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3 years ago

would you expect the air pressure in a valley thats below sea level to be higher or lower than the air pressure at sea level

Physics

1 answer:

mixer [17]3 years ago

4 0

It would depend on how deep the water or pressure is but I persanlly be higher

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Maya uses a tuning fork of frequency 250 Hz to produce transverese waves on a stretched string. Her friend justin measure the di

Goshia [24]

Answer:

The velocity of the wave is 12.5 m/s

Explanation:

The given parameters are;

he frequency of the tuning fork, f = 250 Hz

The distance between successive crests of the wave formed, λ = 5 cm = 0.05 m

The velocity of a wave, v = f × λ

Where;

f = The frequency of the wave

λ = The wavelength of the wave - The distance between crests =

Substituting the known values gives;

v = 250 Hz × 0.05 m = 12.5 m/s

The velocity of the wave, v = 12.5 m/s.

6 0

2 years ago

Six friends are at a pizzeria. They want to order enough pizza so that each person can eat at least 2/5 . start fraction, 2, d

Lyrx [107]

Friend #1 gets at least 2/5 of a pizza.

Friend #2 gets at least 2/5 .

Friend #3 gets at least 2/5 .

Friend #4 gets at least 2/5 .

Friend #5 gets at least 2/5 .

Friend #6 gets at least 2/5 .

Sum . . . . . . . . . at least 12/5 of a pizza.

Simplify . . . . . . at least 2.4 pizzas.

-- If pizzas can be bought by the half, they should order at least <em>2-1/2 pizzas.</em>

-- If only whole pizzas have to be ordered, then they should order at least <em>3 pizzas.</em>

4 0

3 years ago

The even distribution of weight in a compostion is known as

mojhsa [17]

The answer is Balance. hopei helped? ahahahaha

3 0

3 years ago

Example 3 :

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0

1 year ago

To what temperature will a 79.0 g piece of glass raise if it absorbs 9457 calories of heat and its specific heat capacity is 0.5

ruslelena [56]

Answer:

T_final = 279.4 [°C]

Explanation:

In order to solve this problem, we must use the following equation of thermal energy.

$Q=m*C_{p}*(T_{f}-T_{i})$

where:

Q = heat = 9457 [cal]

m = mass = 79 [g] = 0.079 [kg]

Cp = specific heat = 0.5 [cal/g*°C]

T_initial = initial temperature = 40 [°C]

T_final = final temperature [°C]

$9457 = 79*0.5*(T_{f}-40)\\239.41=T_{f}-40\\\\T_{f}=279.4[C]$

8 0

3 years ago