Is quartzite a mineral, a rock, or neither?

The two forces acting on a skydiver are __________ and air _____________.

Stolb23 [73]

Answer:

Gravity and

Air resistance

Explanation:

The two forces acting on a skydiver are gravitational force and air resistance.

Gravitational force is a force that tends to pull all massive bodies towards the center of the earth. It works on all bodies that has mass. The larger or bigger the mass, the more the pull of gravity on the body.

Air resistance is the drag of air on a body as it passes to it. It is resisting force.

When a sky diver jumps out of a plane, he/she encounters both gravity and air resistance.
It soon balances both force and attain terminal velocity.
Air resistance is a frictional force that opposes motion.
This frictional force pushes in the opposite direction of motion
Motion direction is downward due to the celerity caused by gravity.

3 0

3 years ago

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Two drums of the same size and same height are taken.

Gelneren [198K]

Answer:

i) The pressure acting on the base of B will be half the pressure acting on the base of A

ii) The pressure acting on the base of B will be the same as the pressure acting on the base of A

iii) The pressure on the base of drum A will be slightly less than the pressure on the base of drum B

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

$h_{max}$ = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If A is completely filled, we have $h_A$ = $h_{max}$

Therefore, $P_A$ = $h_{max}$ × $\rho_{liquid}$ ×g

If B is half filled, we have, $h_B$ = (1/2)· $h_{max}$

$P_B$ = (1/2) × $h_{max}$ × $\rho_{liquid}$ ×g

Therefore, $P_B$ = (1/2) × $P_A$

The pressure acting on the base of B will be half the pressure acting on the base of A

ii) If both A and B are each filled with water (the same liquid), then the pressure on their bases will be $P_A$ = $h_{max}$ × $\rho_{water}$ ×g = $P_B$ , the same, given that the acceleration due to gravity, g, is constant and the same in Nepal and India

iii) If A is filled with water, and B is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

$P_A$ = $h_{max}$ × $\rho_{water}$ ×g < $P_B$ =

The pressure on the base of drum A will be less than the pressure on the base of drum B.

3 0

3 years ago

How are magnetic poles and electrical charges similar?

exis [7]

I believe the answer is a

3 0

2 years ago

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A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str

ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

m = m

=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0

3 years ago

5.54 Two kilograms of air within a piston–cylinder assembly WP execute a Carnot power cycle with maximum and minimum temper atur

Step2247 [10]

Answer:

A) 60%

B) p2 = 1237.2 kPa

v2 = 0.348 m^3

C) w1-2 = w3-4 = 1615.5 kJ

Q2-3 = 60 kJ

Explanation:

A) calculate thermal efficiency

Л = 1 - $\frac{Tl}{Th}$

where Tl = 300 k

Th = 750 k

hence thermal efficiency ( Л ) = [1 - ( 300 / 750 )] * 100 = 60%

B) calculate the pressure and volume at the beginning of the isothermal expansion

calculate pressure ( P2 ) :

= P3v3 = mRT3 ----- (1)

v3 = 0.4m , mR = 2* 0.287, T3 = 750

hence P3 = 1076.25

next equation to determine P2

Qex = p3v3 ln( p2/p3 )

60 = 1076.25 * 0.4 ln(p2/p3)

hence ; P2 = 1237.2 kpa

calculate volume ( V2 )

p2v2 = p3v3

v2 = p3v3 / p2

= (1076.25 * 0.4 ) / 1237.2

= 0.348 m^3

C) calculate the work and heat transfer for each four processes

work :

W1-2 = mCv( T2 - T1 )

= 2*0.718 ( 750 - 300 ) = 1615.5 kJ

W3-4 = 1615.5 kJ

heat transfer

Q2-3 = W2-3 = 60KJ

Q3-4 = 0

D ) sketch of the cycle on p-V coordinates

attached below

6 0

3 years ago