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coldgirl [10]
3 years ago
13

What are the blood components?

Physics
1 answer:
Alinara [238K]3 years ago
5 0
Plasma

Red blood cells

White blood cells

Platelets

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Plisssss you can help me pliss
Nesterboy [21]
For the first one it’s the guh sound
And the second one it’s the kuh sound
6 0
3 years ago
A 1.25 m long pendulum on Mars
Whitepunk [10]

Answer: 3.71

Explanation:

Just have trust bro

6 0
3 years ago
A cord of negligible mass runs around two massless, frictionless pulleys. A canister with mass m = 20 kg hangs from one pulley.
photoshop1234 [79]

(a) 196 N

The equation of the forces on the side of the cord where the force F is applied is:

F-T=0 (1)

where T is the tension in the cord.

On the other side of the cord, the equation of the forces on the canister is

T-mg = ma (2)

where

m = 20 kg is the mass of the canister

g=9.8 m/s^2 is the acceleration of gravity

a is the acceleration

From (1),

T=F

Substituting into (2),

F-mg = ma\\F=m(g+a)

We want the canister to move at constant speed, so

a = 0

And therefore:

F=mg=(20)(9.8)=196 N

b) 2.0 cm

The cord is inextensible, this means that the acceleration of its parts are the same. Therefore, the acceleration of the free end must be the same as the acceleration of the canister: and this means that the two parts also cover the same distance in the same time.

Therefore, the free end of the cord must be moved exactly the same as the canister, by 2.0 cm.

c) 3.92 J, the same

The work done by the tension in the cord is

W_T = T d

where

T is the tension

d = 2.0 cm = 0.02 m is the displacement

As we said in part (a), the tension in the cord is equal to the force applied to the free end:

T = F

So

T = 196 N

Therefore, the work done by the tension is

W=(196)(0.02)=3.92 J

And since the force applied (F) is the same, then the work done by you when pulling the cord is exactly the same.

(d) -3.92 J

The weight of the canister is

F_g = mg =(20 kg)(9.8 m/s^2)=196 N

However, the direction of the force of gravity is opposite to the displacement. Therefore, the work done by gravity is negative:

W_g = - F_g d

And substituting,

W_g=-(196)(0.02)=-3.92 J

(e) Zero

The net work done on the canister can be simply calculated by adding the work done by the tension in the cord and the weight of the canister:

W=W_T+W_g = 3.92 + (-3.92 ) = 0

This is in agreement with the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. In this situation, the canister is moving at constant speed, so its kinetic energy is not change: therefore,

\Delta K = 0 (change in kinetic energy = 0)

and so, the work done on it is also zero.

(f) The pulley system changes the direction of the force applied

This is a simple pulley system, which means  that the system does not multiply the force applied in input. In fact, the mechanical advantage of the system is

MA=\frac{F_{out}}{F_{in}}

where:

F_{out} is the output force, which is the weight of the canister

F_{in} is the force in input, which is F

So, the mechanical advantage is 1:

MA=\frac{196 N}{196 N}=1

From a point of view of energy, therefore, there is no advantage in this system.

However, the advantage offered by the pulley system concerns the direction of the force: in fact, it changes the direction of the applied force (which is F, downward) into the tension of the cord (which is upward on the canister).

6 0
3 years ago
(a) State and explain which of the arrangements would have the greater extension of spring(s). (b) Explain if there are any chan
swat32

Answer:
arrangement 2

Explanation:
arrangement 1's spring would broke idek

3 0
2 years ago
Read 2 more answers
Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.
Fudgin [204]

Answer:

g'_h=1.096\times 10^{-5}\ m.s^{-2}

Explanation:

We know that the gravity on the surface of the moon is,

  • g'=\frac{g}{6}
  • g'=1.63\ m.s^{-2}

<u>Gravity at a height h above the surface of the moon will be given as:</u>

g'_h=\frac{G.m}{(r+h)^2} ..........................(1)

where:

G = universal gravitational constant

m = mass of the moon

r = radius of moon

We have:

  • G=6.67\times 10^{-11}\ m^3.s^{-2}.kg^{-1}
  • m=7.35\times 10^{22}\ kg
  • r=1.74\times 10^6\ m
  • h=384.4\times 10^6\ m is the distance between the surface of the earth and the moon.

Now put the respective values in eq. (1)

g'_h=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(1.74\times 10^6+384.4\times 10^6)^2}

g'_h=1.096\times 10^{-5}\ m.s^{-2} is the gravity on the moon the earth-surface.

4 0
3 years ago
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