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3 years ago

What is the number “4” in SiCl4? a subscript a coefficient a product a reactant

Physics

2 answers:

Bad White [126]3 years ago

6 0

It is a) a subscript

Svetlanka [38]3 years ago

4 0

Answer: 4 is the subscript in $SiCl_{4}$ .

Explanation:

We know that,

Subscript: The number of atom present in a given compound.

In $SiCl_{4}$ , 4 represents the number of chlorine atom in 1 mole of Silicon tetra chloride.

Coefficient: Number of atom which is used in reaction.

For example: In $3SiCl_{4}$ , 3 is the coefficient of silicon tetra chloride.

A Product : In the balance equation, right hand side is called product.

A reactant : In the balance equation, left hand side is called reactant.

For example: the balance equation is

$2Mg+O_{2}\rightarrow 2MgO$

In this equation, $2Mg+O_{2}$ is the reactants and $2MgO$ is products.

Hence, 4 is the subscript in $SiCl_{4}$ .

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What is the marble's range if it is fired horizontally from 1.6 m above the ground?

cupoosta [38]

Answer:

The answer is 6.40 meters.

Explanation:

The speed v = √(2gh)

v = √( 2×9.8×6.4) = 11.2 m/s

After, finding the time it takes to hit the ground from a height of 1.6 meters.

time = √(2H÷g)

time = √(2×1.6÷9.8)

time = 0.5714 seconds.

Horizontal distance is speed × time = 11.2 × 0.5714 = 6.40 meters.

5 0

2 years ago

timama [110]

Answer:

Workdone = 1960 Joules.

Explanation:

Given the following data;

Mass = 5kg

Force = 49N

Height (distance) = 40m

To find the workdone;

Workdone = force * distance

Substituting into the equation, we have;

Workdone = 49*40

Workdone = 1960 Joules.

Therefore, the amount of work done on the bowling ball to lift it is 1960 Joules.

7 0

3 years ago

A block, M1=10kg, slides down a smooth, curved incline of height 5m. It collides elastically with another block, M2=5kg, which i

erma4kov [3.2K]

Answer:

2.86 m

Explanation:

Given:

M₁ = 10 kg

M₂ = 5 kg

$\mu_k$ = 0.5

height, h = 5 m

distance traveled, s = 2 m

spring constant, k = 250 N/m

now,

the initial velocity of the first block as it approaches the second block

u₁ = √(2 × g × h)

u₁ = √(2 × 9.8 × 5)

u₁ = 9.89 m/s

let the velocity of second ball be v₂

now from the conservation of momentum, we have

M₁ × u₁ = M₂ × v₂

on substituting the values, we get

10 × 9.89 = 5 × v₂

v₂ = 19.79 m/s

now,

let the velocity of mass 2 when it reaches the spring be v₃

from the work energy theorem, we have

Work done by the friction force = change in kinetic energy of the mass 2

$0.5\times5\times9.8\times2 = \frac{1}{2}\times5\times( v_3^2-19.79^2)$

v₃ = 20.27 m/s

now, let the spring is compressed by the distance 'x'

therefore, from the conservation of energy

we have

Energy of the spring = Kinetic energy of the mass 2

$\frac{1}{2}kx^2=\frac{1}{2}mv_3^2$

on substituting the values, we get

$\frac{1}{2}\times250\times x^2=\frac{1}{2}\times5\times20.27^2$

x = 2.86 m

8 0

3 years ago

The force exerted by the earth on a body is called ______of a body.

AVprozaik [17]

Answer:

The force exerted by the earth on a body is called gravitational of a body.

i think this is the answer

4 0

3 years ago

Read 2 more answers

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

2 years ago