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Shkiper50 [21]
2 years ago
10

_________ liquids and gases exert a buoyant force.

Physics
2 answers:
NikAS [45]2 years ago
7 0

Answer:

Both liquids and gases exert a bouyant force when gravity is acting on them

Eduardwww [97]2 years ago
5 0

Answer:

C. Both

Explanation:

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Modern scientists say that electrons are found in a ...
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I think it might be a or d i hope i have helped :)
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3 years ago
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what happens to the current in a circuit if the resitance of the components in the circuit is increased​
Anit [1.1K]

Answer:

The current decreases.

Explanation:

Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is V = IR, where V is voltage, I is current and R is resistance.

Rearranging this equation, you get:

I = \frac{V}{R}

and

R = \frac{V}{I}

If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation R = \frac{V}{I} , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.

Thus, it can be seen that the current will decrease as resistance increases and vice versa.

7 0
3 years ago
What happens in a global convection cell? apex​
zysi [14]

Answer:

In a global convection cell less –dense air at the equator rises and flows towards the poles. And from pole, the dense air sinks down and flows back towards the equator.... This movement of air is also supported by the Earth's rotation known as Coriolis Effect.

4 0
3 years ago
A convex security mirror has a radius of curvature of 12.0 cm. What is the magnification of a pare 3.0 m from the mirror?
Makovka662 [10]

Answer:

magnification will be -0.025

Explanation:

We have given the radius of curvature = 12 cm

And object distance = 3 m

So focal length f=\frac{R}{2}=\frac{12}{2}=6cm

Now for mirror we know that \frac{1}{f}=\frac{1}{u}+\frac{1}{v}

So \frac{1}{0.06}=\frac{1}{3}+\frac{1}{v}

16.66-0.333=\frac{1}{v}

v = 0.750 m

Now magnification of the mirror is m=\frac{-v}{u}=\frac{-0.750}{3}=-0.025

5 0
3 years ago
A small grinding wheel has a moment of inertia of 4.0*10-5kgm2. What net torque must be applied to the wheel for its angular acc
kvv77 [185]

Hi there!

We can use the rotational equivalent of Newton's Second Law:

\huge\boxed{\Sigma \tau = I \alpha}

Στ = Net Torque (Nm)

I = Moment of inertia (kgm²)

α = Angular acceleration (rad/sec²)

We can plug in the given values to solve.

\Sigma \tau = (4 * 10^{-5})(150) = \boxed{0.006 Nm}

4 0
2 years ago
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