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3 years ago

During a neap tide, which phase could the moon be in? select all that apply

Physics

2 answers:

Ne4ueva [31]3 years ago

8 0

Answer:

b. third quarter

c. first quarter

Explanation:

Neap tides happen in the third and first quarter of the moon phase, when the moon appears to be half full, this is because of the position of the moon and the gravitational force of the moon attracting the tides along with the gravitational force of the sun, tha causes high tides on new, and full phases, but the third quarter and first quarter, the moon makes a 90º line with the solar gravitational forces, cancelling out both forces making it low tide or neap tide.

snow_lady [41]3 years ago

4 0

Hello,

Here is your answer:

The proper answer to this question is option B "<span>third quarter".

Here is how:

Neap tides are caused when the moon is three thirds full which causes the current to up rise therefore making a neap tide.

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps!

</span>

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What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport

ad-work [718]

Answer: $1.13(10)^{3} Pa$

Explanation:

This problem can be solved by the following equation:

$\Delta P=\frac{8 \eta L Q}{\pi r^{4}}$

Where:

$\Delta P$ is the pressure difference between the two ends of the pipe

$\eta=0.20 Pa.s$ is the viscosity of oil

$L=2.6 km=2600 m$ is the length of the pipe

$Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s}$ is the Rate of flow of the fluid

$d=36 cm=0.36 m$ is the diameter of the pipe

$r=\frac{d}{2}=0.18 m$ is the radius of the pipe

Soving for $\Delta P$ :

$\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}$

Finally:

$\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa$

7 0

3 years ago

Which one of the following is a correct statement......

IceJOKER [234]

Answer:

Just 3

Explanation:

I believe the other two are incorrect

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3 years ago

Can someone pls match these

Lostsunrise [7]

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3 years ago

Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same

Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

$\underset{V_{A}}{\rightarrow}$ = velocity of car A before collision = 0 i - $V_{A}$ j

$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$ $\underset{V_{A}}{\rightarrow}$ + $M_{B}$ $\underset{V_{B}}{\rightarrow}$ = ( $M_{A}$ + $M_{B}$ ) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ( $V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600 and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h and $V_{A}$ = 34.4 km/

6 0

3 years ago

Nerve impulses in a human body travel at a speed of about 100 m/s. Suppose a person accidentally steps barefoot on a pebble. Abo

PilotLPTM [1.2K]

Answer:

The time required by the impulse to travel from foot to brain equals 0.019 seconds

Explanation:

For uniform motion the distance, speed, time are related by the equation

$Distance=Speed\times time$

In our case since the person is 1.90 meters tall so the nerve impulse will have to cover a distance of 1.90 meters at a speed of 100 m/s.

Hence the time required for the impulse to travel from foot to the brain can be calculated as

$time=\frac{Distance}{speed}\\\\\therefore time=\frac{1.90m}{100m/s}=0.019seconds$

4 0

3 years ago