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2 years ago

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For nitrogen to be a liquid, its temperature must be within 12.78 °f of –333.22 °f. the equation |x 333.22 | = 12.78 can be used

to find x, which represents the maximum and minimum temperatures at which nitrogen is a liquid. is the maximum temperature at which nitrogen is a liquid. is the minimum temperature at which nitrogen is a liquid.

2 answers:

otez555 [7]2 years ago

8 0

Answer:

max=320.44 min=346

Explanation:

Y_Kistochka [10]2 years ago

4 0

The maximum temperature is -320.44 and the minimum temperature is -333.22.

<h3>How is the temperature variation calculated?</h3>

In this case, there was an increase of 50 °C in the temperature of this object. We say that the temperature variation suffered by the body was 50°C, and this can be determined mathematically through the difference between the final value and the initial temperature value (60 – 10 = 50).

In this case, we have:

$x+ 333.22 = 12.78\\x=12.78-333.22\\x=-320.44$

See more about temperature at brainly.com/question/11464844

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Determine the nuclear radius (in fm) for each of the following nuclei.

tekilochka [14]

Answer:

(a) 2.75 fm

(b) 2.89 fm

(c) 4.70 fm

(d) 7.12 fm

Explanation:

For a given element, the radius r of its nuclei is given by;

r = r₀ $A^{(1/3)}$

Where;

A = Atomic mass of the element

r₀ = 1.2 x 10⁻¹⁵m = 1.2fm

Now let's solve for the given elements

(a) ¹²₆C

Carbon element => This has an atomic mass number of 12

Therefore its radius is given by;

r = 1.2 x $12^{1/3}$

r = 1.2 x 2.29

r = 2.75 fm

(b) ¹⁴₇N

Nitrogen element => This has an atomic mass number of 14

Therefore its radius is given by;

r = 1.2 x $14^{1/3}$

r = 1.2 x 2.41

r = 2.89 fm

(c) ⁶⁰₂₇Co

Cobalt element => This has an atomic mass number of 60

Therefore its radius is given by;

r = 1.2 x $60^{1/3}$

r = 1.2 x 3.92

r = 4.70 fm

(d) ²⁰⁸₈₂Pb

Lead element => This has an atomic mass number of 208

Therefore its radius is given by;

r = 1.2 x $208^{1/3}$

r = 1.2 x 5.93

r = 7.12 fm

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2 years ago

A camcorder has a power rating of 13 watts. If the output voltage from its battery is 6 volts, what current does it use?

leva [86]

Power = (voltage) x (current)

13 watts = (6 volts) x (current)

Divide each side by (6 volts):

Current = (13 watts) / (6 volts)

Current = (13/6) Amperes

<em>Current = 2.17 Amperes</em>

5 0

3 years ago

Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?

irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

$M=Fd$

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 500 N$ is the weight of the person and $d_2 = 2 m$ is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

$M_a = W_1 d_1$

where $W_1$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the person on the left:

$W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N$

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 400 N$ is the weight of the person and $d_2$ is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 300 N$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum.

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the distance of the person on the right:

$W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m$

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

$M_c = W_2 d_2$

where $W_2$ is the weight of the seesaw and $d_2 = 3 m$ is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 600 N$ is his weight and $d_1 = 1 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the seesaw:

$W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N$

#LearnwithBrainly

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2 years ago

A cannon fires a 0.652 kg shell with initial

laila [671]

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u can simplify this and u will get the answer

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