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arsen [322]
3 years ago
6

A white blood cell has a diameter of approximately 12 micrometers, or 0.012 um. A model represents its diameter as 24 um. What i

s the ratio of model size to actual size?
Physics
1 answer:
GuDViN [60]3 years ago
8 0

Answer:

2:1

Explanation:

It is given that,

The diameter of a white blood cell is 12 micrometers

The diameter of white blood cell according to a model is 24 micrometers.

We need to find the ratio of model size to actual size. It can be done as follows :

R=\dfrac{\text{model size}}{\text{actual size}}\\\\R=\dfrac{24\ \mu m}{12\ \mu m}\\\\R=\dfrac{2}{1}

Hence, the ratio of model size to actual size is 2:1.

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The average speed of a nitrogen molecule in air is about 6.70×102 m/s, and its mass is 4.68×10-26 kg.
Otrada [13]

Answer:

a)   a = 3.06 10¹⁵ m / s , b)    F= 1.43  10⁻¹⁰ N, c)    F_total = 14.32 10⁻²⁶ N

Explanation:

This exercise will average solve using the moment relationship.

a ) let's use the relationship between momentum and momentum

          I = ∫ F dt = Δp

          F t = m v_{f} - m v₀

          F = m (v_{f} -v₀o) / t

 in the exercise indicates that the speed module is the same, but in the opposite direction

          F = m (-2v) / t

if we use Newton's second law

          F = m a

we substitute

            - 2 mv / t = m a

            a = - 2 v / t

let's calculate

            a = - 2 4.59 10²/3 10⁻¹³

            a = 3.06 10¹⁵ m / s

b)      F= m a

        F= 4.68 10⁻²⁶ 3.06 10¹⁵

        F= 1.43  10⁻¹⁰ N

c) if we hit the wall for 1015 each exerts a force F

            F_total = n F

            F_total = n m a

            F_total = 10¹⁵  4.68 10⁻²⁶ 3.06 10¹⁵

            F_total = 14.32 10⁻²⁶ N

8 0
3 years ago
After rubbing a balloon on your shirt, your hair sticks up when the balloon is near your head. Explain how tiny particles on you
skad [1K]
Almost right. protons are positive and electrons are negative. so when you run the balloon on your hair, electrons are transferred between them (i’m not sure which direction) and now one is positively charged as it lost negative particles and one is negative as it gained negative particles (electrons). opposite charges attract
3 0
2 years ago
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pol
kozerog [31]

Answer:

F=133N

Explanation:

From the question we are told that:

Length l=3.0m

Mass m=24kg

Distance from Tip d=35cm

Generally, the equation for Torque Balance is mathematically given by

mg(l/2)=F(l-d)

2*9.81(3/2)=F(3-35*10^-2)

Therefore

F=133N

8 0
3 years ago
The circumference of an orbit for a toy on a string is 18 m and the centripetal force is 12 N. Does the centripetal force do any
Snezhnost [94]

Answer:

Work done is 0.

Explanation:

Given that,

The circumference of an orbit for a toy on a string is 18 m, r = 18 m

Centripetal force, F = 12 N

In the circular path, the centripetal force is always perpendicular to the motion of the object. Thus it makes an angle of 90 degrees with the force and displacement. Hence, we can say that the centripetal force does not do any work on the toy when it follows its orbit for one cycle.

5 0
3 years ago
Read 2 more answers
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
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