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never [62]
3 years ago
11

As the mass of an object increases, the weight of the object will ______?

Physics
1 answer:
Sergio039 [100]3 years ago
4 0

Answer:

also increase

Explanation:

If the mass increases so will the weight.

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Determine the frequency of a sound wave if it has a speed of 350 m/s and a wavelength of 3.80 m.
Eva8 [605]
Since we have , v=f×lambda (wavelength). Where v equals 350m/s and wavelength equals 3.80. so it will become f = v/lambda=350/3.80=92.1052Hz
7 0
3 years ago
PLEASE HELP WILL GIVE 25 POINTS!!!!!
zimovet [89]

The answer is B. 1,4, and 6

8 0
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What do we call a substance in<br> which two or more elements are<br> chemically bonded
Goryan [66]

Answer:

A compound

Explanation:

A compound is a substance formed when two or more elements are chemically joined

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3 years ago
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A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
Cindy saw her mother heating a wet pan on the stove. As the pan got hotter, the water on the outside began to dry. Why?
dlinn [17]

Answer:

As the temperature of materials increase, the objects find a phenomenon called change of phase.

This means that if you give enough heat to a liquid, this can change of state from liquid state to gas state (the water evaporates)

So the water in the pan reaches the evaporation temperature (around 100°C) and it starts to evaporate, this is why the water on the outside begins to "dry"

7 0
3 years ago
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