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3 years ago

Can someone explain how they got their answer or how I get the change in number? :(

Physics

1 answer:

enyata [817]3 years ago

5 0

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

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Substitution solve for x & y<br>-4x - 2y equals 14<br>-10x+7y=-25

den301095 [7]

Answer:

-4×-2y=14 (1)

-10×+7y=-25 (2)

multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2

-28×-14y=98

-20×+14y=-50

___________

-28×=48

×=48/-28

×=-12/7

now

-4×-2y=14

-4*-12/7-2y=14

48/7-2y=14

-2y=14-48/7

-2y=(98-48)/7

-2y=50/7

y=-50/14

y=-25/7

8 0

3 years ago

A light ray traveling through a material with an index of refraction of 1.2 is incident on a material that has an index of refra

zzz [600]

Answer:

compared to the incident angle, the refracted angle is 45.56⁰

Explanation:

From Snell's law;

n₁sin(I) = n₂sin(r)

Where;

n₁ is the refractive index of light in medium 1 = 1.2

n₂ is the refractive index of light in medium 2 = 1.4

I is the incident angle

r is the refractive angle

$n = \frac{1}{sin(I)}\\\\sin(I) = \frac{1}{n}\\\\sin(I) =\frac{1}{1.2}\\\\sin(I) =0.8333\\\\I = sin^-{(0.8333)$

I = 56.439⁰

Applying snell's law

$n_1sin(I) = n_2sin(r)\\\\sin(r) = \frac{n_1sin(I) }{n_2}\\\\sin(r) = \frac{1.2*sin(56.439) }{1.4}\\\\sin(r) = 0.714\\\\r = sin^-(0.714)\\\\r = 45.56^o$

Therefore, compared to the incident angle, the refracted angle is 45.56⁰

3 0

3 years ago

A force of 85 N is used to push a box along the floor a distance of 15 m. How much work was done?

denpristay [2]

Answer:

1275J

Explanation:

Given parameters:

Force on box = 85N

Distance moved = 15m

Unknown:

Work done = ?

Solution:

Work done is the amount of force applied on a body to move it through a specific distance.

Work done = Force x distance

Now insert the parameters and solve;

Work done = 85 x 15 = 1275J

4 0

3 years ago

Read 2 more answers

A force F is used to raise a 4-kg mass M from the ground to a height of 5 m. What is the work done by the force F? (Note: sin 60

miskamm [114]

Answer:

Answer:196 Joules

Explanation:

Hello

Note: I think the text in parentheses corresponds to another exercise, or this is incomplete, I will solve it with the first part of the problem

the work is the product of a force applied to a body and the displacement of the body in the direction of this force

assuming that the force goes in the same direction of the displacement, that is upwards

W=F*D (work, force,displacement)

the force necessary to move the object will be

$F=mg(mass *gravity)\\F=4kgm*9.8\frac{m}{s^{2} }\\ F=39.2 Newtons\\replace\\\\W=39.2 N*5m\\W=196\ Joules$

Answer:196 Joules

I hope it helps

5 0

3 years ago

20 points and brainiest

DIA [1.3K]

Answer:

electrical energy sometimes.

Explanation:

125.0m

300 degree Fahrenheit

7 0

3 years ago