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3 years ago

Water is returned from earth’s surface to the atmosphere by

Physics

1 answer:

Tems11 [23]3 years ago

4 0

Answer:

Evaporation

Explanation:

Evaporation is a form of mass tranfer phenomena where by water are moved from the earth surface into the atmosphere as vapours,it is path of the water cycle a decription of the path moved by land water until it turns into rain, humidity,air and temperature are factors that influence evaporation though evaporation can happen at all temperature

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A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simp

Wittaler [7]

Answer:

l = 10.16 m

Explanation:

In this case, we have the period of oscillation of the pendulum is 6.4 s. It is required to find the height of the tower.

We know that the pendulum executes SHM. Let l is the height of the tower. The time period of simple pendulum is given by :

$T=2\pi \sqrt{\dfrac{l}{g}}$

g is acceleration due to gravity

We need to rearrange the above equation such that,

$l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(6.4)^2\times 9.8}{4\pi ^2}\\\\l=10.16\ m$

So, the height of the tower is 10.16 m.

4 0

4 years ago

Queremos un cilindro de simple efecto que utilice en su funcionamiento un volumen de aire a presión atmosférica de 13,122 litros

zhuklara [117]

Answer:

1) El diámetro es de aproximadamente 913,987 cm.

2) La fuerza del cilindro es 5576850 kgf

Explanation:

1) Los parámetros dados son;

El volumen del aire = 13,122 litros = 13122000 cm³

La presión de trabajo = 8.5 kgf / cm²

La longitud del cilindro = 20 cm.

Por lo tanto, tenemos;

El área de la base del cilindro = π · r² = 13122000 cm³ / (20 cm) = 656100 cm²

r = √ (656100 / π) ≈ 456,994 cm

El diámetro = 2 × r ≈ 2 × 456.994 ≈ 913.987 cm

El diámetro ≈ 913,987 cm

2) La fuerza del cilindro = El área de la base del cilindro × La presión de trabajo

∴ La fuerza del cilindro = 656100 cm² × 8.5 kgf / cm² = 5576850 kgf

La fuerza del cilindro = 5576850 kgf

3 0

3 years ago

Describe how cool and warm air move in the atmosphere.

ruslelena [56]

Answer:

i think cool air moves slow and hot air moves fast

8 0

3 years ago

A river flows with a uniform velocity vr. A person in a motorboat travels 1.22 km upstream, at which time she passes a log float

storchak [24]

Answer:

t ’= $\frac{1450}{0.6499 + 2 v_r}$ , v_r = 1 m/s t ’= 547.19 s

Explanation:

This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.

By the time the boat goes up the river

v_b - v_r = d / t

By the time the boat goes down the river

v_b + v_r = d '/ t'

let's subtract the equations

2 v_r = d ’/ t’ - d / t

d ’/ t’ = 2v_r + d / t

$t' = \frac{d'}{ \frac{d}{t}+ 2 v_r }$

In the exercise they tell us

d = 1.22 +1.45 = 2.67 km= 2.67 10³ m

d ’= 1.45 km= 1.45 1.³ m

at time t = 69.1 min (60 s / 1min) = 4146 s

the speed of river is v_r

t ’= $\frac{1.45 \ 10^3}{ \frac{ 2670}{4146} \ + 2 \ v_r}$

t ’= $\frac{1450}{0.6499 + 2 v_r}$

In order to complete the calculation, we must assume a river speed

v_r = 1 m / s

let's calculate

t ’= $\frac{ 1450}{ 0.6499 + 2 \ 1}$

t ’= 547.19 s

8 0

3 years ago

. Emily pushes a 38.8 kg grocery cart of groceries by exerting a 76.0 N force on the handle inclined at 40.0 degrees below the h

Lana71 [14]

Answer:

a) $F_{x} = 58.2 N$

$F_{y} = 48.9 N$

b) a = 1.5 m/s²

Explanation:

a) The horizontal and vertical components of Emily's force can be found knowing the angle and the exerted force.

Since the handle is inclined at 40.0° below the horizontal we have:

$F_{x} = |F|*cos(\theta) = 76.0 N*cos(40) = 58.2 N$

$F_{y} = |F|*sin(\theta) = 76.0 N*sin(40) = 48.9 N$

b) The acceleration of the car can be calculated as follows:

$F_{x} = ma$

We used the horizontal component of the force because the cart is moving in that direction.

$a = \frac{F_{x}}{m} = \frac{58.2 N}{38.8 kg} = 1.5 m/s^{2}$

Hence, the acceleration of the car is 1.5 m/s².

I hope it helps you!

3 0

3 years ago