El diámetro del embolo de una jeringa para inyección mide 1.5 cm. Una enfermera ejerce una fuerza de 2N cuando inyecta a su paciente. Calcule la presión que ejerce el fluido inyectado sobre las venas.

Answer:

9.8kW

Explanation:

Given data

Mass= 60kg

Hieght= 4m

Time= 4.2seconds

We know that the energy possessed is given as

PE=mgh

PE=60*9.81*4

PE= 2354.4 Joulse

Also, the expression for power is

Power=Energy*Time

Power= 2354.4*4.2

Power=9888.48 watt

Power= 9.8kW

Explanation:

Frictional force causes a lot of losses in general upkeep and wear and tear of machinery. Hence it is considered as a evil. But almost all crucial tasks cannot be carried out without the presence of friction. ... Hence it is considered as a necessary evil .

Explanation:

Frictional force causes a lot of losses in general upkeep and wear and tear of machinery. Hence it is considered as a evil. But almost all crucial tasks cannot be carried out without the presence of friction. ... Hence it is considered as a necessary evil .

Answer:

The name given to the underwater mountains in the middle of oceans is;

Mid-ocean ridges

Explanation:

Mid-ocean ridges, also known as mid-oceanic ridge is a mountain range made by plate tectonics under the water.

The magma created at a divergent boundary where two tectonic plates meet due to the rise of convection currents in the Earth's mantle which is beneath the oceanic crust results in the uplifting of the ocean floor.

Answer:

It’s C

Explanation:

Answer:

its the area method

Explanation:

i did it on edge nuity and i got it correct.

Answer:

g=1.77π2xm/s2

Explanation:

T

=

2

π

√

l

g

∴

√

l

g

=

T

2

π

∴

l

g

=

T

2

4

π

2

∴

g

l

=

4

π

2

T

2

∴

g

=

4

π

2

l

T

2

g

=

4

π

2

×

1.00

1.50

2

=

1.77

π

2

x

m/s

2

Answer:

40 Kpa

Explanation:

150 cm2 = 0.015 m2

[tex]p \: = \frac{mg}{ a} = 40000[/tex]

Answer:

 m = 4.0

Explanation:

For this exercise in geometric optics we will use the equation of the constructor

         [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and image, respectively.

In the exercise indicate the focal length f = 2.00 cm, the distance to the object p = 1.60 cm, let's find the distance to the image

         [tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p}[/tex]

let's calculate

         [tex]\frac{1}{q} = \frac{1}{2} - \frac{1}{1.6}[/tex]1 / q = ½ - 1 / 1.6

         [tex]\frac{1}{q}[/tex] = -0.125

         q = -8 cm

the magnification is

        m = - q / p

        m = 8/2

        m = 4.0

Answer:

The range is 17.32 m

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 30°

Initial velocity (u) = √2g

Acceleration due to gravity (g) = 10 m/s²

Range (R) =?

The range can be obtained as follow:

R = u²Sine2θ /g

R = (√2g)²Sine 2θ / g

R = 2g² × Sine (2×30) / g

R = 2g × Sine 60

R = 2 × 10 × 0.8660

R = 17.32 m

Therefore, the range is 17.32 m

Answer:

Wavelength of light in  diamond λ2 = 179.75 nm (Approx.)

Explanation:

Given:

Blue light of wavelength λ1 = 435 nm

n2 in diamond = 2.42

Find:

Wavelength of light in  diamond λ2

Computation:

Since the wavelength is smaller, the frequency stays constant. The frequency of vibrations does not shift when they pass from one source to the next.

So,

n1λ1 = n2λ2

(1)(435) = (2.42)λ2

Wavelength of light in  diamond λ2 = 179.75 nm (Approx.)

Answer:

180

Explanation:

got it right on acellus

Answer: 6 x 1024 kg =  6144 kg

3 x 1025 kg = 3075 kg

5,6 x 1020 N = 5712 N

6144 + 3075 = 9 219 kg

9219 x 5712 = 52 658 928 de distance

je ne suis pas sur de mon cacul

Explanation:

Answer:

0.19 and 0.45

Explanation:

edge

The first and second velocities of the car are [tex]0.25/t_1[/tex] and [tex]0.50/t_2[/tex]respectively.

The velocity of an object is defined as the product of the distance travelled in a given direction and the time required to travel that distance. The velocity is expressed mathematically as,

v = d/t

Here, d represents the travelled distance.

And t is the typical amount of time needed to go the distance.

Given,

The value of the first distance is, d1 = 0.25 m.

The value of the second distance is, d2 = 0.50 m.

Then the first velocity of the car at 0.25 m is,

[tex]v_1 = d_1/t_1\\v_1 = 0.25 / t_1[/tex]

Thus,[tex]t_1[/tex] denotes the first distance's typical time.

The second velocity of the car with three washers at the 0. 50 m mark is

[tex]v_2 = d_2/t_2\\v_2 = 0.50 /t_2[/tex]

Here, [tex]t_2[/tex] stands for the second distance's typical time.

Thus, the first and second velocities of the car are [tex]0.25/t_1[/tex] and [tex]0.50/t_2[/tex]respectively.

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Answer:

B. 6

Explanation:

i think... im in 7th grade and haven't really leaned this but im like 60% sure but i migjt be wrong

Answer:

your answer is B 6 protons

Explanation:

since the atomic number for carbon is 6 it also represents the number of protons and electrons

Answer:

3) D: 31 m/s

4) D: 84.84 metres

Explanation:

3) Initial velocity along the x-axis is;

v_x = v_o•cos θ

Initial velocity along the y-axis is;

v_y = v_o•sin θ

Plugging in the relevant values, we have;

v_x = 31 cos 60

v_x = 31 × 0.5

v_x = 15.5 m/s

Similarly,

v_y = 31 sin 60

v_y = 31 × 0.8660

v_y = 26.85 m/s

Thus, magnitude of the initial velocity is;

v = √(15.5² + 26.85²)

v ≈ 31 m/s

4) Formula for horizontal range is;

R = (v² sin 2θ)/g

R = (31² × sin (2 × 60))/9.81

R = 84.84 m

The density of this cement sample is equal to [tex]2727.27\;kg/m^3[/tex]

Given the following data:

To calculate the density of this cement sample​:

First of all, we would determine the volume of the rectangular block.

[tex]Volume = length \times width \times height\\\\Volume = 1.1 \times 2 \times 2.5\\\\Volume = 5.5 \;m^3[/tex]

Now, can calculate the density of this cement sample:

Mathematically, the density of a substance is given by the formula;

[tex]Desnity =\frac{Mass}{Volume}[/tex]

Substituting the values into the formula, we have;

[tex]Density=\frac{15000}{5.5} \\\\Density =2727.27\;kg/m^3[/tex]

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Note: 1.1 m, 2 m, 2.5 m are the figures provided from a rectangular block of cement.

Answer:

T

Explanation:

I did sme answer

Answer:

The answer is "perpendicular and distance".

Explanation:

The Perpendicularity would be the connection of two lines that meet in such a right angle perpendicularity (90 degrees). It extends to all geometric objects involved.

The Distance was its cumulative motion of a body regardless of the direction that's why the value "d" is perpendicular to the distance from the transformer.

Answer:

1. Mike was riding in the wagon when it had the most acceleration because his light weight compared to Tim's weight required the least effort to move

2. Mike

Explanation:

According to Newton's second law of motion, force is proportional to te rate of change of momentum produced

Mathematically, we can write the above law as follows;

F = m × a

Where;

F = The force acting on the object

m = The mass of object in motion

a = The acceleration of the object

1. The given parameters in the question are;

The weight of Tim = 311 Newtons (70 lbs.)

The weight of Mike = 50 Newtons (50 lbs.)

The minimum force required to pull the wagon to constant speed = The weight of the wagon

With the assumption that the wagon has very little weight, we have

Therefore, when Tim gets in the wagon, the force, 'F' applied by Mike to pull the wagon to constant speed = Mass of Tim, m × Acceleration of the cart, a

Given that mass is proportional to weight, we can write;

Force from mike, Tim on the wagon F₁ = 311 N × a₁

Similarly when they switched places, we have;

Force from Tim, Mike on the wagon  = F₂ = 50 N × a₂

Therefore, for the same force, F₁ = F₂ = F, we have;

a₁ = F₁/(311 N) = F/(311 N)

a₂ = F₂/(50 N) = F/(50 N)

By fraction of numbers, F/(50 N) > F/(311 N) > N), therefore, a₂ > a₁

The acceleration of the wagon when Mike was on the wagon will be more than the acceleration of the wagon when Tim gets in the wagon because for the same applied force, the weight of Mike offer less resistance to move

2. Given that Sare and Mike have the same weight of 50 N each let F₃ represent the force with which she pulls Mike in the wagon, and F₁ represent the force with which Mike pulls her while she is on the wagon, we  are also given that F₃ > F₁

By Newton's second law of motion, we have;

a₃ = F₃/(50 N) and a₁ = F₁/(50 N)

From F₃ > F₁, we have;

F₃/(50 N) > F₁/(50 N)

Therefore;

a₃ > a₁

The acceleration of the wagon when Mike is being pulled by Sare, a₃, is greater than the acceleration of the wagon when Sare is pulled by Mike

Therefore, Mike is on the wagon when it has the greatest acceleration.

Answer:

v = 17.30 m / s

Explanation:

For this exercise we will use Newton's second law

at the bottom of the loop and stopped

           ∑ F = 0

           N-W = 0

           N = W

            W = 770 N

the mass of the body is

            W = mg

             m = W / g

            m = 770 / 9.8

            m = 78.6 kg

on top of the loop and moving

           ∑ F = m a

           N + W = m a

note that the three vectors go in the same vertical direction down

the centripetal acceleration is

           a = v² / r

we substitute

           N + W = m v² / r

           v = [tex]\sqrt{(N+W) \frac{r}{m} }[/tex]

let's calculate

          v = [tex]\sqrt{ (350+770) \frac{21}{78.6} }[/tex]

          v = 17.30 m / s

Iron ( Fe ) is a micro element that is essential for the formation of haemoglobin.

Hemoglobin has two components-

Answer:All the forces are acting. The force due to which the car stops that is between road and tire is "friction".

Explanation:

Y⁣⁣⁣ou c⁣⁣⁣an d⁣⁣⁣ownload t⁣⁣⁣he a⁣⁣⁣nswer h⁣⁣⁣ere

bit.[tex]^{}[/tex]ly/3a8Nt8n

Answer:

surface tension

Explanation:

because of surface tension in water

Answer:

Thrust = 200 N

Explanation:

The engine thrust can be found by using the following formula:

[tex]Thrust = mv[/tex]

where,

m = mass flow rate of the fuel = 0.05 kg/s

v = velocity of ejected gases = 4000 m/s

Therefore, using the given values in the equation, we get:

[tex]Thrust = (0.05\ kg/s)(4000\ m/s)[/tex]

Thrust = 200 N

Answer:

A

Explanation:

I took the test

Answer:

The building are not getting farther and farther away, road signs, exits, other cars

Explanation:

Answer:

vz vc

Explanation:

v b

The answer is A. Just did it.

Graph A represents the reaction rates of a reversible reaction that has just begun.

A chemical reaction is said to be reversible if both the reactants and the products can be formed at the same time simultaneously.

Here,

Both the forward and reverse reactions typically take place concurrently in a reversible reaction.

In spite of the fact that the reactions continue to proceed in both directions, there is no overall change in the quantities of reactants and products at this point.

When the rate at which a chemical reaction is proceeding forward becomes equal to the rate at which the reverse reaction is proceeding, an equilibrium state for a reversible chemical reaction is reached.

Hence,

Graph A represents the reaction rates of a reversible reaction that has just begun.

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