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3 years ago

what is the acceleration of an object that moves at a constant velocity of 2.0 meters per second for 5 seconds?

1 answer:

3 0

Constant velocity means moving in a straight line at a speed that doesn't change. If the object is moving with constant velocity then its acceleration is zero. Acceleration is the rate at which velocity is changing.

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Which objects are opaque? Check all that apply.

Vinvika [58]

<span>textbook
track shoes
</span><span>basketball</span>

7 0

3 years ago

Read 2 more answers

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is $V=\frac{1}{4\pi\epsilon} \frac{q}{d}$

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0

3 years ago

What happens to the volume of a gas when pressure is applied? A. increases B. decreases C. Stays the same D. evaporates

BaLLatris [955]

Answer:

Decreases

Explanation:

8 0

3 years ago

Two parallel slits are illuminated by light composed of two wavelengths. One wavelength is λA = 622nm. The other wavelength is λ

Sergio039 [100]

Answer:

$\lambda_{B}=414.67 nm$

Explanation:

In this question we have given

$\lambda_{A}=622nm$

we have to find

$\lambda_{B}=?$

We know that

optical path difference for bright fringe is given as $=n\lambda$

Here,

n is order of fringe

and optical path difference for dark fringe is given as $=(n+.5)\lambda$

since the light with wavelength $\lambda_{A}$ produces its third-order bright fringe at the same place where the light with wavelength $\lambda_{B}$ produces its fourth dark fringe

it means

optical path difference for 3rd order bright fringe= optical path difference for forth order dark fringe

Therefore,

$3\lambda_{A}=(4+.5)\lambda_{B}$ ...............(1)

Put value of $\lambda_{A}$ in equation (1)

$3 \times 622=(4+.5)\lambda_{B}$

$1866=4.5\lambda_{B}$

$\lambda_{B}=414.67 nm$

3 0

3 years ago

Create the following matrix H:

sladkih [1.3K]

Answer:

a) {[1.25 1.5 1.75 2.5 2.75]

[35 30 25 20 15] }

b) {[1.5 2 40]

[1.75 3 35]

[2.25 2 25]

[2.75 4 15]}

Explanation:

Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]

[1 2 3 1 2 3 4]

[45 40 35 30 25 20 15]}

Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.

Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).

For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.

For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:

{[1.5]

[2]

[40]}

Transposing this column will make our first row of K look like this:

{[1.5 2 40]}

Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.

4 0

3 years ago