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I am Lyosha [343]
4 years ago
13

Which element is found in all the acids

Physics
2 answers:
Rashid [163]4 years ago
6 0
Hydrogen is in all acid
Natali5045456 [20]4 years ago
5 0

All acids contain the element Hydrogen.

Hope this helps.

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At its lowest setting a centrifuge rotates with an angular speed of calculate the angular velocity
I am Lyosha [343]

No answer is possible until we know the number that belongs after the words "... angular speed of ".

8 0
3 years ago
Regardless of their frequency, wavelength, or energy, all electromagnetic waves: A. travel only through the vacuum of space. B.
agasfer [191]
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too. 
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3 years ago
Why can the atmosphere hold on to heat
Naddika [18.5K]

Answer:

A. Air

Explanation:

4 0
3 years ago
A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
Two wires A and B with circular cross-section are made of the same metal and have equal lengths, but the resistance of wire A is
Nesterboy [21]

Answer:

r₁/r₂ = 1/2 = 0.5

Explanation:

The resistance of a wire is given by the following formula:

R = ρL/A

where,

R = Resistance of wire

ρ = resistivity of the material of wire

L = Length of wire

A = Cross-sectional area of wire = πr²

r = radius of wire

Therefore,

R = ρL/πr²

<u>FOR WIRE A</u>:

R₁ = ρ₁L₁/πr₁²   -------- equation 1

<u>FOR WIRE B</u>:

R₂ = ρ₂L₂/πr₂²   -------- equation 2

It is given that resistance of wire A is four times greater than the resistance of wire B.

R₁ = 4 R₂

using values from equation 1 and equation 2:

ρ₁L₁/πr₁² = 4ρ₂L₂/πr₂²

since, the material and length of both wires are same.

ρ₁ = ρ₂ = ρ

L₁ = L₂ = L

Therefore,

ρL/πr₁² = 4ρL/πr₂²

1/r₁² = 4/r₂²

r₁²/r₂² = 1/4

taking square root on both sides:

<u>r₁/r₂ = 1/2 = 0.5</u>

6 0
4 years ago
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