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amid [387]
3 years ago
12

How much Work is done when a 5 kg object is picked up a distance of 8 m.

Physics
1 answer:
tigry1 [53]3 years ago
4 0

Answer:

392J?

Explanation:

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Whats the term for climate near equator
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The closer to the equator, the hotter the climate will be.
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4 years ago
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As an interstellar cloud of hydrogen gas shrinks in size, its rate of rotation
FinnZ [79.3K]

Answer:

INCREASES, BECAUSE ITS ANGULAR MOMENTUM IS CONSERVED.

Explanation: Interstellar cloud of Hydrogen is an accumulation of Hydrogen gas in the cloud.

As the Interstellar cloud of Hydrogen shrinks (reduces) in size,the rate of rotation of the shrinked Interstellar cloud Increases because its angular momentum is conserved. GASEOUS MOLECULES MAKE UP ABOUT 99% OF THE INTERSTELLAR CLOUD WITH HYDROGEN HAVING ABOUT 90% OF THE VOLUME OF GASES IN THE INTERSTELLAR CLOUD.

4 0
3 years ago
If the frequency of the motion of a simple harmonic oscillator is doubled, by what factor does the maximum speed of the oscillat
Maksim231197 [3]

Answer:

If the frequency of the motion of a simple harmonic oscillator is doubled , then maximum speed of the oscillator changes by the factor 2

Explanation:

We know that in a simple harmonic oscillator the maximum speed is given by

    v_{max} = Aw

  Here A is amplitude which is constant , so from above equation we see that maximum speed is directly proportional to w\\ of the oscillation .

  Since  w = 2 \pi f

      v_{max}^{|}/v_{max} = 2f/f = 2

  Where v_{max}^{|} is the maximum speed when frequency is doubled .

6 0
3 years ago
Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

Thus for the exercise we will have

\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

Working with the partial derivatives we get the curl

\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

6 0
4 years ago
T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?
Debora [2.8K]

Answer:

P_J=195N

Explanation:

From the question we are told that

Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\

Generally the equation for momentum is mathematically given by

P=mv

Therefore

T-Joe momentum P_J

P_J=65*3

P_J=195N

5 0
3 years ago
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