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Pie
3 years ago
12

Steve and Jose drove Jose's car to the beach during Spring Break. The 495 kilometer drive took them 8 hours and 30 minutes. What

was their average speed for the trip?
Physics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

58.24 Km/h.

Explanation:

From the question given above, the following data were obtained:

Distance (d) = 495 Km

Time (t) = 8 h 30 mins

Speed (S) =?

Next, we shall express 8 hours 30 mins to hours.

We'll begin by convert 30 mins to hour.

60 mins = 1 h

Therefore,

30 mins = 30 mins × 1 h/ 60 mins

30 mins = 0.5 hour.

Thus,

8 h 30 min = 8 + 0.5 = 8.5 hours

Speed is define as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance /time

With the above formula, we can obtain the speed as shown below:

Distance (d) = 495 Km

Time (t) = 8.5 hour

Speed (S) =?

Speed = Distance /time

Speed = 495 Km / 8.5 hour

Speed = 58.24 Km/h

Thus, the speed is 58.24 Km/h.

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A metal wire has a resistance of 14.00 Ω at a temperature of 25.0°C. If the same wire has a resistance of 14.55 Ω at 90.0°C, wha
aliina [53]

Answer:

13.52 Ω

Explanation:

coefficient of thermal resistance be α

R₀ , R₂₅ , R₉₀ and R₋₃₂ be resistances at 0 , 25 , 90 , and - 32 degree

R₂₅ = R₀ + α x 25

R₉₀ = R₀ + α x 90

R₉₀ - R₂₅ = 65 x α

α = (R₉₀ - R₂₅ )/ 65

= (14.55 - 14) / 65

=   .55 / 65 Ω per °C,

R₂₅ = R₀ + α x 25

14 = R₀ + (.55 / 65 )x 25

=  R₀ + .2115

R₀ = 13.7885 Ω

R₋₃₂ = R₀ - α x 32

= 13.7885 -(  .55 / 65) x 32

=  13.7885 - .27077

= 13.51773 Ω

= 13.52 Ω

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4 years ago
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3 years ago
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A 2-m long string is stretched between two supports with a tension that produces a wave speed equal to vw=50.00m/s. What are the
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Answer

given,

Length of the string, L = 2 m

speed of the wave , v = 50 m/s

string is stretched between two string

For the waves the nodes must be between the strings

the wavelength  is given by

           \lambda = \dfrac{2L}{n}

where n is the number of antinodes; n = 1,2,3,...

the frequency expression is given by

            f = n\dfrac{v}{2L}

now, wavelength calculation

      n = 1

           \lambda_1 = \dfrac{2\times 2}{1}

                    λ₁ = 4 m

      n = 2

           \lambda_2 = \dfrac{2\times 2}{2}

                   λ₂ = 2 m

      n =3

           \lambda_3 = \dfrac{2\times 2}{3}

                    λ₃ = 1.333 m

now, frequency calculation

      n = 1

            f = n\dfrac{v}{2L}

            f_1 =1\times \dfrac{50}{2\times 2}

                    f₁ = 12.5 Hz

      n = 2

            f = n\dfrac{v}{2L}

            f_2 =2\times \dfrac{50}{2\times 2}

                    f₂= 25 Hz

      n = 3

            f = n\dfrac{v}{2L}

            f_3 =3\times \dfrac{50}{2\times 2}

                    f₃ = 37.5 Hz

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