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1 year ago

PLEASE HELP ME!!

Physics

1 answer:

statuscvo [17]1 year ago

4 0

The balanced chemical equation is 2 AlI₃ + 3 HgCl₂ $\rightarrow$ 2 AlCl₃ + 3 HgI₂ for the given chemical reaction.

<h3>What is chemical equation?</h3>

Chemical equation is a symbolic representation of a chemical reaction which is written in the form of symbols and chemical formulas.The reactants are present on the left hand side while the products are present on the right hand side.

A plus sign is present between reactants and products if they are more than one in any case and an arrow is present pointing towards the product side which indicates the direction of the reaction .There are coefficients present next to the chemical symbols and formulas .

The first chemical equation was put forth by Jean Beguin in 1615.By making use of chemical equations the direction of reaction ,state of reactants and products can be stated. In the chemical equations even the temperature to be maintained and catalyst can be mentioned.

Learn more about chemical equation,here:

brainly.com/question/28294176

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A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten

babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

$P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2$

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0

3 years ago

Suppose that the angular separation of two stars is 0.1 arcseconds, and you photograph them with a telescope that has an angular

Crank

Answer: the photograph will likely show only one star.

Explanation:

Since their angular separation is smaller than the telescope's angular resolution, the picture will apparently show only one star rather than two.

4 0

3 years ago

A spaceship ferrying workers to Moon Base I takes a straight-line path from the earth to the moon, a distance of 384,000 km. Sup

Tema [17]

Answer:

a) v = 19,149.6 m/s

b) f = 95%

c) t = 346.5min

Explanation:

First put all values in metric units:

$15.8 min*\frac{60s}{1min}=948s$

The equation of motion you need is:

$v_f = a*t+v_0$

where $v_f$ is the final velocity, a is acceleration and t is time in hours.

Since the spaceship starts from 0 velocity:

$v_f = a*t = 20.2*948 = 19,149.6 m/s$

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

$x=\frac{v_0+v_f}{2}t$

solving for first and last interval:

Since the spaceship starts and finish with 0 velocity:

$x=\frac{v}{2}t=\frac{19,149.6}{2}948=9,076,910.4m=9,076.9104km$

Then the ship traveled $384,000-9,076.9104*2 = 361,846.1792km$ at constant speed, which means that it traveled:

$f_{constant_speed} =\frac{ x_{constant_speed}}{x_total} =\frac{361,846.1792}{380,000} =0.95$

Which in percentage is 95% of the trip.

to calculate total time you need to calculate the time used during constant speed:

$t = \frac{361,846,179.2}{19,149.6} = 18,895.75s = 314min$

That added to the other interval times:

$t_{total} = t_1+t_2+t_3=15.8+314.93+15.8=346.5min$

5 0

3 years ago

HELP WILL GIVE BRAINLIEST IF CORRECT

anastassius [24]

It’s B i literally jus learned this

6 0

3 years ago

Read 2 more answers

a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?

Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

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3 0

1 year ago