All categories

2 years ago

The magnitude of perpendicular vectors can be determined using ....

Physics

1 answer:

maxonik [38]2 years ago

8 0

Determined by cross product or ( vector product )

You might be interested in

A 1200-kg car initially at rest undergoes constant acceleration for 9.4 s, reaching a speed of 11 m/s. It then collides with a s

Natalka [10]

To solve this problem we will apply the principle of conservation of energy and the definition of kinematic energy as half the product between mass and squared velocity. So,

$KE_i = KE_f$

$KE_f = \frac{1}{2} mv^2$

Here,

m = Mass

V = Velocity

Replacing,

$KE_f = \frac{1}{2} (12000)(11)^2$

$KE_f = 72600J$

Therefore the final kinetic energy of the two car system is 72.6kJ

8 0

2 years ago

If two vectors are perpendicular to each other, how should you add them?

Aleksandr [31]

Let us consider two vectors A and B.

As per the question, the two vectors are perpendicular to each other.

Hence the angle between them $\theta =90 degree$

We are asked to calculate the resultant of these two vectors.

As per parallelogram law of vector addition, the resultant of two vectors are-

$R=\sqrt{A^{2}+ B^{2}+2ABcos\theta$

$=\sqrt{ A^{2}+ B^{2}+2AB*cos90}$ [cos90=0]

$=\sqrt{ A^{2}+ B^{2}$

This is the way by which we can add two perpendicular vectors.

8 0

2 years ago

Read 2 more answers

Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0

GuDViN [60]

Answer:

20 hertz of frequency produced.

Explanation:

$\boxed{frequency = \frac{1}{period} }$

Here we will find frequency and period should be in second, here given: 0.05 seconds

using the formula:

$\boxed{frequency = \frac{1}{0.05} }$

$\boxed{frequency =20}$

8 0

2 years ago

A very long straight wire has charge per unit length 1.44×10-10C/m.

4vir4ik [10]

Answer:

Distance of the point where electric filed is 2.45 N/C is 1.06 m

Explanation:

We have given charge per unit length, that is liner charge density $\lambda =1.44\times 10^{-10}C/m$

Electric field E = 2.45 N/C

We have to find the distance at which electric field is 2.45 N/C

We know that electric field due to linear charge is equal to

$E=\frac{\lambda }{2\pi \epsilon _0r}$ , here $\lambda$ is linear charge density and r is distance of the point where we have to find the electric field

So $2.45=\frac{1.44\times 10^{-10} }{2\times 3.14\times 8.85\times 10^{-12}\times r}$

r = 1.06 m

So distance of the point where electric filed is 2.45 N/C is 1.06 m

3 0

2 years ago

Is this a testable question?

faltersainse [42]

This is a non testable question because it cannot be answered by doing an experiment. But it could be modified for example Dogs are more obedient then cats.

8 0

2 years ago

Read 2 more answers