what does loudness of a sound depend on?​

Answer:

the force of attraction between the two charges is 4.2 x 10⁹ N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

[tex]F = \frac{Kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.

Answer:

The car will stop moving

Explanation:

If a balanced force is applied to the car in motion, the car will stop accelerating and remain stationary because all the forces acting on the car are equal.

Also, you can say, since a balanced force is applied, the net force or resultant force on the car is zero. According to Newton's second law of motion, an object will move in the direction of the applied force. When the resultant force on the object is zero, it means the object will not move.

Answer:

Electron

Explanation:

An object can become electrically charged when it gains or loses an electron. Because an electron is negatively charged, when an object gains an electron it becomes negatively charged. Also, when it gives up an electron, it becomes positively charged. This positive charge is because the atom has one proton more than electron. In a neutral atom, the number of the proton is equal to the number of the electron. An electron is negatively charged, and a proton is positively charged.

This question is incomplete, the complete question is;

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer:

the required wavelength of light is 161.9 nm

Explanation:

Given the data in the question;

Let us represent work function of the metal by W.

Now, using Einstein photoelectric effect equation;

[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]

hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1

we solve for W

W = hc/λ - [tex]K_{max[/tex]

given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J

we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.

so we substitute

W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] -  3.28 × 10⁻¹⁹

W =  8.99457 × 10⁻¹⁹ -   3.28 × 10⁻¹⁹

W = 5.71457 × 10⁻¹⁹ J

Now, to determine λ for which maximum kinetic energy is double

so;

[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J

from from equation 1

we solve for λ'

λ' = hc / W + [tex]K'_{max[/tex]

we substitute

λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))

λ' = 1.9878 × 10⁻²⁴ /  1.227457 × 10⁻¹⁸

λ' =  1.619 × 10⁻⁷ m

λ' =  161.9 nm

Therefore, the required wavelength of light is 161.9 nm

Answer:

the red pointer on the magnet ( grey region) : points towards north

red pointer outside the magnet ( white region) is pointing towards south

Explanation:

please see the attached image

Solution :

In the question, it is given that the collision is inelastic and the blocks stick together.

In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.

The linear momentum is given by :

[tex]$\vec p = m \vec v$[/tex] (mass x velocity)

So according to the conservation of linear momentum,

[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]

Let the velocity after the collision is [tex]$v_F$[/tex]

[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]

Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]

[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]

∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.

and  [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.

Answer:

The acceleration of 2.0 kg block is 2.7 m/s²

Explanation:

Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:

Force down the plane on the system

= (4 + 2) g sin30°

= (6)(10)(½)

= 30N

Force up the plane on the system

= µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°

= (2µ₁ + 4µ₂) g cos30°

= (2 × 0.2 + 4 × 0.3)(10)(√3/2)

≈ 13.76 N

∴ Net force down the plane is F

F = 30 - 13.76

F = 16.24 N

∴Acceleration of both the blocks down the

plane will b a

a = F ÷ (4 + 2)

a = 16.24 ÷ 6

a = 2.7 m/s²

Thus, The acceleration of 2.0 kg block is

2.7 m/s²

-TheUnknownScientist

Explanation:

2.7m/s2

I hope its helpful

Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!  

bit.[tex]^{}[/tex]ly/3a8Nt8n

Answer:

b

Explanation:

Answer:

C. Slippery feel

Explanation:

Answer:

Resistance of the light bulb = 1.33 ohm (Approx.)

Explanation:

Given:

Current in electric bulb = 2 Amp

Voltage in battery = 1.5 Volt

Find:

Resistance of the light bulb

Computation:

Resistance = Voltage / Current

Resistance of the light bulb = Voltage in battery / Current in electric bulb

Resistance of the light bulb = 2 / 1.5

Resistance of the light bulb = 1.33 ohm (Approx.)

Answer:

C: Ignoring

Explanation:

On edge 2021

Answer:

It's C.

Explanation:

Answer:  

a) 1m

b) 2μs

c) 3mm

Explanation:

Explanation:

I think you forgot to add the other part!

Answer:

uhm what plant...

Explanation:

Answer:

The appropriate solution is "23.87 rev".

Explanation:

The given values are:

Initial angular velocity,

[tex]\omega_i=20 \ rad/s[/tex]

Final angular velocity,

[tex]\omega_f=40 \ rad/s[/tex]

Time taken,

[tex]t = 5.0 \ s[/tex]

If, α be the angular acceleration, then

⇒  [tex]\omega_f=\omega_i+\alpha t[/tex]

or,

⇒  [tex]\alpha t=\omega_f-\omega_i[/tex]

⇒  [tex]\alpha=\frac{\omega_f-\omega_i}{t}[/tex]

On substituting the values, we get

⇒     [tex]=\frac{40-20}{5.0}[/tex]

⇒     [tex]=\frac{20}{5.0}[/tex]

⇒     [tex]=4 \ rad/s^2[/tex]

If, ΔФ be the angular displacement, then

⇒  [tex]\Delta \theta=\omega_i t+\frac{1}{2} \alpha t^2[/tex]

On substituting the values, we get

⇒        [tex]=[(20\times 5.0)+(\frac{1}{2})\times 4\times (5.0)^2][/tex]

⇒        [tex]=100+50[/tex]

⇒        [tex]=150[/tex]

On converting it into "rev", we get

⇒  [tex]\Delta \theta=(\frac{150}{2 \pi} )[/tex]

⇒        [tex]=23.87 \ rev[/tex]

Answer:

C

Explanation:

a=f/m

10Kgm/s2/4kg

2.5m/s2

When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².

The rate at which an item changes its velocity is known as acceleration, a vector quantity. If an object's velocity is changing, it is acceleration. The net acceleration that objects get as a result of the combined action of gravity and centrifugal force is known as the Earth's gravity, or g. It is a vector quantity whose strength or magnitude is determined by the norm and whose direction correlates with a plumb bob.

Given in the question, force 10 N and mass 4.0 Kg the acceleration is,

a = 10/4 = 2.5 m/sec²

When an unbalanced force of 10 N is applied to an object whose mass is 4.0 kg, the acceleration of the object will be 2.5 m/s².

To learn more about acceleration refer to the link:

brainly.com/question/12550364

#SPJ2

Answer:

[tex]12\; \rm N[/tex].

[tex]3\; \rm m\cdot s^{-2}[/tex].

Explanation:

By Newton's Second Law, the acceleration of an object is proportional to the size of the resultant force on it, and inversely proportional to the mass of this object.

[tex]\displaystyle \text{acceleration} = \frac{\text{resultant force}}{\text{mass}}[/tex].

Rearrange this equation for the resultant force on the object:

[tex]\text{resultant force} = \text{acceleration} \cdot \text{mass}[/tex].

For the [tex]6\; \rm kg[/tex] object in this question:

[tex]\begin{aligned} F &= m \cdot a \\ &= 6\; \rm kg \times 2\; \rm m \cdot s^{-2} \\ &=12\; \rm N\end{aligned}[/tex].

When the resultant force on the [tex]4\; \rm kg[/tex] object is also [tex]12\; \rm N[/tex], the acceleration of that object would be:

[tex]\begin{aligned} a &= \frac{F}{m} \\ &= \frac{12\; \rm N}{4\; \rm kg} = 3\; \rm m \cdot s^{-2}\end{aligned}[/tex].

Answer:

C

Explanation:

O has 8   protones,S  tiene 16, Se 34 y Te 52.

Answer:

Work input = Work output * Work against friction is your answer so C

Explanation:

I hope this helps you :)

Answer:

A) Work input = Work output + Work against friction

Explanation:

Answer:

use the formula to calculate acceleration and you'll get the answers

Answer:

T = 0.71 seconds

Explanation:

Given data:

mass m = 1Kg, spring constant K = 78 N/m, time period of oscillation T = 0.71 seconds.

We have to calculate time period when this same spring-mass system oscillates vertically.

As we know

[tex]T = 2\pi \sqrt{\frac{m}{K} }[/tex]

This relation of time period is true under every orientation of the spring-mass system, whether horizontal, vertical, angled or inclined. Therefore, time period of the same spring-mass system oscillating  vertically too remains the same.

Therefore, T = 0.71 seconds

Answer:

its in the picture hope it helps make brainlliest ty

Answer:

the lowest possible frequency of the emitted tone is 404.79 Hz

Explanation:

   Given the data in the question;

S₁ ←  5.50 m → L

↑

2.20 m

↓

S₂

We know that, the condition for destructive interference is;

Δr = ( 2m + [tex]\frac{1}{2}[/tex] ) × λ

where m = 0, 1, 2, 3 .......

Path difference between the two sound waves from the two speakers is;

Δr = √( 5.50² + 2.20² ) - 5.50

Δr = 5.92368 - 5.50

Δr = 0.42368 m

v = f × λ

f = ( 2m + [tex]\frac{1}{2}[/tex])v / Δr

m = 0, 1, 2, 3, ....

Now, for the lowest possible frequency, let m be 0

so

f = ( 0 + [tex]\frac{1}{2}[/tex])v / Δr

f = [tex]\frac{1}{2}[/tex](v) / Δr

we know that speed of sound in air v = 343 m/s

so we substitute

f = [tex]\frac{1}{2}[/tex](343) / 0.42368

f = 171.5 / 0.42368

f = 404.79 Hz

Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz

Answer:

Attached below is the required diagram related to the question

answer :

q'3 = WбT^4

engineer's proposal has merit

Explanation:

Let : A'3 represent the deep space

      A'1 represent the surface area , F13 and  F23  represent the view factors

      T1 , T2, T3 ; represent temperatures

      q'3 represent net rate of heat radiation

 Derive the expression for the rate per unit length at which radiation is transferred from the surfaces to deep space

derived expression ; q'3 = WбT^4

attached below is a detailed solution

Given that The emission is proportional to the area of the opening and the surfaces ( 1 and 2 ) have the same temperature hence this problem can be treated as a two surface enclosure. hence the engineer's proposal have merit .

attached below is a prove ( b )

Answer:

. always start on the north pole and terminate (end) on the South Pole

Explanation:

Answer:

Explanation:

Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.

Answer:

A. 3.9 V B. 1.9 fA

Explanation:

Part A

What emf is induced in the ring as the field changes?

Express your answer to two significant figures and include the appropriate units.

The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time

ε = ΔΦ/Δt

ε = ΔABcos0°/Δt

ε = AΔB/Δt

A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T.  ΔB = B₁ - B₀  = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.

So, ε = AΔB/Δt

ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s

ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s

ε = 3.926 V

ε ≅ 3.9 V

Part B

If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.

Express your answer to two significant figures and include the appropriate units.

Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².

So, i =  ε/R

=  ε/ρl/A

= εA/ρl

= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)

= ‭15.704‬ × 10⁻⁶ V-m²/(82.9356‬ × 10⁸ Ωm²

= 0.1894 × 10⁻¹⁴ A

= 1.894 × 10⁻¹⁵ A

≅ 1.9 fA

Answer:

The answer is "1.26".

Explanation:

[tex]D=18^{\circ}[/tex]

The refractive index is:

[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]

       [tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]

Answer:

hope you can understand