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3 years ago

How are airplanes artificially pressurized?

1 answer:

8 0

They are built to be airtight so thsy sir can't get in or out, and then, as they climb up to where the air is thin, they use an air pump to pimp air into the inside.

Just like a party balloon.

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The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th

Lady_Fox [76]

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

$=10^{-5.9}$

intensity of sound of 5 persons

$I=5\times 10^{-5.9}$

$dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}$

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

4 0

2 years ago

Which state of matter has a definite volume but no definite shape

a_sh-v [17]

The answer is liquid.

6 0

3 years ago

Need Help With Physical Ed

Dimas [21]

which of the following is not a barrier to physical activity it is fear of injury I think.

8 0

2 years ago

Read 2 more answers

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:

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:

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

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:

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

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

2 years ago

What is the direction of the normal contact force of the road on the wheels?

Gekata [30.6K]

Answer:

The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.

Explanation:

3 0

2 years ago