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kherson [118]
4 years ago
12

A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to

the nearest tenth of a kilometer?
Physics
2 answers:
harkovskaia [24]4 years ago
8 0

Answer:

The work for the previous answer is correct, however, since the final velocity is in the southward direction, the answer is -4.3.

Explanation:

Edgenuity will tell you this is the correct answer as well.

liberstina [14]4 years ago
6 0
60 ° is the angle between W- direction and the run direction.

You need the angle betwee S-direction and the run direction. This is 90° - 60° = 30 °.

By geometry you can trace a right triangle, where the S-component is the adyacent side and the run  is the hypotenuse.

Then cos 30° = adyacent side / hypotenuse = S-component / run

Then S-component = run * cos 30° = 5.0 km * 0.866 = 4.3 km

Answer: 4.3 km
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Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the sp
Alekssandra [29.7K]

Answer:

2.92682 m

1.5\times 10^{18}\ Hz

250000000000 Hz

2.88462\times 10^{-8}\ m

0.21126 m

0.12244 m

Explanation:

c = Speed of light = 3\times 10^8\ m/s

\lambda = Wavelength

f = Frequency

Wavelength is given by

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{102.5\times 10^6}\\\Rightarrow \lambda=2.92682\ m

The wavelength is 2.92682 m

Frequency is given by

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{0.2\times 10^{-9}}\\\Rightarrow f=1.5\times 10^{18}\ Hz

The frequency is 1.5\times 10^{18}\ Hz

f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{1.2\times 10^{-3}}\\\Rightarrow f=250000000000\ Hz

The frequency is 250000000000 Hz

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.04\times 10^{16}}\\\Rightarrow \lambda=2.88462\times 10^{-8}\ m

The wavelength is 2.88462\times 10^{-8}\ m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.42\times 10^{9}}\\\Rightarrow \lambda=0.21126\ m

The wavelength is 0.21126 m

\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{2.45\times 10^9}\\\Rightarrow \lambda=0.12244\ m

The wavelength is 0.12244 m

8 0
3 years ago
A student has a displacement of 304 m north in 180 s. What was the student's average velocity?
DanielleElmas [232]

Answer:

v = 1.69 m/s

Explanation:

Given that,

Displacement of the student is 304 m due North and it takes 180 s.

We need to find the student's average velocity. Using formula of velocity.

Velocity = displacement/time

v=\dfrac{304\ m}{180\ s}\\\\v=1.69\ m/s

Hence, the student's average velocity is 1.69 m/s.

6 0
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A student examines the effect of the number of D batteries in a closed circuit on the brightness of a light bulb. In the experim
xxTIMURxx [149]

Answer:

Placing cells in series increases the

voltage in the circuit by 1.5 V for each

cell. Increasing the voltage increases the

brightness of the bulb

Explanation:

I. MAKING THE CONNECTION

How many terminals are located on the battery? 2

How many terminals are located on the bulb? 2

II. PLACING CELLS IN SERIES

What is the effect on the brightness of the bulb by increasing the number of cells?

The bulbs become brighter when increasing the number of cells.

What changes occur in the current in the circuit when increasing the number of cells?

Increasing the number of cells increases the current in the circuit.

What changes occur in the voltage in the circuit when increasing the number of cells?

Increasing the number of cells increases the voltage (for cells in series the voltage is additive).

What changes occur in the resistance in the circuit when increasing the number of cells?

The resistance is determined by the number of bulbs. The resistance in the circuit remains unchanged.

III. PLACING BULBS IN SERIES

What is the effect on the brightness of the bulbs by increasing the number of bulbs?

Increasing the number of bulbs decreases the brightness of the bulbs.

What changes occur in the resistance in the circuit as more bulbs are added?

The resistance increases. In a series circuit, adding bulbs increases the resistance in the circuit.

What changes occur in the current in the circuit as more bulbs are added?

Increasing the resistance decreases the current.

Observations on unscrewing one bulb. Explain your observations.

A complete circuit requires the electrons to move from the negative terminal of the battery to the

positive terminal. When one bulb is unscrewed it opens the circuit preventing a complete circuit and

the electrons cannot return to the battery.

IV. PLACING BULBS IN PARALLEL

Compare the brightness with two bulbs in parallel with two bulbs in series.

Two bulbs in parallel are brighter than two bulbs in series.

How does increasing the number of circuits (bulbs) change the current and resistance?

In a parallel circuit each bulb is in its own circuit. As bulbs are added the resistance in the circuit

decreases since each circuit is another pathway for electrons to move from one end of the circuit

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