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ArbitrLikvidat [17]
2 years ago
13

Which is greater, the force exerted by the Earth on the Sun, or the force exerted by the Sun on the Earth? Why?

Physics
1 answer:
LekaFEV [45]2 years ago
3 0

Answer:

There is no great force, the force exerted by the Earth on the Sun, and the force exerted by the Sun on the Earth are equal

Explanation:

By definition...

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The sensitivity of a measuring instrument is the value of the smallest quantity that can be read or estimated with it. What is t
nirvana33 [79]

Answer:

The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.

6 0
3 years ago
The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc
prohojiy [21]

Answer:

The answer is 218

Explanation:

Weight = mass * gravitational acceleration

weight is represented by F

F = 25kg (8.7)

(I'm pretty sure that you don't have to include the meters per second/per second thing)

4 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
3 years ago
Your high beam headlights illuminate the road in front of you for __________ feet. A. 150 B. 450 C. 650
ASHA 777 [7]

Answer:

B. 450 feet

Explanation:

Due to the angle at which high beam headlights illuminate, they can illuminate the road for about 450 feet.

8 0
3 years ago
The muzzle velocity of a rifle bullet is 709 m s−1along the direction of motion. If the bullet weighs 35 g, and the uncertainty
nydimaria [60]

Answer:

Uncertainty in position of the bullet is \Delta x=1.07\times 10^{-33}\ m

Explanation:

It is given that,

Mass of the bullet, m = 35 g = 0.035 kg

Velocity of bullet, v = 709 m/s

The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :

p=mv

p=0.035\times 709=24.81\ kg-m/s

Uncertainty in momentum is,

\Delta p=0.2\%\ of\ 24.81

\Delta p=0.049

We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :

\Delta p.\Delta x\geq \dfrac{h}{4\pi}

\Delta x=\dfrac{h}{4\pi \Delta p}

\Delta x=\dfrac{6.62\times 10^{-34}}{4\pi \times 0.049}

\Delta x=1.07\times 10^{-33}\ m

Hence, this is the required solution.

7 0
3 years ago
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