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3 years ago

what happens to the density, volume, mass, and shape of an ice cube put in a glass jar and left out in the Sun.

Physics

1 answer:

Elden [556K]3 years ago

3 0

Density I'm not sure
volume unchanged
mass unchanged
shape- water

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A tube is sealed at both ends and contains a 0.0100-m long portion of liquid. The length of the tube is large compared to 0.0100

Ahat [919]

Answer:

31.321 rad/s

Explanation:

L = Tube length

A = Area of tube

$\rho$ = Density of fluid

v = Fluid velocity

m = Mass = $\rho Al$

Centripetal force is given by

$F=\dfrac{mv^2}{L}\\ F=\dfrac{m(\omega L)^2}{L}\\ F=m\omega^2\\ F= 0.01A\rho\omega^2L$

Pressure is given by

$P=\dfrac{F}{A}=\rho gL\\\Rightarrow \dfrac{0.01A\rho\omega^2L}{A}=\rho gL\\\Rightarrow 0.01\omega^2=g\\\Rightarrow \omega^2=\dfrac{g}{0.01}\\\Rightarrow \omega=\sqrt{\dfrac{g}{0.01}}\\\Rightarrow \omega=\sqrt{\dfrac{9.81}{0.01}}\\\Rightarrow \omega=31.321\ rad/s$

The angular speed of the tube is 31.321 rad/s

5 0

3 years ago

Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he

storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0

3 years ago

A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 GHz. Two small boats

Anna35 [415]

Answer:

d = 76.5 m

Explanation:

To find the distance at which the boats will be detected as two objects, we need to use the following equation:

$\theta = \frac{1.22 \lambda}{D} = \frac{d}{L}$

<u>Where:</u>

θ: is the angle of resolution of a circular aperture

λ: is the wavelength

D: is the diameter of the antenna = 2.10 m

d: is the separation of the two boats = ?

L: is the distance of the two boats from the ship = 7.00 km = 7000 m

To find λ we can use the following equation:

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

<u>Where:</u>

c: is the speed of light = 3.00x10⁸ m/s

f: is the frequency = 16.0 GHz = 16.0x10⁹ Hz

$\lambda = \frac{c}{f} = \frac{3.00 \cdot 10^{8} m/s}{16.0 \cdot 10^{9} s^{-1}} = 0.0188 m$

Hence, the distance is:

$d = \frac{1.22 \lambda L}{D} = \frac{1.22*0.0188 m*7000 m}{2.10 m} = 76.5 m$

Therefore, the boats could be at 76.5 m close together to be detected as two objects.

I hope it helps you!

7 0

3 years ago

What is the frequency of a microwave of wavelength 3cm?

navik [9.2K]

Frequency = (speed) / (wavelength)

Speed = 3 x 10⁸ m/s

Wavelength = 3 cm = 0.03 m

Frequency = (3 x 10⁸ m/s) / (0.03 m)

Frequency = (3 x 10⁸ / 0.03) (m / m-s)

Frequency = 1 x 10¹⁰ Hz (10 Gigahertz)

5 0

3 years ago

Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you f

DedPeter [7]

Answer:

d = 142.5 m

Explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

d₀ = $v_{b}$ t

d₀ = 0.75 40

d₀ = 30 m

Let's write the kinematic equations

Boat

x = d₀ + $v_{b}$ t

x = 0 + $v_{h}$ t

At the meeting point the coordinate is the same for both

d₀ + $v_{b}$ t = $v_{h}$ t

t ( $v_{h}$ - $v_{b}$ ) = d₀

t = d₀ / ( $v_{b}$ - $v_{h}$ )

The two go in the same direction therefore the speeds have the same sign

t = 30 / (0.95-0.775)

t = 150 s

The distance traveled by man is

d = $v_{h}$ t

d = 0.95 150

d = 142.5 m

3 0

4 years ago