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2 years ago

12

PLEASE HELP When astronauts eliminate waste in space, which is likely to be used? [A] showers and faucets, spraying hot water..

... [B] vacuum to secure solid and liquid waste.... [C]wet towels that are later laundered.... [D] radiation to disinfect their bodies. WILL GIVE 9000000000000000000000 POINTS :) :) :)..................

1 answer:

lesya [120]2 years ago

3 0

Answer:

B. vacuum to secure solid and liquid waste

Explanation:

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Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre

VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers $d=1.6\ m$

Distance of person from one speaker $x_1=3\ m$

Distance of person from second speaker $x_2=3.5\ m$

Path difference between the waves is given by

$x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}$

for destructive interference m=0 I.e.

$x_2-x_1=\frac{\lambda }{2}$

$3.5-3=\frac{\lambda }{2}$

$\lambda =0.5\times 2$

$\lambda =1\ m$

frequency is given by

$f=\frac{v}{\lambda }$

where $v=velocity\ of\ sound\ (v=343\ m/s)$

$f=\frac{343}{1}=343\ Hz$

For next frequency which will cause destructive interference is

i.e. $m=1$ and $m=2$

$3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda$

$\lambda =\frac{1}{3}\ m$

frequency corresponding to this is

$f_2=\frac{343}{\frac{1}{3}}=1029\ Hz$

for $m=2$

$3.5-3=\frac{5}{2}\cdot \lambda$

$\lambda =\frac{1}{5}\ m$

Frequency corresponding to this wavelength

$f_3=\frac{343}{\frac{1}{5}}$

$f_3=1715\ Hz$

8 0

3 years ago

he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As

k0ka [10]

Answer:

$1.54481175\times 10^{-12}\ C$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area = $6\times 10^{-9}\ m^2$

d = Thickness = $1.6\times 10^{-8}\ m$

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

$Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C$

The charge on the outer surface is $1.54481175\times 10^{-12}\ C$

4 0

2 years ago

Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o

Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

$F_{23} = \frac{k*q_{2}*q_{3}}{x^2}$ (Electric force of q₂ over q₃)

$F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

$\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2}$ (We cancel k and q₃)

$\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}$

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

6 0

2 years ago

Read 2 more answers

Calculate the speed of a car that travels a distance of 2.4km in 2 minutes

hammer [34]

Answer:

72 kilometres per hour

Explanation:

The formula for calculating speed is distance/time.

So to work this out you would convert 2 minutes into hours. You would divide 2 by 60 to convert it into hours. This is because the standard unit for speed with kilometres is kilometres per hour. Then you would divide 2.4 by that.

1)Divide 2 by 60:

$2/60=0.0333333$

2) Divide 2.4 by 0.0333333.

$2.4/0.0333333=72.00007$

3) Round it.

$72km/h$

3 0

2 years ago

write a one or two summary paragraph discussing this experiment and the results use tye following questions 1 according to your

Monica [59]

Answer:heed

Explanation:

heed knows how to divide.

8 0

2 years ago