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KonstantinChe [14]

3 years ago

11

Since vectors always have positive magnitudes, the sum of two vectors must have a magnitude greater than the magnitude of either

one of them. O True O False

1 answer:

Natasha2012 [34]3 years ago

6 0

Answer:

False.

Explanation:

Yes the magnitude of a vector is always positive , but a vector consists of

magnitude

direction

when two vectors are added their direction may be opposite to each other For example-

$a=3i+3j$

$b=3i-6j$ ,

then their resultant

$r=b+a\\r=3i+3j+3i-6j\\r=6i-3j$

This resultant vector's x and y component equal to y and x component of vector b so its magnitude will be equal to magnitude of vector b.

Therefore, the resultant magnitude not necessary equal to the magnitude of either vector.

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pls help quick. the number line shows the starting and ending velocities for ball 1 what's the change in velocity of ball 1 calc

pentagon [3]

Answer:

The starting velocity for ball 1 is 1.00 meter/second. Its ending velocity is 0.25 meter/second.

The change in velocity for ball 1 is 0.25 – 1.00 = -0.75 meter/seconds

7 0

3 years ago

A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is th

IceJOKER [234]

Answer: $E=75\ N/m$

Explanation:

Given

First charge of $q_1=15\ nC$ is placed at $x=1.5\ m$

Second charge $q_2=-20\ nC$ is placed at $y=-2\ m$

Electric field is given by

$E=\frac{kq}{r^2}$

Electric field due to $q_1$ is away from it

$E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}$

$E_1=60\ N/m$

Electric field due to $q_2$

$E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}$

$E_2=45\ N/m$

Net electric field will be vector addition of two

$\vec{E_{net}}=\vec{E_1}+\vec{E_2}$

$\vec{E_{net}}=-60\hat{i}-45\hat{j}$

Magnitude of Electric field is

$E=\sqrt{60^2+45^2}$

$E=75\ N/m$

8 0

3 years ago

How many electrons must be added to an object to give it an overall net charge of -3.0 x 10^-7 C? The charge of an electron is -

katovenus [111]

The number of electrons is $1.9\cdot 10^{12}$

Explanation:

The total net charge on a negatively-charged object is given by

$Q=Ne$

where:

N is the number of excess electrons

$e$ is the charge of a single electron

In this problem, we have

$Q=-3.0\cdot 10^{-7}C$ is the overall charge of the object

$e=-1.602\cdot 10^{-19}C$ is the charge of one electron

Solving the equation for N, we find:

$N=\frac{Q}{e}=\frac{-3.0\cdot 10^{-7}}{-1.602\cdot 10^{-19}}=1.9\cdot 10^{12}$

Learn more about electrons and other particles:

brainly.com/question/2757829

#LearnwithBrainly

6 0

3 years ago

A 68.2 kg base runner begins his slide into second base while moving at a speed of 4.33 m/s. He slides so that his speed is zero

qwelly [4]

Answer:

$147.653J = W_f$

Explanation:

We know that:

$E_i - E_f = W_f$

where $E_i$ is the initial energy, $E_f$ is the final energy and $W_f$ is the energy lost due to friction.

so:

$\frac{1}{2}MV^2-0 = W_f$

Where M is the mass and V is the velocity of the runner, so:

$\frac{1}{2}(68.2kg)(4.33m/s)^2 = W_f$

$147.653J = W_f$

7 0

4 years ago

german

Answer:

The time taken by the ship to cover the distance, t = (x/25) s

Explanation:

Given data,

The initial velocity of the ship, u = 20 m/s

The final velocity of the ship, v = 30 m/s

The average velocity of the ship, V = (20 + 30)/ 2

= 25 m/s

The distance covered by the ship, d = x m

Then the time taken by the ship to cover the displacement is,

d = V x t

∴ t = d / V

Substituting the values,

t = x / 25

t = (x/25) s

Hence, the time taken by the ship to cover the distance, t = (x/25) s

4 0

3 years ago