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2 years ago

If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

2 answers:

6 0

If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.

sveticcg [70]2 years ago

3 0

It would be half as strong

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Estimate the change in the equilibrium melting point of copper caused by a change in pressure of 10 kbar. The molar volume of co

mr Goodwill [35]

Answer:

The change in the equilibrium melting point is 4.162 K.

Explanation:

Given that,

Pressure = 10 kbar

Molar volume of copper $V=8.0\times10^{-6}\ m^3$

Volume of liquid $V=7.6\times10^{-6}\ m^3$

Latent heat of fusion $L= 13.05 kJ$

Melting point =1085°C

We need to calculate the change temperature

Using Clapeyron equation

$\dfrac{\Delta P}{\Delta T}=\dfrac{\Delta H}{T\Delta V}$

Put the value into the formula

$\dfrac{1000\times10^{5}}{\Delta T}=\dfrac{13050}{(1085+273)\times(8.0-7.6)\times10^{-6}}$

$\Delta T=\dfrac{1000\times10^{-5}\times(1085+273)\times(8.0-7.6)\times10^{-6}}{13050}$

$\Delta T=4.162\ K$

Hence, The change in the equilibrium melting point is 4.162 K.

5 0

2 years ago

Two long conducting cylindrical shells are coaxial and have radii of 20 mm and 80 mm. The electric potential of the inner conduc

xxMikexx [17]

Answer: 14.52*10^6 m/s

Explanation: In order to explain this problem we have to consider the energy conservation for the electron within the coaxial cylidrical wire.

the change in potential energy for the electron; e*ΔV is equal to energy kinetic gained for the electron so:

e*ΔV=1/2*m*v^2 v^=(2*e*ΔV/m)^1/2= (2*1.6*10^-19*600/9.1*10^-31)^1/2=14.52 *10^6 m/s

3 0

2 years ago

Drying of Cassava (Tapioca) Root. Tapioca flour is used in many countries for bread and similar products. The flour is made by d

svet-max [94.6K]

Yea it would be 500 minus 10 is 490

5 0

2 years ago

The drawing shows a person (weight W = 588 N, L1 = 0.838 m, L2 = 0.398 m) doing push-ups. Find the normal force exerted by the f

zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

8 0

2 years ago

A block rests on a flat plate that executes vertical simple harmonic motion with a period of 0.74 s. What is the maximum amplitu

Mumz [18]

Answer:

maximum amplitude = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

$T=\dfrac{2\pi}{\omega}$

$0.74=\dfrac{2\pi}{\omega}$

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude = 0.13 m

8 0

2 years ago