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4 years ago

لماذا يتعذر احيانا سير الاشخاص على سطح حليدي افقي

Physics

1 answer:

Sergio [31]4 years ago

5 0

حسنا ، قد يكون أحد الأمثلة على ذلك أن السطح مبلل أو جليدي ، لذلك لا يوجد احتكاك كافٍ للمشي.

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During which phase of a four stroke engine are waste gases removed?

Fed [463]

At the phase of D. Exhaust Stroke, the waste gases are removed during a four stroke engine. It is one of the phases of the four stroke engine. The cylinder from the fuel ignited during the compression step and removed from the cylinder.

4 0

3 years ago

Plasmodium belong to which kingdom

musickatia [10]

Answer:

plasmodium belongs to kingdom protista

4 0

3 years ago

Two atoms with different mass number but the same atomic number are called A) elements. B) ions. C) isotopes. D) nuclei.

saw5 [17]

C isotopes is the correct answer. Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number.All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.

Hope this helps! :)

4 0

3 years ago

Read 2 more answers

During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.

Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

$Q = \frac{\Delta P \pi r^4}{8\eta l}$

Where:

$\eta_i =$ are the viscosities of the concrete before and after the increase

l = Length of the vessel

$r_1, R_2$ = Radio of the vessel before and after the increase

$\Delta P$ = Change in the pressure

$Q_{1,2} =$ The rates of flow before and after he increase

Our values are given as:

$Q_2 = 10Q_1 \rightarrow$ 10 times her resting rate

$\eta_2 = 0.95\eta_1$ 95% of its normal value

$\Delta P_2 = 1.5\Delta P_1$ Increase of 50%

Plugging known information to get

$Q_1 = \frac{\Delta P \pi r^4}{8\eta l}$

$Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4$

$r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}$

$r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}$

$r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}$

$r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}$

$r_2 = 1.586r_1$

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

7 0

4 years ago

Two objects that are not initially in thermal equilibrium are placed in close contact. After a while, the temperature of the cod

Dima020 [189]

Answer:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then the specific heats of both objects will be equal.

Explanation:

If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two objects of same mass are brought into contact, then their specific heat capacity is equal.

We can prove this by the equation of heat for the two bodies:

According to given condition,

$\Delta T_1=\Delta T_2$

$\frac{Q_1}{m_1.c_1} = \frac{Q_2}{m_2.c_2}$

when there is no heat loss from the system of two bodies then $Q_1=Q_2$

$\frac{1}{m.c_1} =\frac{1}{m.c_2}$

$\Rightarrow c_1=c_2$

Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.

The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:

when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.

the mass of colder object is half the mass of the hotter object while their specific heat is same.

3 0

3 years ago