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horsena [70]
3 years ago
15

A human-operated spaceship reaches the moon in 3 days. The moon is about 386,400 km from Earth. Mars, our closest planetary neig

hbor, is, at its closest, about 140 times farther away from us than the moon if Mars stays in place. Assuming a very simple model of the solar system, about how long would it take that same spaceship to reach Mars?(1 point) about 22.5 years about 1 year about 2.5 years about 7.5 years
Physics
1 answer:
krok68 [10]3 years ago
8 0

Answer:

About 1 year.

Explanation:

Since the spacecraft travels the 386,400 kilometers of distance between the Earth and the Moon in 3 days, and the planet Mars is 140 times further from the Earth than the Moon, to obtain the time it will take for the same spacecraft to reach Mars must perform the following calculation:

140 x 3 = 420 days

If a year has 365 days, the travel time that said spacecraft would take to reach Mars would be just under 1 year and 2 months.

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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

8 0
3 years ago
An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
Aleksandr-060686 [28]

The tension in the upper rope is determined as 50.53 N.

<h3>Tension in the upper rope</h3>

The tension in the upper rope is calculated as follows;

T(u) = T(d)+ mg

where;

  • T(u) is tension in upper rope
  • T(d) is tension in lower rope

T(u) = 12.8 N + 3.85(9.8)

T(u) = 50.53 N

Thus, the tension in the upper rope is determined as 50.53 N.

Learn more about tension here: brainly.com/question/918617

#SPJ1

6 0
2 years ago
What formula relstes work n power
777dan777 [17]
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8 0
3 years ago
Recall that the spring constant is inversely proportional to the number of coils in the spring, or that shorter springs equate t
ruslelena [56]

Answer:

x_1= 0.0425m

Explanation:

Using the tension in the spring and the force of the tension can by describe by

T = kx

, T = mg

Therefore:

m*g = k*x

With two springs, let, T1 be the tension in each spring,  x1 be the extension of each spring.  The spring constant of each spring is 2k so:

T_1 = 2k*x_1

2T_1 = m*g=4k x_1

Solve to x1

x_1=\frac{m*g}{4k}

x_1=\frac{k*x}{4*k}

x_1=\frac{x}{4}

x_1 = 0.170 / 4

x_1= 0.0425m

7 0
3 years ago
A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the c
puteri [66]

Correct question:

A solenoid of length 0.35 m and diameter 0.040 m carries a current of 5.0 A through its windings. If the magnetic field in the center of the solenoid is 2.8 x 10⁻² T, what is the number of turns per meter for this solenoid?

Answer:

the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

Explanation:

Given;

length of solenoid, L= 0.35 m

diameter of the solenoid, d = 0.04 m

current through the solenoid, I = 5.0 A

magnetic field in the center of the solenoid, 2.8 x 10⁻² T

The number of turns per meter for the solenoid is calculated as follows;

B =\mu_o  I(\frac{N}{L} )\\\\B =  \mu_o  I(n)\\\\n = \frac{B}{\mu_o I} \\\\n = \frac{2.8 \times 10^{-2}}{4 \pi \times 10^{-7} \times 5.0} \\\\n = 4.5 \times 10^3 \ turns/m

Therefore, the number of turns per meter for the solenoid is 4.5 x 10³ turns/m.

3 0
3 years ago
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