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3 years ago

Please help with this!!!!!

Physics

1 answer:

34kurt3 years ago

4 0

(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K) I hoped that helped

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NEED HELP!!!!

melamori03 [73]

Answer:

The correct choice is D. Concentric exercise can be hard on joints, so she should be very careful.

4 0

2 years ago

What is the magnitude of the electric field at the point if the electric potential in the region is given by V 2.00xyz2, where V

Hatshy [7]

Answer:

Electric field at a point ( x , y , z) is $E=-2yz^2-2xz^2-4xyz$ .

Explanation:

Given :

Electric potential in the region is , $V = 2xyz^2\ .$

We need to find the electric field .

We know , electric field , $E=-\dfrac{dV}{dr}$ { Here r is distance }

In coordinate system ,

$E=-\dfrac{dV}{\delta x }-\dfrac{dV}{\delta y }-\dfrac{dV}{\delta z }$ { $\delta$ is partial derivative }

Putting all values we get ,

$E=-\dfrac{2xyz^2}{\delta x }-\dfrac{2xyz^2}{\delta y }-\dfrac{2xyz^2}{\delta z }\\\\E=-2yz^2-2xz^2-4xyz$

Hence , this is the required solution.

3 0

3 years ago

Tony ran 600 meters in 60 seconds. What was Tony's speed during the<br> race?

NARA [144]

10 meters per second.

4 0

3 years ago

Read 2 more answers

The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

irina [24]

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

$m\propto \dfrac{1}{a}$

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

$\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g$

So, the car's acceleration would be 15.1 g.

6 0

3 years ago

The equation r (t )=(2t + 4)⋅i + (√ 7 )t⋅ j + 3t ²⋅k the position of a particle in space at time t. Find the angle between the v

velikii [3]

Answer:

$\theta = n\pi/2, {\rm where~n~is~an~integer.}$

Explanation:

We should first find the velocity and acceleration functions. The velocity function is the derivative of the position function with respect to time, and the acceleration function is the derivative of the velocity function with respect to time.

$\vec{v}(t) = \frac{d\vec{r}(t)}{dt} = (2)\^i + (\sqrt{7})\^j + (6t)\^k$

Similarly,

$\vec{a}(t) = \frac{d\vec{v}(t)}{dt} = (6)\^k$

Now, the angle between velocity and acceleration vectors can be found.

The angle between any two vectors can be found by scalar product of them:

$\vec{A}.\vec{B} = |\vec{A}|.|\vec{B}|.\cos(\theta)$

So,

$\vec{v}(t).\vec{a}(t) = |\vec{v}(t)|.|\vec{a}(t)|.\cos(\theta)\\36t = \sqrt{4 + 7 + 36t^2}.6.\cos(\theta)$

At time t = 0, this equation becomes

$0 = 6\sqrt{11}\cos(\theta)\\\cos(\theta) = 0\\\theta = n\pi/2, {\rm where~n~is~an~integer.}$

7 0

3 years ago