ANSWER ASAPPPPPPPP BE 100% CORRECT

Displacement vector points due east and has a magnitude of 3.35 km. Displacement vector points due north and has a magnitude of

tekilochka [14]

Answer:

(a) 6.56 km

(b) $59.68^\circ$ north of west

Explanation:

Given:

$\vec{d}_1= 3.35\ km\ east = 3.35\ km\ \hat{i}\\\vec{d}_2= 9.31\ km\ north = 9.31\ km\ \hat{j}\\\vec{d}_3= 6.66\ km\ west = -6.61\ km\ \hat{i}\\\vec{d}_4= 3.65\ km\ south = -3.65\ km\ \hat{j}\\$

Let the resultant displacement vector be $\vec{D}$ .

As the resultant is the vector sum of all the vectors.

$\therefore \vec{D}=\vec{d}_1+\vec{d}_2+\vec{d}_3+\vec{d}_4\\\Rightarrow \vec{D} =(3.35\ km\ \hat{i})+(9.31\ km\ \hat{j})+(-6.61\ km\ \hat{i})+(-3.65\ km\ \hat{j})\\\Rightarrow \vec{D} =-3.31\ km\ \hat{i}+5.66\ km\ \hat{j}\\\textrm{Magnitude of the resultant displacement vector} = \sqrt{(-3.31)^2+(5.66)^2}\ km= 6.56\ km\\$

$\textrm{Angle with the positive west} = \theta = \tan^{-1}(\dfrac{5.66}{3.31})= 59.68^\circ$

7 0

4 years ago

HELPPP MEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

almond37 [142]

Answer:

It involves all three methods as the handle of the pot is conducting heat via touching the pan which gets it's heat from the burner that is radiating heat to the entire pan through waves of heat from the burner. There is then convection as the heat is being transferred as the liquids temperature changes.

This should be good if I remember it correctly which I hope so much I do.

6 0

3 years ago

Read 2 more answers

The measurement of an exoplanet radius is measured in units compared to

maksim [4K]

<span>After an exoplanet has been identified using a given detection method, scientists attempt to identify the basic properties of the planet which can tell us what it might be made of, how hot it might be, whether or not it contains an atmosphere, how that atmosphere might behave, and finally, whether the planet may be suitable for life. It is often useful to first determine basic properties of the parent star (such as mass and distance from the Earth). This is then followed by the use of planetary detection methods to calculate planetary mass, radius, orbital radius, orbital period, and density. The density calculation will provide clues as to what the planet is made of and whether or not it contains a significant atmosphere. Mass and Distance of Parent Star The mass and distance of an exoplanet's parent star must often be calculated first, before certain measurements of the exoplanet can be made. For example, determining the star's distance is an important step in determining a star's mass (see below). Knowing the mass of a star then allows the mass of the planet to be measured, for example when using the Radial Velocity Method.</span>

7 0

3 years ago

A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t

lutik1710 [3]

a) E = 0

b) $3.38\cdot 10^6 N/C$

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

$\int EdS=\frac{q}{\epsilon_0}$

where

E is the electric field

q is the charge contained by the Gaussian surface

$\epsilon_0$ is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

$r_1 = 10 cm\\r_2 = 15 cm$

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

$E=\frac{Q}{4\pi \epsilon_0 r^2}$