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3 years ago

Systemic circulation requires help from which of the following bones, lungs, arteries and veins, or muscles

Physics

1 answer:

ahrayia [7]3 years ago

4 0

Systemic circulation is between the heart and the rest of the body. Therefore arteries and veins are needed.

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The probability which is based on the observations of an experiment is called: A. Theoretical Probability B. Axiomatic Probabili

FromTheMoon [43]

Answer:

C. Experimental Probability

Explanation:

The empirical (or experimental) probability means the event that arise and depend how the event arise when the data is collected from an experiment in a more no of trials. It would be depend upon the direct observation. Here each and every observation in an experiment is known as trial

So the probability that depend upon the experiment observation is known as the experimental probability

Hence, the option c is correct

8 0

3 years ago

A closed box is filled with dry ice at a temperature of -94.7°C, while the outside temperature is 26.8°C. The box is cubical, me

Nikitich [7]

Answer:

Explanation:

3.64 x 10⁶ J passes through 6 walls

heat energy passing through 1 wall = 0.606 x 10⁶ J

Surface Area of 1 wall A = .285² = 0.081225 m²

Temperature Difference = T₁ - T₂ = 26.8 + 94.7 = 121.5

Thickness of wall d = 3.75 x 10⁻² m

Rate of heat flow per second R = $\frac{0.606 \times10^6}{24\times60\times60}$

=7.01 J per s.

Formula for rate of heat flow

R = $\frac{KA(T_1-T_2)}{d}$

Where K is thermal conductivity.

7.01 = $\frac{K\times121.5\times.081225}{3.75\times10^{-2}}$

K = 2.66 X 10⁻² W m⁻¹s⁻¹

8 0

3 years ago

Which of the following is a true statement for a child's toy spinning in a circle at constant speed?

exis [7]

Answer:

The answer is C.

Explanation:

4 0

3 years ago

Read 2 more answers

A sphere of mass m" = 2 kg travels with a velocity of magnitude υ") = 8 m/s toward a sphere of mass m- = 3 kg initially at rest,

aleksklad [387]

a) 6.4 m/s

b) 2.1 m

c) $61.6^{\circ}$

d) 14.0 N

e) 4.6 m/s

f) 37.9 N

Explanation:

Since the system is isolated (no external forces on it), the total momentum of the system is conserved, so we can write:

$p_i = p_f\\m_1 u_1 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 2 kg$ is the mass of the 1st sphere

$m_2 = 3kg$ is the mass of the 2nd sphere

$u_1 = 8 m/s$ is the initial velocity of the 1st sphere

$v_1$ is the final velocity of the 1st sphere

$v_2$ is the final velocity of the 2nd sphere

Since the collision is elastic, the total kinetic energy is also conserved:

$E_i=E_k\\\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

Combining the two equations together, we can find the final velocity of the 2nd sphere:

$v_2=\frac{2m_1}{m_1+m_2}u_1=\frac{2(2)}{2+3}(8)=6.4 m/s$

Now we analyze the 2nd sphere from the moment it starts its motion till the moment it reaches the maximum height.

Since its total mechanical energy is conserved, its initial kinetic energy is entirely converted into gravitational potential energy at the highest point.

So we can write:

$KE_i = PE_f$

$\frac{1}{2}mv^2 = mgh$

where

m = 3 kg is the mass of the sphere

v = 6.4 m/s is the initial speed of the sphere

$g=9.8 m/s^2$ is the acceleration due to gravity

h is the maximum height reached

Solving for h, we find

$h=\frac{v^2}{2g}=\frac{(6.4)^2}{2(9.8)}=2.1 m$

Here the 2nd sphere is tied to a rope of length

L = 4 m

We know that the maximum height reached by the sphere in its motion is

h = 2.1 m

Calling $\theta$ the angle that the rope makes with the vertical, we can write

$h = L-Lcos \theta$

Which can be rewritten as

$h=L(1-cos \theta)$

Solving for $\theta$ , we can find the angle between the rope and the vertical:

$cos \theta = 1-\frac{h}{L}=1-\frac{2.1}{4}=0.475\\\theta=cos^{-1}(0.475)=61.6^{\circ}$

The motion of the sphere is part of a circular motion. The forces acting along the centripetal direction are:

- The tension in the rope, T, inward

- The component of the weight along the radial direction, $mg cos \theta$ , outward

Their resultant must be equal to the centripetal force, so we can write:

$T-mg cos \theta = m\frac{v^2}{r}$

where r = L (the radius of the circle is the length of the rope).

However, when the sphere is at the highest point, it is at rest, so

v = 0

Therefore we have

$T-mg cos \theta=0$

So we can find the tension:

$T=mg cos \theta=(3)(9.8)(cos 61.6^{\circ})=14.0 N$

We can solve this part by applying again the law of conservation of energy.

In fact, when the sphere is at a height of h = 1 m, it has both kinetic and potential energy. So we can write:

$KE_i = KE_f + PE_f\\\frac{1}{2}mv^2 = \frac{1}{2}mv'^2 + mgh'$

where:

$KE_i$ is the initial kinetic energy

$KE_f$ is the kinetic energy at 1 m

$PE_f$ is the final potential energy

v = 6.4 m/s is the speed at the bottom

v' is the speed at a height of 1 m

h' = 1 m is the height

m = 3 kg is the mass of the sphere

And solving for v', we find:

$v'=\sqrt{v^2-2gh'}=\sqrt{6.4^2-2(9.8)(1)}=4.6 m/s$

Again, since the sphere is in circular motion, the equation of the forces along the radial direction is

$T-mg cos \theta = m\frac{v^2}{r}$

where

T is the tension in the string

$mg cos \theta$ is the component of the weight in the radial direction

$m\frac{v^2}{r}$ is the centripetal force

In this situation we have

v = 4.6 m/s is the speed of the sphere

$cos \theta$ can be rewritten as (see part c)

$cos \theta = 1-\frac{h'}{L}$

where in this case,

h' = 1 m

L = 4 m

And $r=L=4 m$ is the radius of the circle

Substituting and solving for T, we find:

$T=mg cos \theta + m\frac{v^2}{r}=mg(1-\frac{h'}{L})+m\frac{v^2}{L}=\\=(3)(9.8)(1-\frac{1}{4})+(3)\frac{4.6^2}{4}=37.9 N$

4 0

3 years ago

When an object is fully converted into energy the amount of energy liberated is

EleoNora [17]

Answer:

Mass, m = 4 kg

Explanation:

<u>Given the following data;</u>

Energy = 3.6 * 10^17 Joules

We know that the speed of light is equal to 3 * 10⁸ m/s.

To find the mass of the substance;

The theory of special relativity by Albert Einstein gave birth to one of the most famous equation in science.

The equation illustrates, energy equals mass multiplied by the square of the speed of light.

Mathematically, the theory of special relativity is given by the formula;

$E = mc^{2}$

Where;

E is the energy possessed by a substance.
m is the mass.
c is the speed of light.

Substituting into the formula, we have;

$3.6 * 10^{17} = m * 300000000^{2}$

$3.6 * 10^{17} = m * 9*10^{16}$

$m = \frac {3.6 * 10^{17}}{9*10^{16}}$

Mass, m = 4 kg

8 0

3 years ago