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ad-work [718]
4 years ago
15

In a 5.00 km race, one runner runs at a steady 11.4 km/h and another runs at 14.7 km/h . How long does the faster runner have to

wait at the finish line to see the slower runner cross?
Physics
1 answer:
topjm [15]4 years ago
8 0

Answer:

0.0986 h or 5 minutes 55 seconds.

Explanation:

Speed: This can be defined as the rate of change of distance of a body. The S.I unit of speed is m/s. Speed is a scalar quantity, because it can only be represented by magnitude alone.

Mathematically,

Speed = distance/time.

S = d/t ........................... Equation 1

making t  the subject  of the equation

t = d/S ......................... Equation 2

Form the question,

Time taken for the faster runner to reach the finish line

t₁ = d/S₁................... Equation 3

Where t₁ = time taken for the faster runner to reach the finish line, d = distance, S₁ = speed of the faster runner.

Given: d = 5.0 km, S₁ = 14.7 km/h.

Substituting into equation 3

t₁ = 5/14.7

t₁ = 0.340 h

Also,

t₂ = d/S₂................... Equation 4

Where t₂ = time taken for the slower runner to reached the finished line, d = distance, S₂ = speed of the slower runner.

Given: d = 5 km, S₂ = 11.4 km/h.

Substitute into equation 4,

t₂ = 5/11.4

t₂ = 0.4386 h.

The time the faster runner have to wait at the finish line to see the slower runner cross = t₂ - t₁ = 0.4386-0.340

The time the faster runner have to wait at the finish line to see the slower runner cross = 0.0986 h = 5 mins 55 s.

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A 2.2 m long simple pendulum oscillates with a period of 4.8 s on the surface of unknown planet. What is the surface gravity of
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Answer:

g=3.76\ m/s^2

Explanation:

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We need to find the surface gravity of the planet. The time period of the planet is given by the relation as follows :

T=2\pi \sqrt{\dfrac{l}{g}} \\\\T^2=4\pi ^2\times \dfrac{l}{g}\\\\g=\dfrac{4\pi ^2 l}{T^2}

Put all the values,

g=\dfrac{4\pi ^2 \times 2.2}{(4.8)^2}\\\\=3.76\ m/s^2

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3 years ago
What might cause a shotgun to explode? <br> this is due in like 10 mins so help me please
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At which temperature do the lattice and conduction electron contributions to the specific heat of Copper become equal.
sp2606 [1]

Answer:

At 3.86K

Explanation:

The following data are obtained from a straight line graph of C/T plotted against T2, where C is the measured heat capacity and T is the temperature:

gradient = 0.0469 mJ mol−1 K−4 vertical intercept = 0.7 mJ mol−1 K−2

Since the graph of C/T against T2 is a straight line, the are related by the straight line equation: C /T =γ+AT². Multiplying by T, we get C =γT +AT³ The electronic contribution is linear in T, so it would be given by the first term: Ce =γT. The lattice (phonon) contribution is proportional to T³, so it would be the second term: Cph =AT³. When they become equal, we can solve these 2 equations for T. This gives: T = √γ A .

We can find γ and A from the graph. Returning to the straight line equation C /T =γ+AT². we can see that γ would be the vertical intercept, and A would be the gradient. These 2 values are given. Substituting, we f ind: T =

√0.7/ 0.0469 = 3.86K.

4 0
3 years ago
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