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juin [17]
2 years ago
8

Which of the following relationships determines an object’s ability to float in a liquid?

Physics
1 answer:
azamat2 years ago
7 0

Answer:

C

Explanation:

The density is what determines if it will float. Lower densities float, while higher densities sink. Density takes mass and volume into account.

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Ceres, Pluto, and Eris are all round in shape and classified as:_________ A) Leftover planetesimals that formed inside the frost
motikmotik

Answer

Ceres, Pluto, and Eris are classified as DWARF PLANET.

A) Leftover planetesimals inside the frost line are known as ASTEROIDS.

B) METEORITES are the pieces of Asteroids which are fallen on the earth's surface.                                                

C) COMETS are the objects which are visible with long tails.            

D) COMETS are also the leftover planetesimals that are occupied by the jovian planets and are formed in the solar system.                

E) Meteor showers are associated with debris from COMETS                      

5 0
3 years ago
liquid helium has a very low boiling point, 4.2 k, as well as a very low latent heat of vaporization, 2.00 104 j/kg. if energy i
aksik [14]

4.80 \times 10^3 \text { seconds }  long does it take to boil away 2.40 kg of the liquid.

Boiling point of He is $T=4.2 \mathrm{k}$

Latent heat of vapourization $L=2.00 \times 10^4 \mathrm{~J} / \mathrm{kg}$

Power of electrical heater $P=30 \mathrm{w}$

mass of liquid is $m=2.40 \mathrm{~kg}$

amount of heat required to boil

$$\begin{aligned}&Q=m L \\&Q=2.40 \times 2 \times 10^4 \mathrm{~J} \\&Q=4.80 \times 10^4 \mathrm{~J}\end{aligned}$$

Power $p=\frac{\text { work }}{\text { time }}=\frac{\text { Energy }}{\text { Time }}$

$$\begin{aligned}P &=\frac{Q}{t} \\\text { tine } t &=\frac{Q}{P}=\frac{4.80 \times 10^4 \mathrm{~J}}{10} \\t &=4.80 \times 10^3 \text { seconds }\end{aligned}$$

The heat or energy that is absorbed or released during a substance's phase shift is known as latent heat. It could go from a solid to a liquid or from a liquid to a gas, or vice versa. Enthalpy, a characteristic of heat, is connected to latent heat.

The heat that is used or lost as matter melts and transitions from a solid to a fluid form at a constant temperature is known as the latent heat of fusion.

Due to the fact that during softening the heat energy anticipated to transform the substance from solid to fluid at air pressure is the latent heat of fusion and that the temperature remains constant during the process, the "enthalpy" of fusion is a latent heat. The enthalpy change of any quantity of material during dissolution is known as the latent heat of fusion.

For learn more about Latent heat of vaporization, visit: brainly.com/question/14980744

#SPJ4

3 0
1 year ago
What is the frequency of this wave?
Anastasy [175]
In telecommunication systems, Carrier frequency is a technical term used to indicate: ... The frequency of the unmodulated electromagnetic wave at the output of a conventional amplitude-modulated (AM-unsupressed carrier), or frequency-modulated (FM), or phase-modulated (PM) radio transmitter.
6 0
3 years ago
Read 2 more answers
Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each
jarptica [38.1K]

Answer with Explanation:

We are given that

I_1=29 A

I_2=78 A

d=38 cm=\frac{38}{100}=0.38 m

1 m=100 cm

a.Length of segment,l=20 m

Magnetic force ,F=\frac{2\mu_0I_1I_2 l}{4\pi d}

\frac{\mu_0}{4\pi}=10^{-7}

Substitute the values

F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

4 0
3 years ago
Read 2 more answers
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
4 years ago
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