Magnetic fields go from the north poles of magnets to the south poles of magnets. true or false

Two friends are playing a version of proton golf where the hole is marked by a single proton. The first friend reads his meter,

yarga [219]

Answer:

the electric field strength on the second one is 2.67 N/C.

Explanation:

the electric fiel on the first one is:

E1 = k×q/(r^2)

r^2 = k×q/(E1)

= (9×10^9)×(q)/(24.0)

= 375000000q

then the electric field on the second one is:

E2 = k×q/(R^2)

we know that R = 3r

R^2 = 9×r^2

E2 = k×q/(9×r^2)

= k×q/(9×375000000q)

= k/(9×375000000)

= (9×10^9)/(9×375000000)

= 2.67 N/C

Therefore, the electric field strength on the second one is 2.67 N/C.

5 0

2 years ago

An entertainer pulls a table cloth off a table leaving behind the plates and sliverware undisturbed is an example of

umka2103 [35]

Answer:

D. Newton's first law

Explanation:

Newton's first law of inertia says that an object will remain how it is, unless affected by an outside force. In this case, the plates want to remain stationary(not moving). Therefore, if you pull the table cloth fast enough, the force of friction produced will be small enough so that the Inertia of the plates will overcome the force of friction.

5 0

2 years ago

The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called lase

emmainna [20.7K]

The answer is photocoagulation.

The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called laser photocoagulation.

What is photocoagulation?

A minimally invasive method used to treat numerous retinal illnesses is photocoagulation of the retina, also known as retinal laser photocoagulation. The retina may expand due to aberrant leaky blood vessels developing across it in a number of disorders. Laser photocoagulation uses thermal energy above 65 °C to burn the retinal tissue by creating thermal burns. This can prevent the retina from being damaged by the bleeding blood vessels. In addition to causing fibrosis, laser photocoagulation can also seal retinal tears. Laser photocoagulation is typically unable to recover already lost vision in cases of retinal disease, but it can slow the progression of the condition, lower the chance of further vision loss, and preserve residual vision. The likelihood of problems following the operation is quite minimal.

To learn more about photocoagulation click on the link below:

brainly.com/question/16016898

#SPJ4

6 0

6 months ago

Read 2 more answers

Suppose that 4.4 moles of a monatomic ideal gas (atomic mass = 7.9 × 10-27 kg) are heated from 300 K to 500 K at a constant volu

timurjin [86]

Answer:

1) ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2) W₁ = 0 J

3) P = 41.66 x 10³ Pa = 41.66 KPa

4) v = 1618.72 m/s

5) ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

6) W₂ = - 7.33 KJ

Explanation:

1)

The heat transfer for a constant volume process is given by the formula:

ΔQ₁ = ΔU = n Cv ΔT

where,

ΔQ₁ = Heat transfer during constant volume process

ΔU = Change in internal energy of gas

n = No. of moles = 4.4 mol

Cv = Molar Specific Heat at Constant Volume = 12.47 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 500 k - 300 k = 200 k

Therefore,

ΔQ₁ = (4.4 mol)(12.47 J/mol.k)(200 k)

ΔQ₁ = 10.97 x 10³ J = 10.97 KJ

2)

Since, work done by gas is given as:

W₁ = PΔV

where,

ΔV = 0, due to constant volume

Therefore,

W₁ = 0 J

4)

The average kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

but, K.E is also given by:

K.E = (1/2)mv²

Comparing both equations:

(1/2)mv² = (3/2)KT

mv² = 3KT

v = √(3KT/m)

where,

v = average speed of gas molecue = ?

K = Boltzman Constant = 1.38 x 10⁻²³ J/k

T = Absolute Temperature = 500 K

m = mass of a molecule = 7.9 x 10⁻²⁷ kg

Therefore,

v = √[(3)(1.38 x 10⁻²³ J/k)(500 k)/(7.9 x 10⁻²⁷ kg)]

v = 1618.72 m/s

3)

From kinetic molecular theory, we know that or an ideal gas:

P = (1/3)ρv²

where,

P = pressure of gas = ?

m = Mass of Gas = (Atomic Mass)(No. of Atoms)

m = (Atomic Mass)(Avogadro's Number)(No. of Moles)

m = (7.9 x 10⁻²⁷ kg/atom)(6.022 x 10²³ atoms/mol)(4.4 mol)

m = 0.021 kg

ρ = density = mass/volume = 0.021 kg/0.44 m³ = 0.0477 kg/m³

Therefore,

P = (1/3)(0.0477 kg/m³)(1618.72 m/s)²

P = 41.66 x 10³ Pa = 41.66 KPa

5)

The heat transfer for a constant pressure process is given by the formula:

ΔQ₂ = n Cp ΔT

where,

ΔQ₂ = Heat transfer during constant pressure process

n = No. of moles = 4.4 mol

Cp = Molar Specific Heat at Constant Pressure = 20.79 J/mol.k

ΔT = Change in Temperature = T₂ - T₁ = 300 k - 500 k = -200 k

Therefore,

ΔQ₂ = (4.4 mol)(20.79 J/mol.k)(-200 k)

ΔQ₂ = - 18.29 x 10³ J = - 18.29 KJ

Negative sign shows heat flows from system to surrounding.

6)

From Charles' Law, we know that:

V₁/T₁ = V₂/T₂

V₂ = (V₁)(T₂)/(T₁)

where,

V₁ = 0.44 m³

V₂ = ?

T₁ = 500 K

T₂ = 300 k

Therefore,

V₂ = (0.44 m³)(300 k)/(500 k)

V₂ = 0.264 m³

Therefore,

ΔV = V₂ - V₁ = 0.264 m³ - 0.44 m³ = - 0.176 m³

Hence, the work done , will be:

W₂ = PΔV = (41.66 KPa)(- 0.176 m³)

W₂ = - 7.33 KJ

Negative sign shows that the work is done by the gas

4 0

2 years ago

Identified __________ should be tracked until they are corrected

Reika [66]

<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ It would be A. Hazards

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

3 0

2 years ago