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3 years ago

5

Which of these is a process in which the atmosphere does not play any role? a. formation of snow b. evaporation of water c. tran

spiration in plants d. erosion of soil by water

1 answer:

iris [78.8K]3 years ago

7 0

I'd say the atmosphere plays a role in all of these but if I had to choose one it would be D

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A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500

kenny6666 [7]

Answer:

6 m/s

Explanation:

Given that :

mass of the block m = 200.0 g = 200 × 10⁻³ kg

the horizontal spring constant k = 4500.0 N/m

position of the block (distance x) = 4.00 cm = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e $\frac{1}{2} kx^2 = \frac{1}{2} mv^2$

$kx^2 = mv^2$

$4500* 0.04^2 = 200*10^{-3} *v^2$

$7.2 =200*10^{-3}*v^{2}$

$v^{2} =\frac{7.2}{200*10^{-3}}$

$v =\sqrt{\frac{7.2}{200*10^{-3}}}$

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is 6 m/s

5 0

3 years ago

One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and

taurus [48]

Answer:

(a) Initial volume will be 7.62 L

(b) Final temperature will be 303.85 K

Explanation:

We have given one mole of ideal gas done 3000 J

So work done W = 3000 J

Let initial volume is $V_1$ and initial pressure $P_1=1atm$ ( As pressure is constant )

Final volume $V_2=25L$ = 0.025 $m^3$

Number of moles n = 1

(B) From ideal gas of equation we know that $PV=nRT$

So $1.01\times 10^5\times0.025=1\times 8.31\times T$

T = 303.85 Kelvin

(B) For isothermal process work done is equal to

$W=nRTln\frac{V_2}{V_1}$

$3000=1\times 8.314\times 303.85\times ln\frac{0.025}{V_1}$

$ln\frac{0.025}{V_1}=1.1881$

$\frac{0.025}{V_1}=3.2808$

$V_2=0.00846m^3=7.62L$

So initial volume will be 7.62 L

5 0

2 years ago

WILL MAKE BRAINLYEST!!!!!! PLZ HELP

mylen [45]

Answer:

D.

Explanation:

4 0

3 years ago

How much work does a 65 kg person climbing a 2000 m high cliff do?

qaws [65]

The answer for the following question is explained below.

<u><em>Therefore the work done is 130 kilo Joules.</em></u>

Explanation:

Work:

A force causing the movement or displacement of an object.

Given:

mass of the person (m) = 65 kg

height of the cliff (h) = 2000 m

To calculate:

work done (W)

We know;

According to the formula:

<u>W = m × g × h</u>

Where;

m represents mass of the person

g represents the acceleration due to gravity

where the value of g is;

<u> g = 10 m/ s²</u>

h represents the height of the cliff

From the above formula;

W = 65 × 10 × 2000

W = 130,000 J

W = 130 Kilo Joules

<u><em>Therefore the work done is 130 kilo Joules.</em></u>

3 0

2 years ago

After an afternoon party, a small cooler full of ice is dumped onto the hot ground and melts. If the cooler contained 5.50 kg of

zysi [14]

Answer:

$Q = 4.40 \times 10^5 Cal$

Explanation:

Here we know that initial temperature of ice is given as

$T = 0^o C$

now the latent heat of ice is given as

$L = 80 Cal/g$

now we also know that the mass of ice is

$m = 5.50 kg$

so here we know that heat required to change the phase of the ice is given as

$Q = mL$

$Q = (5.50 \times 10^3)(80)$

$Q = 4.40 \times 10^5 Cal$

3 0

3 years ago