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3 years ago

PLEASE HELP!!!

Physics

1 answer:

Simora [160]3 years ago

8 0

I only know what number 1. is and its Mechanical Energy.

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Find the angle between forces of 41 pounds and 68 pounds given a magnitude of 87 pounds for the resultant force. (Hint: Write fo

dedylja [7]

Answer:

$\theta = 76.9 degree$

Explanation:

As we know that the resultant of two vectors is given as

$R = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 cos\theta$

here we know that

$R = 87 Lb$

$F_1 = 41 Lb$

$F_2 = 68 Lb$

now we have

$87 = \sqrt{41^2 + 68^2 + 2(41)(68)cos\theta$

$87^2 = 6305 + 5576 cos\theta$

$\theta = cos^{-1}(\frac{1264}{5576})$

$\theta = 76.9 degree$

7 0

3 years ago

A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is

erica [24]

The correct answer for the question that is being presented above is this one: "Schmidt-Cassegrain focus." A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is a Schmidt-Cassegrain focus.
Here are the following choices:
a. Cassegrain focus
b. Newtonian focus
c. Schmidt-Cassegrain focus
<span>d. Schmidt focus</span>

4 0

3 years ago

MAXImum [283]

Answer:

true

Explanation:

Newton is the measure of the force with turns to be gravity multiplying the mass. Thus, the forces acts on the particles in the direction of the movement of the particles

7 0

3 years ago

John goes grocery shopping with his mother. His job is to push the cart. The cart is

lora16 [44]

Answer:

Beacause he has more grocceries and food heavy

Explanation:

7 0

3 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

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

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

#SPJ4

6 0

2 years ago