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hoa [83]
3 years ago
12

Which rock type can be formed by adding heat and pressure to any of the three types of rock?

Physics
1 answer:
vaieri [72.5K]3 years ago
7 0
The answer is to u question is: C
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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
(a) A hole of 1 cm in diameter is drilled in a plate of steel at 20°C. What happens to the diameter of the hole as the steel is
Anestetic [448]

When heat is applied to a substance/material or steel in our case, a series of changes will occur, one of which is Expansion.

Hence the hole will expand from 1cm to a diameter that is greater.

Thermal expansion refers to the expansion or contraction of the dimensions of the solid, liquid or gas when their temperature is changed.

<h3>Types of Expansion</h3>
  • Linear expansion
  • Areal expansion and
  • Volumetric volume.

Learn more about expansion here:

brainly.com/question/1166774

8 0
3 years ago
How does the mass of an object affect the gravitational forces experienced by another object?.
Leviafan [203]

Explanation:

F=(Gmm)/(r^2)

larger the mass stronger the force. An object with a large mass will produce a large force. A small mass will produce a small force.

3 0
2 years ago
What are two factors that determine<br> object's gravitational potential energy?<br> an
Artyom0805 [142]

Answer:

The factors that affect an object's gravitational potential energy are its height relative to some reference point, its mass, and the strength of the gravitational field it is in.

6 0
2 years ago
A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a
Fittoniya [83]

Answer:

F_T=60132.52N

P=15814852.76W

Explanation:

From the question we are told that

Velocity of aircraft  V=263m/s

Engine air intake rate \triangle M_a=85.9kg/s

Fuel burn rate  \triangle M_f =3.92kg/s

Velocity of exhaust gas V_e =921m/s

Generally the Mass change rate of Rocket is mathematically given by

 \triangle M = \triangle M_a+\triangle M_f

 \triangle M= 85.9+3.92

 \triangle M=89.82kg/s

Generally the Trust of the rocket is given mathematically by

 F_T=(\triangle M *V_e)-(dM_a/dt)*(V)

 F_T=(89.82 *921)-(85.9)*(263)

 F_T=60132.52N

Generally the Rocket's delivered power is mathematically given by

Delivered power P

 P=V*F_T

 P=263*60132.52N

 P=15814852.76W.

8 0
3 years ago
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