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3 years ago

Which of the following are places that you can work out

Physics

2 answers:

son4ous [18]3 years ago

4 0

Gym, it has everything you need, your own room, no one would be there to distract u, feel more safe and comfortable

LUCKY_DIMON [66]3 years ago

3 0

Answer:

gym

Explanation:

the gym has everything thing that you need to work out with

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Consider the vector field. f(x, y, z) = xy2z2i x2yz2j x2y2zk (a) find the curl of the vector field?

Marat540 [252]

Observe that the given vector field is a gradient field:

Let $f(x,y,z)=\nabla g(x,y,z)$ , so that

$\dfrac{\partial g}{\partial x} = x y^2 z^2$

$\dfrac{\partial g}{\partial y} = x^2 y z^2$

$\dfrac{\partial g}{\partial z} = x^2 y^2 z$

Integrating the first equation with respect to $x$ , we get

$g(x,y,z) = \dfrac12 x^2 y^2 z^2 + h(y,z)$

Differentiating this with respect to $y$ gives

$\dfrac{\partial g}{\partial y} = x^2 y z^2 + \dfrac{\partial h}{\partial y} = x^2 y z^2 \\\\ \implies \dfrac{\partial h}{\partial y} = 0 \implies h(y,z) = i(z)$

Now differentiating $g$ with respect to $z$ gives

$\dfrac{\partial g}{\partial z} = x^2 y^2 z + \dfrac{di}{dz} = x^2 y^2 z \\\\ \implies \dfrac{di}{dz} = 0 \implies i(z) = C$

Putting everything together, we find a scalar potential function whose gradient is $f$ ,

$f(x,y,z) = \nabla \left(\dfrac12 x^2 y^2 z^2 + C\right)$

It follows that the curl of $f$ is 0 (i.e. the zero vector).

5 0

2 years ago

Which phrases describe the atmosphere? Check all that apply.

vladimir2022 [97]

Answer:

There is no options for this answer

Explanation:

7 0

3 years ago

A load of mass 5kg is raised through a height of 2m. (g=10mls)

s2008m [1.1K]

The gain in gravitational potential energy of the mass = 100 kg m^2 / s^2.

<u>Explanation:</u>

Gravitational potential energy is an energy in which an object possesses because of its position in a gravitational field. The most common use of gravitational potential energy is for an object near the Earth's surface where the gravitational acceleration can be assumed to be constant at about 9.8 m/s2.

The formula for the gravitational potential energy is the product of mass, gravity, and height.

GPE = m *g *h

where m represents mass in kg,

g represents the gravity,

h represents the height.

GPE = 5 * 10 * 2 = 100 kg m^2 / s^2.

7 0

3 years ago

Suppose a star has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 ) but is located at a distance of 300 ligh

iragen [17]

For a star that has the same apparent brightness as Alpha Centauri A ( 2.7×10−8watt/m2 is mathematically given as

L=2.7*10^30w

<h3> What is its luminosity?</h3>

Generally, the equation for the luminosity is mathematically given as

L=4*\pi^2*b

Therefore

L=4*\pi^2*b

L=4* \pi *(2.83*10^{18})*2.7*10^{-8}

L=2.7*10^30

In conclusion, the luminosity

L=2.7*10^30w