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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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The Palo Verde nuclear power generator of Arizona has three reactors that have a combined generating capacity of 3.937×109 W . H

den301095 [7]

Answer:

t = 2.68 x 10¹⁴ years

Explanation:

First we need to find the amount of energy that Sun produce in one day.

Energy = Power * Time

Energy of Sun in 1 day = (3.839 x 10²⁶ W)(1 day)(24 hr/1 day)(3600 s/ 1 hr)

Energy of Sun in 1 day = 3.32 x 10³¹ J

Now, the time required by the nuclear power generator, in years, will be:

Energy of power generator = Energy Sun in 1 day = 3.32 x 10³¹ J

3.32 x 10³¹ J = Power * Time

3.32 x 10³¹ J = (3.937 x 10⁹ W)(t years)(365 days/1 year)(24 hr/1 day)(3600 s/ 1 hr)

t = 3.32 x 10³¹ /1.24 x 10¹⁷

t = 2.68 x 10¹⁴ years

8 0

3 years ago

A student pulls horizontally on a block with a spring scale. The block reads 24.5 Newtons before the block starts to move. It re

Ahat [919]

Answer:

The coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

Explanation:

Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.

Now, since the velocity is constant so, there is no acceleration and thus, the net force is zero.

Let F = force applied and f = frictional force = μN = μW where μ = coefficient of friction and N = normal force. The normal force also equals the weight of the object W.

Now, since F - f = ma and a = 0 where a = acceleration and m = mass of block,

F - f = m(0) = 0

F - f = 0

F = f

Since the force applied equals the frictional force, we have that

F = μW and F = 23.7 N and W = 24.5 N

So, 23.7 N = μ(24.5 N)

μ = 23.7 N/24.5 N

μ = 0.97

Since μ = 0.97 < 1, the coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

7 0

2 years ago

A gray kangaroo can bound across level ground with each jump carrying it 8.7 from the takeoff point. Typically the kangaroo leav

oksano4ka [1.4K]

Answer:

a) The takeoff speed is 10 m/s.

b) The maximum height above the ground is 1.2 m.

Explanation:

The position of the kangaroo and its velocity at any given time "t" can be calculated by the following equations:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

v =(v0 · cos α, v0 · sin α + g · t)

Where:

r = position vector at time "t".

x0 = initial horizontal position.

v0 = initial velocity.

α = jumping angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity vector at time "t"

a) Please see the attached figure for a better understanding of the problem. In red is depicted the position vector at the final time (r final). The components of r final are known:

r final = (8.7 m, 0 m)

Then at final time:

8.7 m = x0 + v0 · t · cos α

0 m = y0 + v0 · t · sin α + 1/2 · g · t²

(notice in the figure that the origin of the frame of reference is located at the jumping point so that x0 and y0 = 0). Then:

8.7 m = v0 · t · cos α

Solving for "v0":

8.7 m /(t · cos α) = v0

Replacing v0 in the equation of the y-component, we can obtain the final time:

0 m = 8.7 m · tan 29° - 1/2 · 9.8 m/s² · t² (remember: sin α / cos α = tan α)

- 8.7 m · tan 29° / -4.9 m/s² = t²

t = 0.99 s

Now, we can calculate the initial speed:

8.7 m /t · cos α = v0

v0 = 8.7 m / (0.99 s · cos 29°)

v0 = 10 m/s

The takeoff speed is 10 m/s

b) When the kangaroo is at its maximum height, the velocity vector is horizontal (see figure). That means that the y-component of the velocity at that time is 0:

0 = v0 · sin α + g · t

Solving for "t":

-v0 · sin α / g = t

t = - 10 m/s · sin 29° / 9.8 m/s²

t = 0.49 s

Notice that we could have halved the final time (0.99 s, calculated above) to obtain the time at which the kangaroo is at its maximum height. That´s because the trajectory is parabolic.

Now, let´s find the height of the kangaroo at that time:

y = y0 + v0 · t · sin α + 1/2 · g · t²

y = 10 m/s · 0.49 s · sin 29° - 1/2 · 9.8 m/s² · (0.49 s)²

y = 1.2 m

The maximum height above the ground is 1.2 m.

4 0

3 years ago

Effect of impurities and temperature to surface tension

oksano4ka [1.4K]

In general, surface tension decreases when temperature increases because cohesive forces decrease with an increase of molecular thermal activity. The influence of the surrounding environment is due to the adhesive action of liquid molecules that they have at the interface.

8 0

3 years ago

Equation of motion description is v=20+2t.How big is the initial speed,acceleration?

trasher [3.6K]

Answer:

the initial velocity is 20 m/s and the acceleration is 2 m/s²

Explanation:

Given equation of motion, v = 20 + 2t

If V represents the final velocity of the object, then the initial velocity and acceleration of the object is calculated as follows;

From first kinematic equation;

v = u + at

where;

v is the final velocity

u is the initial velocity

a is the acceleration

t is time of motion

If we compare (v = u + at) to (v = 20 + 2t)

then, u = 20 and

a = 2

Therefore, the initial velocity is 20 m/s and the acceleration is 2 m/s²

4 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?