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3 years ago

5

A position vector in the first quadrant has an x-component of 18 m and a magnitude of 30 m. What is the value of its y-component

?

1 answer:

Snezhnost [94]3 years ago

7 0

Answer:

The value is 24meters

Explanation:

Using

r= xi+yj

To get the magnitude of vector x

We say

/r/= √x²+y²

So

30²= √18² + y²

y= √576

Y= 24m

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B) Calculate the net work required to bring a 1300 kg car moving at 30 m/s to rest.

prohojiy [21]

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5 0

3 years ago

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl

USPshnik [31]

Given Information:

Wavelength = λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0

3 years ago

What is gamma radiation composed of?

Paha777 [63]

Gamma radiation is composed of the stable form of nuclei and most important Helium Atom is released in this reaction !

gamma radiation is due to fission in which the radioactive substance is bombarded and as a result a stable form of that element is formed and a Helium atom !

3 0

3 years ago

Read 2 more answers

Two charges, X and Y, are placed along the x-axis. Charge X is +18 nC and is placed at x = 0. Charge Y is placed at a location o

Helen [10]

Answer:

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m.

Explanation:

The Coulomb force between two charges, $Q_1$ and $Q_2$ , separated by a distance, $d$ , is given

$F = k\dfrac{Q_1Q_2}{r^2}$

<em>k</em> is a constant.

For the charge Z to be at equilibrium, the force exerted on it by charge X must be equal and opposite to the force exerted on it by charge Y.

It is to be placed along the <em>x</em>-axis. Hence, it is on the same line as charges X and Y.

Let the charge on Z be <em>Q</em>. It is positive.

Let the distance from charge X be <em>x m.</em> Then the distance from charge Y will be (0.60 - <em>x</em>) m.

Force due to charge X

$F_X = k\dfrac{18Q}{x^2}$

Force due to charge Y

$F_Y = k\dfrac{-27Q}{(0.60-x)^2}$

Since both forces are equal and opposite,

$F_X = -F_Y$

$k\dfrac{18Q}{x^2} = -k\dfrac{-27Q}{(0.60-x)^2}$

$\dfrac{2}{x^2} = \dfrac{3}{(0.60-x)^2}$

$2(0.60-x)^2 = 3x^2$

$2(0.36-1.20x+x^2) = 3x^2$

$0.72-2.40x+2x^2 = 3x^2$

$x^2+2.40x-0.72 = 0$

Applying the quadratic formula,

$x = \dfrac{-2.40\pm\sqrt{2.40^2 - (4)(1)(-0.72)}}{2} = \dfrac{-2.40\pm\sqrt{8.64}}{2}$

$x = -2.7$ or $x = 0.27$

Charge Z can be placed at <em>x</em> = -2.7 m or at <em>x</em> = 0.27 m

3 0

3 years ago

We know we have exerted of force even when we have done no work this is called _____

zheka24 [161]

Answer: The correct answer is zero work done.

Explanation:

Work is said to be done when the object moves through a distance when the force is applied to the object.

If the object does not move a distance even the force is exerted on the object then the work done is zero in this case.

Therefore, when the force is exerted even when no work is done then this is called zero work done.

5 0

4 years ago