Which of these Would you include in a free body diagram of a coin sitting on a table

statuscvo [17]

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

#SPJ1

4 0

7 months ago

A worker on the roof of a house drops his 0.58 kg hammer, which slides down the roof at constant speed of 6.69 m/s. The roof mak

shusha [124]

Answer:

17.3 m

Explanation:

Given that,

Mass of a hammer is 0.58 kg

Velocity with which the hammer slides is 6.69 m/s at constant speed.

The roof makes an angle of 18 ◦ with the horizontal, and its lowest point is 18.2 m from the ground. We need to find the horizontal distance traveled by the hammer between the time is leaves the roof of the house and the time it hits the ground. Firstly, we will find the time taken by the hammer when it reaches ground in vertical direction.

$y=y_0+v_ot +\dfrac{1}{2}gt^2$

Putting all the values,

$0=18.2+6.69t-\dfrac{1}{2}\times 9.8t^2\\\\-4.9t^2+6.69t+18.2=0\\\\t=\dfrac{-6.69+\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}, \dfrac{-6.69-\sqrt{6.69^{2}-4\cdot-4.9\cdot18.2}}{2\cdot-4.9}\\\\t=-1.36\ s\ \text{and}\ t=2.72\ s$

Neglecting negative value,

To find horizontal distance, multiply 2.72 s with the horizontal component of velocity.

$d=2.72\times 6.69\times \cos(18)\\\\d=17.3\ m$

3 0

3 years ago

When does it start to cool down in arizona?

KatRina [158]

<span>It starts to cool down in Arizona in mid-October. The weather remains good in October- April and the heat tends to break in mid of October. The time when it is coldest in Arizona in December or January. However, it's too hot in the month of July and Rain becomes common in Spring season.</span>

7 0

2 years ago

A wheel is rotating at 30.0 rpm. The wheel then accelerates uniformly to 50.0 rpm in 10.0 seconds. Determine the – a) rate of an

Mademuasel [1]

Answer:

The angular acceleration is $0.209\ rad/s^2$

Explanation:

Given that,

Angular velocity, $\omega_{i} = 30.0\ rpm$

Angular velocity, $\omega_{f} = 50.0\ rpm$

Time t = 10.0 sec

We need to calculate the angular acceleration

Using formula of angular acceleration

$\alpha=\dfrac{\Delta \omega}{\Delta t}$

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{\Delta t}$

$\alpha=\dfrac{50.0-30.0}{10.0}$

Now, we change the angular velocity in rad/s.

$\omega=20\times\dfrac{2\pi}{60}$

$\omega=2.09\ rad/s$

$\alpha=\dfrac{2.09}{10.0}$

$\alpha=0.209\ rad/s^2$

Hence, The angular acceleration is $0.209\ rad/s^2$

5 0

2 years ago

Read 2 more answers

A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 15.0 cm has 500 turns and carries a current of

SCORPION-xisa [38]

Answer:

a).β=0.53 $x10^{-3}$ T

a).β=0.40 $x10^{-4}$ T

Explanation:

The magnetic field at distance 'r' from the center of toroid is given by:

$\beta =\frac{u_{o}*I*N}{2\pi*r}$

a).

$N=500\\I=0.800A\\r=15cm*\frac{1m}{100cm}=0.15m\\u_{o}=4\pi x10^{-7}\frac{T*m}{A} \\\beta=\frac{4\pi x10^{-7}\frac{T*m}{A}*0.8A*500}{2\pi*0.15m} \\\beta=0.53x10^{-3}T$

b).

The distance is the radius add the cross section so:

$r_{1}=15cm+5cm\\r_{1}=20cm$

$r_{1} =20cm*\frac{1m}{100cm}=0.20m$

$\beta =\frac{u_{o}*I*N}{2\pi*r1}$

$\beta =\frac{4\pi x10^{-7}*0.80A*500 }{2\pi*0.20m} \\\beta=0.4x10^{-3} T$

3 0

3 years ago