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Makovka662 [10]
3 years ago
6

A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i

n terms of the number of seconds, t , since the car started moving. Suppose t varies from t = 2 to t = 5 . Does the car travel at a constant speed over this interval of time? Correct What is the car's average speed over this interval of time? 10 Incorrect feet per second Suppose t varies from t = 1.8 to t = 2.1 . Does the car travel at a constant speed over this interval of time? Correct What is the car's average speed over this interval of time?
Physics
1 answer:
Ede4ka [16]3 years ago
4 0

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2   Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft     and      d(5) = 35 ft

V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft     and      d(2.1) = 8.61 ft

V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s

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Suppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When car
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W=2219pounds

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6 0
3 years ago
Two resistances, R1 and R2, are connected in series across a 9-V battery. The current increases by 0.450 A when R2 is removed, l
Rina8888 [55]

Answer:

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

Explanation:

Since the resistors R1 and R2 are connected in series, the current flowing through them when the 9 V battery is applied is 9/R1 + R2.

When the current increases by 0.450 A wen only R1 is in the circuit, the current is

9/R1 + R2 + 0.450 A = 9/R1       (1)

When the current increases by 0.225 A when only R2 is in the circuit, the current is

9/R1 + R2 + 0.225 A = 9/R2       (2)

equation (1) - (2) equals

9(1/R1 - 1/R2) = 0.450 A - 0.225

9(1/R1 - 1/R2) = 0.125

(1/R1 - 1/R2) = 0.125 A/9 = 0.0138

1/R1 = 0.0138 + 1/R2

R1 = R2/(1 + 0.0138R2)     (3)

From (1)

9/R1 - 9/R1 + R2 = 0.450 A

9R2/[R1(R1 + R2)] = 0.450 A

R2/[R1(R1 + R2)] = 0.450 A/9 = 0.5

R2/[R1(R1 + R2)] = 0.5    (4)

From (3) R2/R1 = (1 + 0.0138R2) and from (4) R2/R1 = 0.5(R1 + R2). So,

(1 + 0.0138R2) = 0.5(R1 + R2)

0.5R1 + 0.5R2 = 1 + 0.0138R2

0.5R1 = 1 + 0.0138R2 - 0.5R2

0.5R1 = 1 - 0.4862R2        (5)

Substituting (3) into (5) we have

0.5R2/(1 + 0.0138R2) = 1 - 0.4862R2

R2 = (1 + 0.0138R2)(1 - 0.4862R2)

R2 = 1 - 0.4724R2 - 0.0067R2²

Collecting like terms, we have

0.0067R2² + 0.4724R2 + R2 - 1 = 0

0.0067R2² + 1.4724R2 - 1 = 0

Using the quadratic formula,

R_{2} = \frac{-1.4724 +/-\sqrt{(1.4724)^{2} - 4 X 0.0067 X -1} }{2 X 0.0067}  \\= \frac{-1.4724 +/-\sqrt{2.1680 + 0.0268} }{0.0268}\\= \frac{-1.4724 +/-\sqrt{2.1948} }{0.0268}\\= \frac{-1.4724 +/- 1.4815 }{0.0268}\\= \frac{-1.4724 + 1.4815 }{0.0268} or \frac{-1.4724 - 1.4815 }{0.0268}\\= \frac{0.0091 }{0.0268} or \frac{-2.9539}{0.0268}\\= 0.340 or -110.22

We choose the positive answer.

So R2 = 0.340 Ω

From (5)

R1 = 0.5 - 0.9931R2

   = 0.5 - 0.9931 × 0.340

   = 0.5 - 0.338

   = 0.162 Ω

a. R1 = 0.162 Ω

b. R2 = 0.340 Ω

5 0
2 years ago
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