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alexgriva [62]
3 years ago
10

How can you use the density of an object to predict whether it will sink or float?

Physics
2 answers:
Bas_tet [7]3 years ago
7 0

Answer:

Explanation:

If an object has a higher density than the fluid it is in (fluid can mean liquid or gas), it will sink. If it has a lower density, it will float. Density is determined by an object's mass and volume. If two objects take up the same volume, but have one has more mass, then it also has a higher density.

Naya [18.7K]3 years ago
5 0
The density of an object determines whether it will float or sink in another substance. An object will float if it is less dense than the liquid it is placed in. An object will sink if it is more dense than the liquid it is placed in.
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How many hydrogen and carbon atoms in a diamond
Wewaii [24]

Answer:

Explanation:

Thus, total 4+4=8 C atoms are present per unit cell of diamond. Carbon has an electronic arrangement of 2,4. In diamond, each carbon shares electrons with four other carbon atoms - forming four single bonds.

4 0
3 years ago
A pendulum of length L=36.1 cm and mass m=168 g is released from rest when the cord makes an angle of 65.4 degrees with the vert
pychu [463]

(a) -0.211 m

At the beginning the mass is displaced such that the length of the pendulum is L = 36.1 cm and the angle with the vertical is

\theta=65.4^{\circ}

The projection of the length of the pendulum along the vertical direction is

L_y = L cos \theta = (36.1 cm)(cos 65.4^{\circ})=15.0 cm

the full length of the pendulum when the mass is at the lowest position is

L = 36.1 cm

So the y-displacement of the mass is

\Delta y = 15.0 cm - 36.1 cm = -21.1 cm = -0.211 m

(b) 0.347 J

The work done by gravity is equal to the decrease in gravitational potential energy of the mass, which is equal to

\Delta U = mg \Delta y

where we have

m = 168 g = 0.168 kg is the mass of the pendulum

g = 9.8 m/s^2 is the acceleration due to gravity

\Delta y = 0.211 m is the vertical displacement of the pendulum

So, the work done by gravity is

W=(0.168 kg)(9.8 m/s^2)(0.211 m)=0.347 J

And the sign is positive, since the force of gravity (downward) is in the same direction as the vertical displacement of the mass.

(c) Zero

The work done by a force is:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement

\theta is the angle between the direction of the force and the displacement

In this situation, the tension in the string always points in a radial direction (towards the pivot of the pendulum), while the displacement of the mass is tangential (it follows a circular trajectory): this means that the tension and the displacement are always perpendicular to each other, so in the formula

\theta=90^{\circ}, cos \theta = 0

and so the work done is zero.

5 0
3 years ago
A magnetic force can act on an electron even when it A) is at rest B) moves parallel to magnetic field lines C) both of these D)
Kobotan [32]

Answer: A)

Explanation: when an electron is placed in a magnetic field, it experiences a force.

This force is given below as

F=qvB*sinθ

F = force experienced by charge.

q = magnitude of electronic charge

v = speed of electron

B= strength of magnetic field

θ = angle between magnetic field and velocity.

What defines the force exerted on the charge is the angle between the field and it velocity.

If magnetic field is parallel to velocity, then it means that θ=0° which means sin 0 = 0, which means

F = qvB * 0 = 0.

The charge being at rest has nothing to do with the angle between magnetic field strength and velocity.

3 0
3 years ago
"One of the main projects being carried out by the Hubble Space Telescope is to measure the distances of galaxies located in gro
pickupchik [31]

Answer:

finding Cepheid variable and measuring their periods.

Explanation:

This method is called  finding Cepheid variable and measuring their periods.

Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.

A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.

5 0
3 years ago
Does anyone know the answer to all of these questions
Rasek [7]

Answer:

i don't understand the hw

5 0
3 years ago
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