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3 years ago

Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?

Physics

2 answers:

olya-2409 [2.1K]3 years ago

6 0

Answer:

Gravity

Explanation:

Gravity is bringing down the rocket and is not air resistance

Wittaler [7]3 years ago

3 0

Answer:

unbalenced force

Explanation:

make Brainliest plz

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Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa

lyudmila [28]

Answer:

No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field $\vec{E}$ is related to the gradient of the electric potential V with :

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})$

This means that for constant electric potential the electric field must be zero:

$V(\vec{r}) = k$

$\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k$

$\vec{E} (\vec{r}) = - (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k$

$\vec{E} (\vec{r}) = - (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})$

$\vec{E} (\vec{r}) = - (0,0,0)$

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

$V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4$

give an electric field of zero at point (0,0,0)

8 0

3 years ago

Hartman value profile

Grace [21]

What is your question exactly? I'm confused?

6 0

3 years ago

nevsk [136]

Answer:

true

Explanation:

<h3><em>A healthy body will keep you in shape and less risk of getting sick</em></h3>

3 0

3 years ago

The initial temperature of a bomb calorimeter is 28.50°

solniwko [45]

The answer is 5,170 J

8 0

3 years ago

Read 2 more answers

The spring of a spring balance is 6.0 in. long when there is no weight on the balance, and it is 8.4 in. long with 4.0 lb hung f

Sergeeva-Olga [200]

Answer:

W = 55.12 J

Explanation:

Given,

Natural length = 6 in

Force = 4 lb, stretched length = 8.4 in

We know,

F = k x

k is spring constant

4 = k (8.4-6)

k = 1.67 lb/in

Work done to stretch the spring to 10.1 in.

$W =k\int_{6}^{10.1} x$

$W = \dfrac{k}{2}[x^2]_6^{10.1}$

$W = \dfrac{1}{2}\times 1.67\times (10.1^2-6.0^2)$

W = 55.12 J

Work done in stretching spring from 6 in to 10.1 in is equal to 55.12 J.

3 0

3 years ago