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3 years ago

How Antman can breathe when his size is smaller then oxygen molecules?

1 answer:

8 0

Antman is fictional. However, in the case we were hypothetically assuming it was possible, he is the size of an ant which is not smaller than an oxygen molecule. When he goes into the quantum realm, he exists between the elemental particles (oxygen) and must use his mask to breathe.

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Will the velocity of the book change as it moves across the surface with NO friction? Explain your answer.

MakcuM [25]

No velocity will not be changed

Why?

According to Newtons 1st law the velocity of a moving object remains unchanged unless a external force affect that.

6 0

3 years ago

Read 2 more answers

If a 1.00 kg body has an acceleration of 2.44 m/s2 at 53Â° to the positive direction of the x axis, then what are (a) the x comp

Ilia_Sergeevich [38]

(a) Fx = 1.464 N

(b) Fy = 1.952 N

Given

Mass = 1kg

Acceleration = 2.44 m/s2

Angle with positive X axis = 53°

As we know

F = ma

By substituting value

F= 1×2.44 N

F= 2.44 N

(a) Component of force in X direction

Fx = F Cosθ

Fx = 2.44 Cos(53°)

Fx = 2.44 × 0.60 = 1.464 N

(b) Component of force in Y direction

Fy = F Sinθ

Fy = 2.44 Sin(53°) = 2.44 × 0.80 = 1.952 N

F(x, y) = 1.464 i + 1.952 j

Thus we got net force.

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6 0

2 years ago

On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass

Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

$\theta=53^{\circ}$

$\mu_k=0.16$

Speed,v=1.5m/s

Net force in x direction must be zero

$F_{net}=0$

$Fsin\theta-f=0$

$Fsin\theta=f$

Net force in y direction

$N-mg-Fcos\theta=0$

$N=mg+Fcos\theta$

$f=\mu_kN=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_k(mg+Fcos\theta)$

$Fsin\theta=\mu_kmg+\mu_kFcos\theta$

$Fsin\theta-\mu_kFcos\theta=\mu_kmg$

$F(sin\theta-\mu_kcos\theta)=\mu_kmg$

$F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}$

Power,P=Fv

$P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v$

Where $g=9.8m/s^2$

B.Substitute the values

$P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5$

$P=40.17W$

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4 years ago

Elliot jumps up and down on a pogo stick. He weighs 600.N, and his pogo stick has a spring with spring constant 1100N/m. What is

tia_tia [17]

From conservation of energy, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

The given weight of Elliot is 600 N

From conservation of energy, the total mechanical energy of Elliot must have been converted to elastic potential energy. Then, the elastic potential energy from the spring was later converted to maximum potential energy P.E of Elliot.

P.E = mgh

where mg = Weight = 600

To find the height Elliot will reach, substitute all necessary parameters into the equation above.

250 = 600h

Make h the subject of the formula

h = 250/600

h = 0.4167 meters

Therefore, the height he will reach when he has gravitational potential energy 250J is 0.42 meters approximately

Learn more about energy here: brainly.com/question/24116470

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3 years ago

Sue and Jenny kick a soccer ball at exactly the same time. Sue’s foot exerts a force of 75.9 N to the north. Jenny’s foot exerts

Lady_Fox [76]

Answer:

Fr^2 = 75.9N+105.8N=181.7

Fr = 181.7N.

6 0

3 years ago

How Antman can breathe when his size is smaller then oxygen molecules?​

How Antman can breathe when his size is smaller then oxygen molecules?