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Mademuasel [1]
3 years ago
10

Which condition would cause salinity of ocean water to decrease

Physics
2 answers:
docker41 [41]3 years ago
5 0
Yupp cause water expands when frozen but extreme heat would cause a lot to evaporate so the water wouldn’t completely disappear but it would lower sea levels and the water would be extra salty
diamong [38]3 years ago
3 0

D. A river flowing into the sea

The freezing of glaciers would do nothing to the salinity. Hot weather on the other hand, and evaporation, would leave behind the salt and thus it Increases the amount of salt to water in the ocean.  

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Initilal velocity of projectile at an angle of 22.5° is 10m/s. Then what is the magnitude of range of projectile?
crimeas [40]

Answer:

0.707m

Explanation:

from formula of range i.e R=Usin2Q/g

3 0
3 years ago
A fluid flows through a pipe whose cross-sectional area changes from 2.00 m2 to 0.50 m2 . If the fluid’s speed in the wide part
borishaifa [10]

Answer:

v₂ = 7/ (0.5)= 14 m/s

Explanation:

Flow rate of the fluid

Flow rate is the amount of fluid that circulates through a section of the pipeline (pipe, pipeline, river, canal, ...) per unit of time.

The formula for calculated the flow rate is:

Q= v*A Formula (1)

Where :

Q is the Flow rate (m³/s)

A is the cross sectional area of a section of the pipe (m²)

v is the speed of the fluid in that section (m/s)

Equation of continuity

The volume flow rate Q for an incompressible fluid at any point along a pipe is the same as the volume flow rate at any other point along a pipe:

Q₁= Q₂

Data

A₁ = 2m² : cross sectional area 1

v₁ = 3.5 m/s : fluid speed through A₁

A₂ = 0.5 m² : cross sectional area 2

Calculation of the fluid speed through A₂

We aply the equation of continuity:

Q₁= Q₂

We aply the equation of Formula (1):

v₁*A₁= v₂*A₂

We replace data

(3.5)*(2)= v₂*(0.5)

7 = v₂*(0.5)

v₂ = 7/ (0.5)

v₂ =  14 m/s

4 0
3 years ago
I drop an egg from a certain distance and it takes 3.74 seconds to reach the ground. How high up was the egg?
OverLord2011 [107]

Answer:

68.5 meters

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 3.74 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0) (3.74) + ½ (9.8) (3.74)²

Δy = 68.5

The egg fell 68.5 meters.

7 0
3 years ago
A spacecraft is fueled using hydrazine (N2H4; molecular weight of 32 grams per mole [g/mol]) and carries 1640 kilograms [kg] of
Sauron [17]

Answer:

The value is t = 689.029 \  hours

Explanation:

From the question we are told that

The molar mass of hydrazine is Z =  32 g/mol = \frac{32}{1000} = 0.032 \  kg/mol

The initial temperature is T_i  =  -186 ^o F = (-186-32) *\frac{5}{9} +273.15 = 152\ K

The final temperature is T_f  =  78 ^o F = (78-32) *\frac{5}{9} +273.15 = 298.7 \ K

The specific heat capacity is c_h  =  0.099 [kJ/(mol K)] = 0.099 *10^3 J/(mol/K)

The power available is P = 300 \ W

The mass of the fuel is m =   1640 \  kg

Generally the number of moles of hydrazine present is

n  =  \frac{m}{Z}

=> n  =  \frac{1640}{= 0.032}

=> n  =  51250 \ mol

Generally the quantity of heat energy needed is mathematically represented as

Q =  n * c_h * (T_f -T_i)

=> Q =  51250  * 0.099 *10^3  * (298.7 - 152)

=> Q =  7.441516913 * 10^{8} \  J

Generally the time taken is mathematically represented as

t =  \frac{Q}{P}

=> t =  \frac{7.441516913 * 10^{8} }{300}

=> t = 2480505.6377 s

Converting to hours

t = \frac{2480505.6377}{3600}

=> t = 689.029 \  hours

6 0
3 years ago
A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.5 s
gulaghasi [49]

(a) 0.96 m/s

The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:

T=2\cdot 2.5 s=5.0 s

The frequency of the waves is the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{5.0 s}=0.20 Hz

The wavelength instead is just the distance between two consecutive crests, so

\lambda=4.8 m

And the wave speed is given by:

v=\lambda f=(4.8 m)(0.20 Hz)=0.96 m/s

(b) 0.265 m

The total distance between the highest point of the wave and its lowest point is

d = 0.53 m

The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is

A=\frac{d}{2}=\frac{0.53 m}{2}=0.265 m

(c) Amplitude: 0.15 m, wave speed: same as before

In this case, the amplitude of the wave would be lower. In fact,

d = 0.30 m

So the amplitude would be

A=\frac{d}{2}=\frac{0.30 m}{2}=0.15 m

Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.

8 0
3 years ago
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