What information can you determine from the coefficients of this balanced chemical equation?

What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t

vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0

3 years ago

What derived unit is used to measure the slope of the line in this graph?

Alexxx [7]

Answer:

g/cm³

Explanation:

From the question given above,

The y-axis is representing mass (g)

The x-axis is representing volume (cm³)

Unit of slope =?

Slope of a graph is simply defined as the change in y-axis divided by the change in x-axis. Mathematically it is expressed as:

Slope = change in y-axis (Δy)/change in x-axis (Δx)

Slope = Δy/Δx

Thus, with the above formula, we can obtain the unit used for measuring the slope as follow:

y-axis = mass (g)

x-axis = volume (cm³)

Slope =.?

Slope = Δy/Δx

Slope = mass (g) /volume (cm³)

Slope = g/cm³

Therefore, the derive unit used for measuring the slope is g/cm³

5 0

3 years ago

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0

Tema [17]

Answer:

841.5 Hz

Explanation:

Given

y = 50 cm = 0.5 m

d = 5.00 m

L = 12.0 m away from the wall

v = speed of sound = 343 m/s

The image of the scenario is presented in the attached image.

When destructive interference is being experienced from 50 cm (0.5 m) parallel to the wall, the path difference between the distance of the two speakers from the observer is equal to half of the wavelength of the wave.

Let the distance from speaker one to the observer's new position be d₁

And the distance from the speaker two to the observer's new position be d₂

(λ/2) = |d₁ - d₂|

d₁ = √(12² + 3²) = 12.3693 m

d₂ = √(12² + 2²) = 12.1655 m

|d₁ - d₂| = 0.2038 m

(λ/2) = |d₁ - d₂| = 0.2038

λ = 0.4076 m

For waves, the velocity (v), frequency (f) and wavelength (λ) are related thus

v = fλ

f = (v/λ) = (343/0.4076) = 841.5 Hz

Hope this Helps!!!

7 0

4 years ago

Read 2 more answers

A force of 6.0 N is applied horizontally to a 3.0 kg crate initially at rest on a horizontal frictionless surface. After the cra

Savatey [412]

Answer:

a. Yes, because the acceleration of the crate is 2.0 m/s².

Explanation:

Given

$Force = 6N$ --- f

$Mass = 3kg$ --- m

$Time = 1.5s$ --- t

$Velocity = 3.0m/s$ --- v

Required

Does the system support $F=ma$

Yes, it does and this is shown below

The crate is initially at rest; so:

$u = 0$

Using the first equation of motion

$v = u + at$

Substitute values for v, u and t

$3 = 0 + a*1.5$

$3 = 1.5a$

Make a the subject

$a = 3/1.5$

$a = 2$

Using $F = ma$

Substitute values for F and m

$6 = 3 * a$

Divide both sides by 3

$6/3 = 3/3 * a$

$2 = a$

$a = 2$

In both cases:

$a = 2$

<em>Hence, option (a) is correct.</em>

6 0

3 years ago

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

Scilla [17]

Answer:

Maximum acceleration will be equal to $3.43m/sec^2$

Explanation:

We have given coefficient of kinetic friction $\mu _k=0.7$

And coefficient of static friction $\mu _s=1$

Acceleration due to gravity $g=9.8m/sec^2$

When truck moves maximum force will be equal to $F=\mu _kmg$

It is given that half of the weight is supported by its drive wheels

So force required $=\frac{\mu _kmg}{2}$

From newtons law maximum acceleration will be equal to $a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2$

8 0

3 years ago