All categories

3 years ago

HELP ME PLEASE!!!!!!!!!

Physics

1 answer:

Stolb23 [73]3 years ago

6 0

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

$F = \frac{kq_1q_3}{d_{13}^2}$

now from the above equation

$F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}$

$F = 0.288 N$

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

$F_{net} = 2Fcos\theta$

here we know that angle is

$\theta = 37 degree$

now we have

$F_{net} = 2\times 0.288 cos37$

$F_{net} = 0.46 N$

so net force on q3 is 0.46 N vertically upwards along +Y axis

You might be interested in

I need help with this

lisabon 2012 [21]

A. Condensation

Hope this helps!!!

5 0

2 years ago

A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to

Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0

3 years ago

The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa

Sonja [21]

Answer:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

$c_{pA} > c_{pB}$

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

$T_{iA}= T_{iB}$

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

$Q = m c_p \Delta T$

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

$Q_A = Q_B$

And if we replace the formula for sensible heat we got:

$m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B$

And if we replace for the change of the temperature we got:

$m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})$

And since $T_{iA}= T_{iB}= T$ we have this:

$m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)$

d) More information is needed

(The relation between the masses is not given)

4 0

3 years ago

The second-order rate constants for the reaction of oxygen atoms ·with aromatic hydrocarbons have been measured (R. Atkinson and

Blizzard [7]

Answer: Frequency factor A = 8 × 10⁹

activation energy Ea = 15.5 KJ/Mol

Explanation: to begin, let us first define the parameters given;

K₁ = 1.44 × 10⁷dm³mol⁻¹s⁻¹

K₂ = 3.03 × 10⁷ dm³mol⁻¹s⁻¹

K₃ = 6.9 × 10 dm³mol⁻¹s⁻¹

also T₁ = 300.3 K

T₂ = 341.2 K

T₃ = 392.2 K

we know that;

㏑ K₂ / K₁ = Ea/R [1/T₁ -1/T₂]

where R is given as 8.314 J/mol-k

Ea = activation energy

K₁, K₂ = rate constant

T₁, T₂ = Temperature

therefore, ㏑ (3.03 × 10⁷/ 1.44 × 10⁷) = Ea / 8.314 [1/300.3 - 1/341.2]

this gives Ea = 15496.16 J/Mol ≈ 15.5 KJ /Mol

∴ Ea = 15.5 KJ/ Mol

also given that K = A e⁻∧Ea/RT

here A = frequency factor

∴ 6.9 × 10⁷ = A e⁻ ∧(15496.16/8.314 × 392.2)

A = 7.99 × 10⁹ = 8 × 10⁹

3 0

3 years ago

When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi

Nikitich [7]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ .

Comparing two wires,

$R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}$

Dividing the above 2 equations,

$\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{A_2 }{A_1} \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{d_2^2 }{d_1^2} \\$

Since $d_2=2d_1$

The above ratio is

$\frac{R_1}{R_2}=\frac{1.68(10^{-8}) }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264$

We also have,

$\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1$

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0

3 years ago