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2 years ago

12

Pls help i’ll give brainliest if you give a correct answer!!

1 answer:

enot [183]2 years ago

4 0

Answer:

the second one

Explanation:

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A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

5 0

2 years ago

What information can be determined by observing dark lines within the spectrum of a given star?

Scrat [10]

I believe the answer is D.

6 0

2 years ago

NEED HELP ASAP!!!!! IN MIDDLE OF TEST RN HELP PLZ

svlad2 [7]

ONEEEEEEEEEEEEEEEEEEEEEEEEE

4 0

2 years ago

Read 2 more answers

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N push

atroni [7]

Answer:

0.8214 m/s^2

Explanation:

Fnet= Fpushed - Ffriction

Fpushed = 12.7N Ffriction = 8.33N

Fnet = 12.7N - 8.33N = 4.37N

Fnet= mass(acceleration)

Fnet = 4.37N mass = 5.32 kg

4.37N = 5.32 kg(acceleration)

acceleration= 0.8214 m/s^2

3 0

2 years ago