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2 years ago

Pls help I give brainLy and a thanks and points

Physics

2 answers:

inna [77]2 years ago

8 0

Yeah yeah I can

But it's you
You're the one that I want
You
Everything that I want
Blue
Nae gyeote naerin blue, modeun ge bakkwieo
Daseotshi oshipsambune i segyeneun areumdaweo

Cuz of imagination
Jeo haneure
Orenjibit mabeobi
Kkeuchi nagi jeone

Cuz of imagination
Geu challae
Can you feel the rush
Can you feel the rush
Shigana meomchweojweo

I wanna stay
(Can you feel the rush)
(Can you feel the rush)

Lina20 [59]2 years ago

5 0

Awww the dog is so cute!

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If a car with a mass of 5Kg, has a momentum of 25kgm/s, how fast is it travelling?

Fantom [35]

Answer:

5m/s

Explanation:

p=mv, or momentum (p) is equal to mass (m) times velocity (v).

so:

m=5Kg

p=25Kgm/s

v=p÷m

v=25÷5

v=5m/s

hoped this helped :)

8 0

2 years ago

Read 2 more answers

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

Irina18 [472]

Answer:

a. $q_C$ experiences the greatest net force and $q_B$ experiences the smallest net force

b. Ratio of the greatest to the smallest net force= 9

Explanation:

<u>Electrostatic Forces </u>

Two point-charges q1 and q2, separated a distance d, exert on each other an electrostatic force of magnitude

$\displaystyle F=K\frac{q_1q_2}{d^2}$

If the charges have the same sign, they repel each other, for different signed charges, they attract. That gives us the direction of each force in the space.

Let's assume all the charges of the problem have a magnitude q, and between two consecutive charges, the distance is d. The proposed layout is shown it the image.

The net force on qA is the sum of those exerted by qB, qC, and qD. But note qB and qC repel qA and qD attracts it, so the total force on qA is

$F_{TA}=-F_B-F_C+F_D$

Computing the individual forces we have

$\displaystyle F_B=\frac{K\ q_A\ q_B}{d^2}=K\ \frac{q^2}{d^2}$

$\displaystyle F_C=\frac{K\ q_A\ q_C}{(2d)^2}=\frac{1}{4}\ \ \frac{K\ q^2}{d^2}$

$\displaystyle F_D=\frac{K\ q_A\ q_D}{(3d)^2}=\frac{1}{9}\ \ \frac{K\ q^2}{d^2}$

The total force on qA is:

$\displaystyle F_{TA}=\frac{K\ q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_{TA}=-\frac{41}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TA}|=\frac{41}{36}\ \frac{K\ q^2}{d^2}$

Charge qA repels qB to the right, qC repels qB to the left, and qD attracts qB to the right, thus

$\displaystyle F_{TB}=F_A-F_C+F_D$

$\displaystyle F_{TB}=\frac{K\ q^2}{d^2}-\frac{K\ q^2}{d^2}+\frac{K\ q^2}{(2d)^2}$

$\displaystyle F_{TB}=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TB}|=\frac{1}{4}\ \frac{K\ q^2}{d^2}$

Charges qA and qb repel qC to the right, and qD attracts qC to the right, thus

$\displaystyle F_{TC}=F_A+F_B+F_D$

$\displaystyle F_{TC}=\frac{K\ q^2}{(2d)^2}+\frac{K\ q^2}{d^2}+\frac{K\ q^2}{d^2}$

$\displaystyle F_{TC}=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TC}|=\frac{9}{4}\ \frac{K\ q^2}{d^2}$

Charge qA and qB attract qD to the left, and qC atracts qD to the left, thus

$\displaystyle F_{TD}=-F_A-F_B-F_C$

$\displaystyle F_{TD}=-\frac{K\ q2}{(3d)^2}-\frac{K\ q2}{(2d)^2}-\frac{K\ q2}{d^2}$

$\displaystyle F_{TD}=-\frac{49}{36}\ \frac{K\ q^2}{d^2}$

$\displaystyle |F_{TD}|=\frac{49}{36}\ \frac{K\ q^2}{d^2}$

Comparing the relative values of all the forces

$\displaystyle |F_{TC}|>|F_{TD}|>|F_{TA}|>|F_{TB}|$

This means that qc experiences the greatest net force and qB experiences the smallest net force

The ratio of the greatest to the smallest forces is

$\displaystyle \frac{|F_{TC}|}{|F_{TB}|}=\frac{\frac{9}{4}}{\frac{1}{4}}=9$

5 0

2 years ago

The weight of a body above sea level varies inversely with the square of the distance from the center of Earth. If a woman weigh

Serjik [45]

Answer:

89.16pounds

Explanation:

The equation that defines this problem is as follows

W=k/X^2

where

W=Weight

K= proportionality constant

X=distance from the center of Earth

first we must find the constant of proportionality, with the first part of the problem

k=WX^2=131x3960^2=2054289600pounds x miles^2

then we use the equation to calculate the woman's weight with the new distance

W=2054289600/(4800)^2=89.16pounds

4 0

2 years ago

580 nm light shines on a double slit

lesantik [10]

Answer:

Angular position of the first minimum is 0.27 degree

Explanation:

As we know by the formula of single slit diffraction pattern

$a sin\theta = m\lambda$

here we have

a = d = 0.000125 m

for m = 1

$\lambda = 580 nm$

now we have

$0.000125 sin\theta = 580 \times 10^{-9}$

$sin\theta = 4.64 \times 10^{-3}$

$\theta = 0.27 degree$

6 0

2 years ago

Find the pressure exerted by a force 8000 newton's on an area of 25m^2. Give your answer in newton's/n^2

Vanyuwa [196]

The pressure exerted by this force is equal to 320 $N/m^2$ .

<u>Given the following data:</u>

Force = 8000 Newton.

Area = 25 $m^2$ .

To calculate the pressure exerted by this force:

<h3>How to calculate pressure.</h3>

Mathematically, pressure is given by this formula:

$P=\frac{F}{A}$

<u>Where:</u>

P is the pressure.

F is the force.

A is the area.

Substituting the given parameters into the formula, we have:

$P = \frac{8000}{25}$

P = 320 $N/m^2$ .

Read more on pressure here: brainly.com/question/8033367

5 0

2 years ago