A star that is several thousand dimmer than the sun with a temperature than 10000 k would be classified as

Given the vector A with components Ax = 2.00, Ay = 6.00, the vector B with components Bx = 2.00, By = 22.00, and the vector D =

nekit [7.7K]

Answer:

<em>The magnitude of vector d is 16 and the angle with the x-axis is 270°</em>

Explanation:

<u>Operations With Vectors</u>

Given two vectors in rectangular components:

$\vec a=(ax,ay)\ ,\ \vec b=(bx,by)$

The sum of the vectors is:

$\vec a+\vec b=(ax+bx,ay+by)$

The difference between the vectors is:

$\vec a-\vec b=(ax-bx,ay-by)$

The magnitude of $\vec a$ is:

$|\vec a|=\sqrt{ax^2+ay^2}$

The angle $\vec a$ makes with the horizontal positive direction is:

$\displaystyle \tan\theta=\frac{ay}{ax}\\$

The question provides the vectors:

$\vec a=(2,6)$

$\vec b=(2,22)$

$\vec d=\vec a-\vec b$

Calculate:

$\vec d=(2,6)-(2,22)=(0,-16)$

The magnitude of $\vec d$ is:

$|\vec d|=\sqrt{0^2+(-16)^2}=\sqrt{0+256}=16$

The angle is calculated by:

$\displaystyle \tan\theta=\frac{-16}{0}$

The division cannot be calculated because the denominator is zero. We need to estimate the correct angle by looking at the components of the vector. Since the x-coordinate is zero and the y-coordinate is negative, the vector points downwards (south), thus the angle must be -90° or 270° if the range goes from 0° to 360°.

The magnitude of vector d is 16 and the angle with the x-axis is 270°

4 0

3 years ago

The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra

Scrat [10]

Answer:

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

Explanation:

Given

Number of protons $= 92$

Radius of nucleus $r_n = 7.4 * 10^{-15} m$

Distance of the electrons $r_1 = 1.0 * 10^ {-10} m$

Part 1

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21} N/C$

Part 2

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13} N/C$

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

7 0

3 years ago

An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)

Alecsey [184]

"60 kg" is not a weight. It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is

   (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

   Weight = (60 kg) x (9.8 m/s²)

      =    588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to 5R from the center, the gravity out there is

(1R/5R)² = (1/5)² = 1/25 = 0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

   (0.04) x (588 Newtons) = 23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink.
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.

6 0

3 years ago

Which of the following is most likely not a case of uniform circular motion

Ivan

Answer:

Motion of a racing car on a circular track

Explanation:

Uniform circular motion means the motion of the object is in a circle with a CONSTANT SPEED.

The racing car will accelerate during its motion. Hence, it is not a uniform circular motion.

7 0

2 years ago

Can someone please help me

pickupchik [31]

Answer:C (198 seconds)

Explanation: The cyclist makes the first lap in (180.00 - 6.00) = 174.00 seconds. The average time per lap for all three circuits is (600.00 - 6.00) = 594/3 = 198 seconds.

7 0

3 years ago