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NISA [10]
2 years ago
7

A 11-inch candle is lit and burns at a constant rate of 0.9 inches per hour. Let t t represent the number of hours since the can

dle was lit, and suppose f f is a function such that f ( t ) f(t) represents the remaining length of the candle (in inches) t t hours after it was lit. Write a function formula for f f.
Physics
1 answer:
Serggg [28]2 years ago
8 0

Answer: F(t) = 11 - 0.9(t)

Explanation:

We know the following:

The candle burns at a ratio given by:

Burning Ratio (Br)  = 0.9 inches / hour

The candle is 11 inches long.

To be able to create a function that give us how much on the candle remains after turning it after a time (t). We will need to know how much of the candle have been burned after t.

Let look the following equation:

Br = Candle Inches (D) / Time for the Candle to burn (T)      (1)

Where (1) is similar to the Velocity equation:

Velocity (V) = Distance (D)/Time(T)

This because is only a relation between a magnitude and time.

Let search for D on (1)

D = Br*T  (2)

Where D is how much candle has been burn in a specif time

To create a function that will tell us how longer remains of the candle after be given a variable time (t) we use the total lenght minus (2):

How much candle remains? ( F(t) ) = 11 inches - Br*t

F(t) = 11 - 0.9(t)

F(t) defines the remaining length of the candle t hours after being lit

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kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from 15 cm to 21 cm will be , E_{1}=27J

The energy used for stretch the spring from 21 cm to 27 cm will be , E_{2}=45J

using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm

Therefore, the natural length of the spring is, 9 cm.

4 0
2 years ago
A car is traveling at an average speed of 70 m/s. How many km would the car travel in 6.5 hrs. ?
docker41 [41]

Answer:

<h2>38,769.23 miles</h2>

Explanation:

given:

A car is traveling at an average speed of 70 m/s.

find:

How many km would the car travel in 6.5 hrs. ?

solution:

distance = velocity over time

let velocity = 70 m/s

           time = 6.5 hrs.

convert velocity 70 m/s into m/h for consistency of units.

<u>70 mi. </u> x   <u>3600 sec.</u>  =  252,000 mi/hour

  sec.          1 hr.

now plugin values into the formula d = v/t

d = <u>252,000 miles/hour </u>

              6.5 hours

d = 38,769.23 miles

therefore, the distance travelled in 6.5 hours with a speed of 70 m/s is 38,769.23 miles

5 0
3 years ago
7. If a person starts from a standstill and the person can accelerate at 1 m/s/s, how long will it take for the person to get up
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Use the first kinematic formula
Vf = Vi + at
10 = 0 + 1(t)
10 = t

10 seconds
6 0
2 years ago
How much power does it take to lift 30.0 N 10.0 m high in 5.00 s?
Rudik [331]

Power = work/time = (Force times distance)/time

= (30N *10.0m)/5.00s = 300/5 = 60 Watts

7 0
3 years ago
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The sport with the fastest moving ball is jai alai, where measured speeds can be 286 km/h. If a professional jai alai player fac
cricket20 [7]

Answer:

d= 794.4 cmExplanation:

Given that

Speed ,V= 286 km/h

=286\times \dfrac{1000}{3600}\ m/s

V=79.44 m/s

Given that time ,t= 100 ms

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d= V t

Now by putting the values in the above equation

d = 79.44 x 0.1 m

d= 7.944 m

We know that 1 m = 100 cm

d= 794.4 cm

5 0
3 years ago
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