Which of the following are likely to form a covalent bond?

What gravitational force does the moon produce on the earth is their centers are 3. 88x10^8 m apart and the moon has a mass of 7

vitfil [10]

The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.

<h3>What kind of gravitational pull does the moon have on the planet?</h3>

On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.

<h3>What does the Earth's center's gravitational pull feel like?</h3>

Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).

<h3>Where is the Earth's and the moon's gravitational centre?</h3>

It is around 1700 kilometres below Earth's surface.

To know more about gravitational force visit:-

brainly.com/question/12528243

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6 0

1 year ago

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains

OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

5 0

3 years ago

Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

4 0

3 years ago

When light is reflected by a mirror, the angle of incidence is always A. equal to the angle of reflection. B. less than the angl

ankoles [38]

When light is reflected by a mirror, the angle of incidence is always <span>A. equal to the angle of reflection. We know this by the Law of Reflection.</span>

6 0

3 years ago

Read 2 more answers

A cylinder of diameter 100 mm rolls from restdown a 5 m long ramp and its center of mass is moving with velocity 2 m/s at the bo

RoseWind [281]

Answer:

(a): a = 0.4m/s²

(b): α = 8 radians/s²

Explanation:

First we propose an equation to determine the linear acceleration and an equation to determine the space traveled in the ramp (5m):

a= (Vf-Vi)/t = (2m/s)/t

a: linear acceleration.

Vf: speed at the end of the ramp.

Vi: speed at the beginning of the ramp (zero).

d= (1/2)×a×t² = 5m

d: distance of the ramp (5m).

We replace the first equation in the second to determine the travel time on the ramp:

d = 5m = (1/2)×( (2m/s)/t)×t² = (1m/s)×t ⇒ t = 5s

And the linear acceleration will be:

a = (2m/s)/5s = 0.4m/s²

Now we determine the perimeter of the cylinder to know the linear distance traveled on the ramp in a revolution:

perimeter = π×diameter = π×0.1m = 0.3142m

To determine the angular acceleration we divide the linear acceleration by the radius of the cylinder:

α = (0.4m/s²)/(0.05m) = 8 radians/s²

α: angular aceleration.

3 0

3 years ago