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Greeley [361]
1 year ago
13

of the following and draw a neat, properly labelled free body diagram. Indicate the size of forces with the lengths of the arrow

s. i.e., if two forces are the same size draw them the same length and if one is much bigger draw it much longer. a) The Space Shuttle accelerating vertically just after leaving the launch pad.
Physics
1 answer:
julsineya [31]1 year ago
4 0

We will have that the diagram will be the following:

Here "F" is the force the shuttle is using to take off, friction is the friction with the air and "W" is the weight of the shuttle.

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Answer:

Each type of electromagnetic radiation have been matched to their descriptions as follows:

Ultraviolet light A) Given off by very hot objects, such as the sun

Gamma rays B) Given off by radioactive substances  

Microwaves C) Used in radar and to heat food

X‒rays D) Used in the medical field to see bone structure

Explanation:

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3 years ago
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natali 33 [55]
As waves get closer to the beach they increase in energy
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3 years ago
A 2 nC point charge is at the origin, and a second 5 nC point charge is on the x-axis at x = 8 m. Find the electric field (magni
dimaraw [331]

Answer:

The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

Explanation:

Given that,

Point charge at origin = 2 nC

Second charge = 5 nC

Distance at x axis = 8 m

We need to calculate the electric field at the point x = 2 m

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}(\dfrac{q_{1}}{r_{1}^2}+\dfrac{q_{2}}{r_{2}^2})

Put the value into the formula

E=9\times10^{9}\times(\dfrac{2\times10^{-9}}{2^2}+\dfrac{5\times10^{-9}}{(8-2)^2})

E=5.75\ N/C

The direction is toward positive x- axis.

Hence, The magnitude of  the electric field is 5.75 N/C towards positive x- axis.

7 0
3 years ago
A force on a particle depends on position such that F(x) = (3.00 N/m2)x2 + (6.00 N/m)x for a particle constrained to move along
Pachacha [2.7K]

Answer:

The work done by a particle from x = 0 to x = 2 m is 20 J.

Explanation:

A force on a particle depends on position constrained to move along the x-axis, is given by,

F(x)=(3\ N/m^2)x^2+(6\ N/m)x

We need to find the work done on a particle that moves from x = 0.00 m to x = 2.00 m.

We know that the work done by a particle is given by the formula as follows :

W=\int\limits {F{\cdot} dx}

W=\int\limits^2_0 {(3x^2+6x){\cdot} dx} \\\\W={(x^3}+3x^2)_0^2\\\\\W={(2^3}+3(2)^2)\\\\W=20\ J

So, the work done by a particle from x = 0 to x = 2 m is 20 J. Hence, this is the required solution.

3 0
3 years ago
Read 2 more answers
Can someone help me ASAP
Phantasy [73]

Well I don't know.  Let's actually LOOK at the picture and see if that helps.

A,  B,  C,  and D all have the same TOTAL length, but  A  has the most waves crammed into that same total length.

By golly, that means the length of <u><em>each</em></u> wave in  A  must be shorter than each wave in  B,  C,  or D.

The correct choice is <em> A </em>.  Looking at the picture did the trick !

7 0
3 years ago
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