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Vikki [24]
3 years ago
10

A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this

time
find distance and acceleration
FAST PLEASE !!
Physics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

-1.2×10^2 m

I hope this helps you

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An ellipse has two focal points. One of the focal points is the _____. earth, moon, or Sun
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Water pollution that elevates the temperature of the water​
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An increase in the air temperature will cause water temperatures to increase as well. As water temperatures increase, water pollution problems will increase, and many aquatic habitats will be negatively affected.

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Lower levels of dissolved oxygen due to the inverse relationship that exists between dissolved oxygen and temperature. As the temperature of the water increases, dissolved oxygen levels decrease.

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4 years ago
PLEASE HELP!!! WOULD REALLY APPRECIATE!
Eva8 [605]

Answer:

a. slope=rise/run

rise=0.02

run=-2

determined using the point (3,0.08) and (1,0.1) on the graph

slope=0.02/-2

= -0.01 or -1/100

b.area= area of trapizoid+ rectangle

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0.36+0.07

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4 years ago
A 200-Ω resistor is connected in series with a 10-µF capacitor and a 60-Hz, 120-V (rms) line voltage. If electrical energy costs
Vika [28.1K]

Answer:

Cost to leave this circuit connected for 24 hours is $ 3.12.

Explanation:

We know that,

\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \pi \mathrm{fc}}

f = frequency (60 Hz)

c= capacitor (10 µF = 10^-6)  

\mathrm{X}_{\mathrm{c}}=\text { Capacitive reactance }

Substitute the given values

\mathrm{X}_{\mathrm{c}}=\frac{1}{2 \times 3.14 \times 10 \times 10^{-6} \times 60}

\mathrm{X}_{\mathrm{c}}=\frac{1}{3.768 \times 10^{-3}}

\mathrm{x}_{\mathrm{c}}=265.39 \Omega

Given that, R = 200 Ω

X^{2}=R^{2}+X c^{2}

X^{2}=200^{2}+265.39^{2}

X^{2}=40000+70431.85

X^{2}=110431.825

x=\sqrt{110431.825}

X = 332.31 Ω

\text { Current }(I)=\frac{V}{R}

\text { Current }(I)=\frac{120}{332.31}

Current (I) = 0.361 amps

“Real power” is only consumed in the resistor,  

\mathrm{I}^{2} \mathrm{R}=0.361^{2} \times 200

\mathrm{I}^{2} \mathrm{R}=0.1303 \times 200

\mathrm{I}^{2} \mathrm{R}=26.06 \mathrm{Watts} \sim 26 \mathrm{watts}

In one hour 26 watt hours are used.

Energy used in 54 hours = 26 × 24 = 624 watt hours

E = 0.624 kilowatt hours

Cost = (5)(0.624) = 3.12  

3 0
3 years ago
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