is in the 3rd and 4th quadrant, when calculated, you would get negative answer.However, area can not be negative because it is a place/ location. It's exactly the same as mass.

4 0
2 years ago
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.
leonid [27]

Answer:

(a): F_e = 8.202\times 10^{-8}\ \rm N.

(b): F_g = 3.6125\times 10^{-47}\ \rm N.

(c): \dfrac{F_e}{F_g}=2.27\times 10^{39}.

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053\times 10^{-9} m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges q_1 and q_2 respectively is given by

F_e = \dfrac{k|q_1||q_2|}{r^2}

where,

  • k = Coulomb's constant = 9\times 10^9\ \rm Nm^2/C^2.
  • r = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, q_1 = +1.6\times 10^{-19}\ C.

The charge on the electron, q_2 = -1.6\times 10^{-19}\ C.

These two are separated by the distance, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.

Part (b):

The gravitational force of attraction between two objects of masses m_1 and m_1 respectively is given by

F_g = \dfrac{Gm_1m_2}{r^2}.

where,

  • G = Universal Gravitational constant = 6.67\times 10^{-11}\ \rm Nm^2/kg^2.
  • r = distance of separation between the masses.

For the given system,

The mass of proton, m_1 = 1.67\times 10^{-27}\ kg.

The mass of the electron, m_2 = 9.11\times 10^{-31}\ kg.

Distance between the two, r = 0.053\times 10^{-9}\ m.

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.

The ratio \dfrac{F_e}{F_g}:

\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.

6 0
3 years ago
Which of the following is an example of how humans can increase biodiversity?
Leto [7]

Answer:

The following is an example of how humans can increase biodiversity

Provide Wildlife Corridors and Connections Between Green Spaces

Use Organic Maintenance Methods and Cut Back On Lawns

Use a Native Plant Palette and Plant Appropriately

Utilize Existing Green Space Connections

Be Mindful of Non-Native Predators

7 0
3 years ago
An aircraft flies at sea level at a speed of 220 m/s. What is the highest pressure that can be acting on the surface of the airc
goldenfox [79]

Answer:

An aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

Explanation:

Applying Bernoulli's equation, we determine the highest pressure on the aircraft.

P = \frac{1}{2} \rho V^2

where;

P is the highest pressure on the aircraft

\rho is the density of air = 1.204 kg/m³ at sea level temperature.

V is the velocity of the aircraft = 220 m/s

P = 0.5*1.204*(220)² = 29136.8 N/m²

Therefore, an aircraft flying at sea level with a speed of 220 m/s, has a highest pressure of 29136.8 N/m²

6 0
3 years ago
Read 2 more answers
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