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4 years ago

Can someone please tell me...... What does it mean if you through up thew your nose and mouth at the same time!?

2 answers:

3 0

if you're worried about youre health you're completely fine. (other than the fact you are throwing up, if your sick please see a doctor) but if the things you're worried about is that it can out of your nose, dont worry it happenes to a lot of people. Your nose and mouth are both connected to your throat and that sudden rush of fluids from your throat can confuse your body, Its like when you laugh while eating and then you accidently get food in your nose. Don't worry this isn't anything that is particulary bad but i will say that the vomiting is not good

belka [17]4 years ago

3 0

Answer:

It does not mean anything bad it is very common and has happened to a lot of people

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Capacitors C1 = 6.45 µF and C2 = 2.50 µF are charged as a parallel combination across a 250 V battery. The capacitors are discon

denpristay [2]

Answer:

For C1, Q = 1.6125×10⁻³ C

For C2, Q = 6.25×10⁻⁴ C

Explanation:

Note: Since the capacitors are connected in parallel, The voltage across each of them is equal.

From the question,

Q = CV........................ Equation 1

Where Q = Charge on the capacitor, V = Voltage across the capacitor, C = Capacitance of the capacitor.

For the first capacitor,

Q = C1V............. Equation 2

Where C1 = 6.45 μF= 6.45×10⁻⁶ F, V = 250 V

Substitute into equation 2

Q = (6.45×10⁻⁶ )(250)

Q = 1.6125×10⁻³ C.

For the the second capacitor,

Q = C2V............. Equation 3

Given: C2 = 2.50 μF = 2.5×10⁻⁶ F, V = 250 V

Q = (2.5×10⁻⁶ )(250)

Q = 6.25×10⁻⁴ C

4 0

3 years ago

an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni

stiv31 [10]

The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N

3 0

3 years ago

A thin uniform rod (mass = 0.53 kg) swings about an axis that passes through one end of the rod and is perpendicular to the plan

gladu [14]

Answer:

(a) L = 0·73 m

(b) 4·39 × $10^{-3}$ J

Explanation:

(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment

Let L m be the length of the rod and β be the angle between the rod and the vertical

Let α be the angular acceleration of the rod

As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ

Assuming that the value of amplitude of this oscillation to be small

As torque = moment of inertia × angular acceleration

0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)

<h3>For small oscillations, α = ω² × β</h3>

After substituting the value of α and solving we get

ω = √((3 × g) ÷ (2 × L))

Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))

∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4

Substituting the value of g as 9·8 m/s² and solving we get

L = 0·73 m

(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving

∴ Taking the reference for finding the potential energy as the lowest point

<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>

Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2

∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get

Maximum potential energy = 4·39 × $10^{-3}$ J

∴ Maximum kinetic energy = 4·39 × $10^{-3}$ J

4 0

3 years ago

A water tank has a mass of 3.64 kg when empty and 51.8 kg when filled to a certain level. What is the mass of the water in the t

mrs_skeptik [129]

Subtract the total mass of the tank and water by the mass of the empty tank. That will yield the mass of the water.

51.8 - 3.64 = 48.16.

Your answer is 48.16 kg

3 0

3 years ago

A wagon weighs 1,800 N and is pulled by a horse at a speed of 0.40 meters/ second. What is the power of this horse

Alona [7]

Answer:

Power=720[watt]

Explanation:

We need to remember the definition of mechanical work which is equal to the product of the force applied by the distance traveled.

In this problem, we have to find the power which is defined as the work divided into the time in which such work is performed. This way if we have the displacement and the time, this will be the speed with which this work is done.

$Power= W/T\\T=time [s]\\W=work [J]\\Power = F*V\\where\\V=velocity [m/s]\\Power=1800 * 0.4 = 720 [watt]$

4 0

3 years ago

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