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Simora [160]
4 years ago
5

which of the following statements is not true Negatively charged objects attract other negatively charged objects. Positively ch

arged objects attract negatively charged objects. Positively charged objects attract neutral objects. Negatively chargers objects attract neutral objects.
Physics
2 answers:
mote1985 [20]4 years ago
4 0

Answer:

negativelycharged objects attract other negatively charged objects

Explanation:

unlike charges attract like charges repel

maxonik [38]4 years ago
3 0

Answer:

negatively charged object attract other negatively objects

Explanation:

opposites attract

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If the car ha a mass of 1000 kilograms, what is its momentum (v=35m/s)
Black_prince [1.1K]

Answer:

<em>The momentum of the car is 35,000 kg.m/s</em>

Explanation:

<u>Momentum</u>

Momentum is often defined as <em>mass in motion.</em>

Since all objects have mass, if it's moving, then it has momentum. It can be calculated as the product of the mass by the velocity of the object:

\vec p = m\vec v

If only magnitudes are considered:

p = mv

The car has a mass of m=1,000 kg and travels at v=35 m/s. Calculating its momentum:

p = 1,000 kg * 35 m/s

p = 35,000 kg.m/s

The momentum of the car is 35,000 kg.m/s

4 0
3 years ago
Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
4 years ago
What are the products of a fusion reaction? Check all that apply. Lighter atoms energy heavier atoms a neutron a proton.
svet-max [94.6K]

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction. Heavier atoms are products of a fusion reaction.

<h3 /><h3>What is nuclear fusion?</h3>

The process by which two or more tiny nuclei unite to generate a bigger nucleus is known as a nuclear fusion reaction.

For example, the fusion of two hydrogen atoms produces more energy than the fusion of one helium atom, and surplus energy is expelled into space upon binding.

Hence heavier atoms are e products of a fusion reaction.

To learn more about nuclear fusion refer to the link;

brainly.com/question/14019172

6 0
2 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
How does metamorphic rock change into another type of metamorphic rock?
Marrrta [24]
Metamorphic rocks are formed by tremendous heat, great pressure, and chemical reactions. To change it into another type of metamorphic rock you have to reheat it and bury it deeper again beneath the Earth's surface.

Hope this helped! :)
4 0
4 years ago
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