Motion is defined by how fast an object is moving? True False

A vector A⃗ has a magnitude of 40.0 m and points in a direction 20.0∘ below the positive x axis. A second vector, B⃗, has a magn

jolli1 [7]

The magnitude of the vector C is 96.32m

<h3>How to solve for the magnitude of vector c</h3>

Ax = AcosθA

= 40 cOS 20

= 37.59

Ay = AsinθA

-40sin20

= -13.68

Bx = B cos θ B

= 75Cos50

= 48.21

By = BsinθB

= 75sin50

= 57.45

Cx = AX + Bx

= 37.59 + 48.21

= 85.8

Cy = Ay + By

= -13.65 + 57.45

= 43.77

The magnitude is solved by

|c| = $\sqrt{Cx^{2}+Cy^{2} }$

= √85.8² + 43.77²

= 96.32m

The magnitude of the vector c is 96.32m

Read more on the magnitude of a vector here:

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6 0

1 year ago

You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh

lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

$F=\frac{kQq}{r^2}$

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

$\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}$

And the force on the charge due to the charge on the north is,

$\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}$

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of $45^\circ$ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

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3 0

1 year ago

Monochromatic light falls on two very narrow slits 0.046 mm apart. Successive fringes on a screen 6.20 m away are 8.9 cm apart n

Elanso [62]

Answer:

λ = 6.602 x 10^(-7) m

Explanation:

In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is given as ;

y = mλD/d

Where;

D is the distance of the screen from the slits = 6.2 m

d is the distance between the two slits = 0.046 mm = 0.046 x 10^(-3) m

The fringes on the screen are 8.9 cm = 0.089 m apart from each other, this means that the first maximum (m=1) is located at y = 0.089 m from the center of the pattern.

Therefore, from the previous formula we can find the wavelength of the light:

y = mλD/d

So, λ = dy/mD

Thus,

λ = (0.046 x 10^(-3) x 0.089)/(1 x 6.2)

λ = 6.602 x 10^(-7) m

8 0

3 years ago

A flatbed truck is carrying an 800-kg load of timber that is not tied down. The maximum friction force between the truck bed and

ycow [4]

Answer:

Acceleration a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

Explanation:

For the truck to accelerate without losing its load.

Acceleration force of truck must be less than or equal to the maximum friction force between the truck bed and the load.

Fa ≤ F(friction)

But;

Fa = mass × acceleration

Fa = ma

ma ≤ F(friction)

a ≤ (F(friction))/m ......1

Given;

Fa = mass × acceleration

Fa = ma

mass m = 800 kg

F(friction) = 2400 N

Substituting the given values into equation 1;

a ≤ F(friction)/m

a ≤ 2400N/800kg

a ≤ 3 m/s^2

the greatest acceleration that the truck can have without losing its load is 3 m/s^2

4 0

3 years ago

Suppose that a comet that was seen in 550 A.D. by Chinese astronomers was spotted again in year 1941. Assume the time between ob

pickupchik [31]

To solve the problem it is necessary to apply the concepts related to Kepler's third law as well as the calculation of distances in orbits with eccentricities.

Kepler's third law tells us that

$T^2 = \frac{4\pi^2}{GM}a^3$