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Luden [163]
3 years ago
9

(b) If the object is at 330 feet and its instantaneous velocity is 3 feet per minute at 30 minutes, what is the approximate posi

tion of the object at 32 minutes
Physics
1 answer:
ELEN [110]3 years ago
8 0

Answer:

The final position is 36 feet.

Explanation:

initial position, d = 330 feet

speed, v = 3 feet per minute

time, t = 30 minute

now the time is 32 minute

time interval = 2 minute

So, the distance in 2 minutes is

d' = 2 x 3 = 6 feet

So, the final position is

D = 30 + 6 = 36 feet

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The speed of car is 100.8km/h

KE =  \frac{1}{2} m(v \: truck ) {}^{2}

= 0.5 \times 18777 \times  \frac{26}{3.6} \times  \frac{26}{3.6}

= 489708.79j

=  \frac{1}{2}  \times 1248 \times( v \: car) {}^{2}

= 624v {}^{2}

so \: v {}^{2}  =  \frac{489708.7}{624}

= 784

v =  \sqrt{784}

v = 28m/s

v car= 28×3.6

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Solve for the unknown in each of these circuits
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<u> Ohms law: </u> This law relates voltage difference between two points. Mathematically, the law states that V=IR;

                Where

                          V = voltage difference ; in volts

                          I = Current ; in Amperes

                          R = Resistance ; in ohms

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From ohms law,    I = V/R

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<u>2. Answer:</u>  given that R = 10 ; V= ? ;  I = 5

From ohms law,    V = IR

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<u>3 . Answer:</u>  given that R = ? ; V= 120 ;  I = 5

From ohms law,    R =  V/I

                                 = 120/5

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From ohms law,    R =  V/I

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<u>5 . Answer:</u>  given that R = 480 ; V= 24 ;  I = ?

From ohms law,  I =  V/R

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From ohms law,  V = IR

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