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Inessa05 [86]
3 years ago
9

Points A (-9,2), B (2,-9), and C (-9,-9) are placed in three different quadrants of a Cartesian coordinate system. Convert each

set of Cartesian coordinates to polar coordinates. The angle should be reported as a positive angle, in degrees, from the positive x-axis. (0° < θ < 360°)
Physics
1 answer:
slavikrds [6]3 years ago
5 0

Answer:

answer

Explanation:

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What type of matter is absorbent
Elanso [62]
Gas, liquid, or solid
8 0
3 years ago
Human reaction times are worsened by alcohol. How much further (in feet) would a drunk driver's car travel before he hits the br
sweet-ann [11.9K]

Answer:

A drunk driver's car travel 49.13 ft further than a sober driver's car, before it hits the brakes

Explanation:

Distance covered by the car after application of brakes, until it stops can be found by using 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,  

Vf = Final Velocity of Car = 0 mi/h

Vi = Initial Velocity of Car = 50 mi/h

a = deceleration of car  

s = distance covered

Vf, Vi and a for both drivers is same as per the question. Therefore, distance covered by both car after application of brakes will also be same.

So, the difference in distance covered occurs before application of brakes during response time. Since, the car is in uniform speed before applying brakes. Therefore, following equation shall be used:

s = vt

FOR SOBER DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 0.33 s

s = s₁

Therefore,

s₁ = (73.33 ft/s)(0.33 s)

s₁ = 24.2 ft

FOR DRUNK DRIVER:

v = (50 mi/h)(1 h/ 3600 s)(5280 ft/mi) = 73.33 ft/s

t = 1 s

s = s₂

Therefore,

s₂ = (73.33 ft/s)(1 s)

s₂ = 73.33 ft

Now, the distance traveled by drunk driver's car further than sober driver's car is given by:

ΔS = s₂ - s₁

ΔS = 73.33 ft - 24.2 ft

<u>ΔS = 49.13 ft</u>

6 0
3 years ago
ANSWERRR PLEASEEEE!!!
GuDViN [60]

Answer:

50N to the left

Explanation:

I'll explain if you ask

6 0
2 years ago
Read 2 more answers
An electron in the beam of a cathod-ray tube is accelerated by a potential difference of 2.12 kV . Then it passes through a regi
son4ous [18]

Answer:

B=9.1397*10^-4 Tesla

Explanation:

To find the velocity first we put kinetic energy og electron is equal to potential energy of electron

K.E=P.E

\frac{1}{2}*m*v^{2}  =e*V

where :

m is the mass of electron

v is the velocity

V is the potential difference

v=\sqrt{\frac{2*e*V}{m} }    eq 1

Radius of electron moving in magnetic field is given by:

R=\frac{m*v}{q*B}       eq 2

where:

m is the mass of electron

v is the velocity

q=e=charge of electron

B is the magnitude of magnetic field

Put v from eq 1 into eq 2

R=\frac{m*\sqrt{\frac{2*e*V}{m} } }{e B}

B=\sqrt{\frac{2*m*V}{e*R^{2} } }

B=\sqrt{\frac{2*(9.31*10^{-31})*(2.12*10^{3})  }{(1.60*10^{-19})*(0.170)^{2}  } }

B=9.1397*10^-4 Tesla

3 0
3 years ago
What is the minimum energy required to lift an object weighing 200 Newtons at a height of 20 meters?
lakkis [162]
Work = force x distance

200 Newtons x 20 meters

= 4,000 Joules
4 0
3 years ago
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