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eimsori [14]
2 years ago
5

Compare the freezing point of water in the aquanaut’s apartment to its value at the surface. Is it higher, lower, or the same?

Physics
1 answer:
Lelu [443]2 years ago
5 0

Answer:

Freezing Point - Lower

Boiling Point - Higher

Solid- liquid transition line in the phase diagram has a negative slope, but the liquid-gas transition line has a positive slope. Since there is more air pressure at 100m it will take less to freeze the water but more to boil it since it requires a larger temperature under larger pressures

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Savatey [412]

Answer:

input process output

Explanation:

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3 years ago
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The change in electric potential energy per unit charge is
klemol [59]

D-Volatge / gravity and current aren’t units and amperes is the loop of electrical current

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ou have designed and constructed a solenoid to produce a magnetic field equal in magnitude to that of the Earth (5.0 10-5 T). If
Nikitich [7]

Answer:

I = 0.0256 A

Explanation:

Given,

Magnetic field, B = 5 x 10⁻⁵ T

Number of turns, N = 450

length = 29 cm

Current,I = ?

Using formula of magnetic field due to solenoid

B = \mu_0 NI

I = \dfrac{B}{\mu_0N}

I = \dfrac{5\times 10^{-5}\times 0.29}{450\times 4\pi \times 10^{-7}}

I = 0.0256 A

Hence, the current obtained is equal to 0.0256 A.

5 0
3 years ago
A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto
Ilya [14]
Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer: 
0.32 m (single amplitude), or
0.64 m (double amplitude)

6 0
2 years ago
Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri
mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o $ = 5.50 m/s

Her speed after entering the water, $V_f$= 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times  dT$

Here, F =  the force ,

       dT =  time interval over which the force is applied for

            = 1.65 s

       dP  = change in momentum

dP = m x dV

    $= m \times [V_f - V_o] $

    = 82 x (1.1 - 5.5)

    = -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

  = 218 N

8 0
3 years ago
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