Select the correct answer.
If you increase the frequency of a sound wave four times, what will happen to its speed?
A.
The speed will increase four times.
B.
The speed will decrease four times.
C.
The speed will remain the same.
D.
The speed will increase twice.
E.
The speed will decrease twice.

Answer:

t = 6 minutes

Explanation:

Given that,

Force,[tex]F=6.8\times 10^5\ N[/tex]

Initial speed of the train, u = 0

Final speed of the train, v = 80 km/h = 22.22 m/s

The mass of the train, [tex]m=1.1\times 10^7\ kg[/tex]

We need to find the time taken by the train to come to rest. We know that,

F = ma

[tex]F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.1\times10^7\times (22.22-0)}{6.8\times 10^5}\\\\t=359.44\ s[/tex]

or

t = 6 minutes (approx)

So, the required time is equal to 6 minutes.

Answer:

C

Explanation:

I think it would be there, it sounds like silicone and thats on the right side

By the work-energy theorem, the total work done on the car is equal to the change in its kinetic energy:

W = ∆K

W = 1/2 (0.34 kg) (22.9 m/s)² - 1/2 (0.34 kg) (6.5 m/s)²

W ≈ 82 J

Answer:

The particle speed is constant.

Explanation:

Particles in gases travel quickly in all directions, frequently clashing with each other and the container's edge. The particles gather kinetic energy and travel faster as the temperature rises. The true average speed of the particles is determined by their mass and temperature; larger particles travel more slower around the same temperature than lighter particles.

Thus, the false statement about a wave moving through a constant speed is that:

The particle speed is constant.

M1 = 15/g = 15/9.8 = 1.53 kg = mass of object in air. M2 = 12/9.8 = 1.22 kg = mass of object immersed. M1-M2 = 1.53-1.22 = 0.31 kg lost by object = mass of water displaced. ... Do = 4.94 g/cm^3 = density of object.

Answer:

The angle is 4.1 rad.

Explanation:

The centripetal acceleration (α) is given by:

[tex] \alpha = \omega^{2} r [/tex]    (1)                  

Where:

ω: is the angular velocity  

r: is the radius

And the tangential acceleration (a) is:                      

[tex] a = \alpha r [/tex]      (2)

Since the magnitude of "α" is 8.2 times the magnitude of "a" (equating (2) and (1)) we have:

[tex] \omega^{2} r = 8.2\alpha r   [/tex]

[tex] \omega^{2} = 8.2\alpha [/tex]    (3)      

Now, we can find the angle with the following equation:

[tex] \omega_{f}^{2} = \omega_{0}^{2} + 2\alpha \Delta \theta [/tex]

Where:

[tex] \omega_{f}[/tex]: is the final angular velocity                                                                              [tex] \omega_{0}[/tex]: is the initial angular velocity = 0 (it starts from rest)

[tex]\Delta \theta[/tex]: is the angle

[tex] \omega^{2} = 2\alpha \Delta \theta [/tex]     (4)    

By entering equation (3) into (4) we can calculate the angle:

[tex] 8.2\alpha = 2\alpha \Delta \theta [/tex]

[tex] \Delta \theta = 4.1 rad [/tex]

Therefore, the angle is 4.1 rad.

I hope it helps you!                  

Answer:

19

Explanation:

Given that:

wavelength = 600 nm

Distance (d) = 2.02 cm = 2.02 × 10⁻² m

refraction index of air (n) = 1.00028

Pressure = 1.00 atm

∴

The number of bright-dark-bright fringe shifts can be determined by using the formula:

[tex]\Delta m = \dfrac{2d}{\lambda} (n -1 ) \\ \\ \Delta m = \dfrac{2\times2.02 \times 10^{-2}}{600\times 10^{-9}} (1.00028 -1 ) \\ \\ \Delta m = 67333.33 \times 10^{-5}(1.00028 -1) \\ \\ \Delta m = 67333.33 \times 10^{-5}(2.8\times 10^{-4}) \\ \\ \Delta m = 18.853 \\ \\ \mathbf{\Delta m = 19}[/tex]

Answer:

Yes

Explanation:

Given that the battery is the same the PD ( potential difference ) in the circuit will also be the same likewise the flow of charge in the circuit,

Hence the same amount of charge flow is delivered to any circuit.

attached below are examples

Answer:

the first answer

Explanation:

(32°F − 32) × 5/9 = 0°C

Answer:

Answer: A

Explanation:

Answer:

v = 6491.94 m/s

Explanation:

We are given;

Frequency; f = 87 Hz

Wavelength;λ = 74.62 m

Formula for velocity(v) of waves from the wave equation is;

v = fλ

Thus;

v = 87 × 74.62

v = 6491.94 m/s

Answer:

The carbon rod in batteries is used as an inert electrode

Explanation:

A battery is considered as a power source that consists of one or more electrochemical cells having an external connections to provide power to the electrical devices such as the lights, bulbs, fans, mobile phones, etc.

It contains a positive terminal and a negative terminal.

The carbon rod in the battery does not help in the electrochemical reactions. It acts as an inert electrode and helps to flow the electrons only.

Thus the true statement is :

The carbon rod in batteries is used as an inert electrode.

Answer:

12

Explanation:

I'm a beginner so am not sureeeeee

Answer:

Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature); dT/dt = -K(T - Tₐ) where T = the temperature of the body (°C), t = time (min), k = the proportionality constant (per minute),

Explanation:

Answer:

speeding up is the answer

Explanation:

from the graph it can be seen that as the time (horizontal axis) increases the speed of vehicle (vertical axis) increases

Answer:

F' = 73.7 N

F = 8.749×10⁻³ N

a' = a =  9.83 m/s²

Explanation:

(a)

For the rock

Applying

F' = Gm'm/r²................... Equation 1

Where F = magnitude of the gravitational force on the rock, G = Gravitational constant, m' = mass of the rock, m = mass of the earth, r = radius of the earth.

From the question,

Given: m' = 7.5 kg

Constant: m = 5.98×10²⁴ kg, G = 6.67×10⁻¹¹ Nm²/kg², r = 6.37×10⁶ m

Substitute these values into equation 1

F' = 6.67×10⁻¹¹ (7.5)(5.98×10²⁴)/(6.37×10⁶)²

F' = 7.37×10¹ N

F' = 73.7 N

Also, For the pebble,

F = GMm/r².............. Equation 2

Where M = mass of the pebble, F = Gravitational force exerted on the pebble by the earth

Given: M = 8.9×10⁻⁴ kg,

Substitute into equation 2

F = 6.67×10⁻¹¹(8.9×10⁻⁴)(5.98×10²⁴)/(6.37×10⁶)²

F = 8.749×10⁻³ N

(b)

For the rock,

a' = F'/m'

Where a' = magnitude of the acceleration of the rock

a' = 73.7/7.5

a' = 9.83 m/s²

For the pebble,

a = F/M

Where a = acceleration of the pebble

a = (8.749×10⁻³)/(8.9×10⁻⁴)

a = 9.83 m/s²

Answer:

Δd = 7.22 10⁻² m

Explanation:

For this exercise we must use the dispersion relationship of a diffraction grating

           d sin θ = m λ

let's use trigonometry

           tan θ = y / L

how the angles are small

           tant θ = sinθ  /cos θ = sin θ

we substitute  

           sin θ = y / L

          d y / L = m λ

          y = m λ L / d

let's use direct ruler rule to find the distance between two slits

If there are 500 lines in 1 me, what distance is there between two lines

         d = 2/500

        d = 0.004 me = 4 10⁻⁶ m

diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1

let's calculate for each wavelength

λ = 656 nm = 656 10⁻⁹ m

         d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶

         d₁ = 2.788 10⁻¹ m

λ = 486 nm = 486 10⁻⁹ m

         d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶

         d₂ = 2.066 10⁻¹ m

the distance between the two lines is

         Δd = d1 -d2

         Δd = (2,788 - 2,066) 10⁻¹

         Δd = 7.22 10⁻² m

Answer:

the rate of heat transfer after the system achieves steady state is -3.36 kW

Explanation:

Given the data in the question;

mass of water m = 50 kg

N = 300 rpm

Torque T = 0.1 kNm

V = 110 V

I = 2 A

Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW

Now, work supplied by paddle wheel W₂ is;

W₂ = 2πNT/60

W₂ = (2π × 0.1 × 300) / 60

W₂ = 188.495559 / 60

W₂ = 3.14 kW

So the total work will be;

W = 0.22 + 3.14

W = 3.36 kW

Hence total work done on the system is 3.36 kW.

At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.

The heat will be rejected by the system so the sign of heat will be negative.

i.e Q = -3.36 kW

Therefore,  the rate of heat transfer after the system achieves steady state is -3.36 kW

Answer:

0.4

Explanation:

[tex] \frac{1}{2} mv ^{2} [/tex]

kinetic energy formula , potential energy is not considered

0.5×0.2×2×2

Answer:

Explanation:

From the given information:

The difference in the maximum energy stored is can be determined by finding the difference in the maximum stored energy in the sprinters and that of the non-athlete:

[tex]\Delta U = \dfrac{1}{2}k(x_2^2 - x_1^2)[/tex]

[tex]\Delta U = \dfrac{1}{2} (32 \ N/mm) (\dfrac{1 \ mm}{10^{-3} \ m}) ((40\times 10^{-3})^2 - (32\times 10^{-3})^2)[/tex]

[tex]\Delta U =16000 \times (5.76\times 10^{-4})[/tex]

[tex]\mathbf{\Delta U =9.216\ J}[/tex]

Answer:

Explanation:

We are looking for final velocity. Since the ball is thrown horizontally, there is no upwards velocity, so the y dimension here is only useful to us for finding how long the ball was in the air. In the y dimension, here's what we know:

a = -9.8 m/s/s

Δx = -59 m

[tex]v_0=0[/tex] (again, initial upwards velocity is 0 because the ball was thrown horizontally)

We can put all that together in the equation:

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and filling in:

[tex]-59=0t+\frac{1}{2}(-9.8)t^2[/tex] which simplifies to

[tex]-59=\frac{1}{2}(-9.8)t^2[/tex] and solving for t:

[tex]t=\sqrt{\frac{2(-59)}{-9.8} }[/tex] and

t = 3.5 sec

Now we can use that time in the d = rt equation, which is all we need for the horizontal dimension (I'll show you why in just a second). In the horizontal dimension, here's what we know:

a = 0 m/s/s

Δx = 65 m

t = 3.5 sec

Putting that all together in our one-dimensional equation for displacement:

Δx = [tex]v_0t+\frac{1}{2}at^2[/tex] and acceleration is 0, we can simplify that down to

Δx = [tex]v_0t[/tex] which is the exact same thing as d = rt where r is the velocity we are looking for. Filling in:

65 = v(3.5) so

v = 18.6 m/s

That's the velocity with which the ball strikes the ground.

Answer:

Use below table to read the color codes.

Answer:

A. 1 m/s²

B. 312.5 m

C. 12.5 m/s

Explanation:

We'll begin by converting the velocity i.e 90 Km/h to m/s. This can be obtained as follow:

Velocity (Km/h) = 90 Km/h

Velocity (m/s) =?

Velocity (m/s) = Velocity (Km/h) × 1000 / 3600

Velocity (m/s) = 90 × 1000 / 3600

Velocity (m/s) = 90000 / 3600

Velocity (m/s) = 25 m/s

A. Determination of the acceleration.

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Time (t) = 25 s

Acceleration (a) =?

v = u + at

25 = 0 + (a × 25)

25 = 0 + 25a

25 = 25a

Divide both side by 25

a = 25/25

a = 1 m/s²

B. Determination of the distance travelled.

Initial velocity (u) = 0 m/s

Final velocity (v) = 25 m/s

Acceleration (a) = 1 m/s²

Distance travelled (s) =?

v² = u² + 2as

25² = 0 + (2 × 1 × s)

625 = 0 + 2s

625 = 2s

Divide both side by 2

s = 625 / 2

s = 312.5 m

C. Determination of the average velocity.

Total distance travelled = 312.5 m

Total time = 25 s

Average velocity =?

Average velocity = Total distance / total time

Average velocity = 312.5 / 25

Average velocity = 12.5 m/s

Answer:

No of proton is 13 and nucleus is 13

Answer:

not using it too much and getting addicted

Explanation:

Answer:

time is 32 s and speed is 304.3 m/s

Explanation:

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

[tex]h = u t +0.5 at^2\\\\- 146 =14 sin 43 t - 4.9 t^2\\\\4.9 t^2 - 9.5 t - 146 =0 \\\\t =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\\\\t=\frac{9.5\pm 54.3}{9.8}\\\\t = 32.05 s, - 22.4 s[/tex]

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is  

v' =  u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

[tex]v=\sqrt{v'^2 + v''^2}\\\\v = \sqrt{304.1^2 +10.3^2 }\\\\v = 304.3 m/s[/tex]  

Answer:

Solid is the state in which Matter maintains a fixed volume

Answer:

The state of matter that has a fixed volume is Solid.

Explanation:

Solid substances will maintain a fixed volume and shape.

Answer:

460.52 s

Explanation:

Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that

dV/dt ∝ V

dV/dt = kV

separating the variables, we have

dV/V = kdt

integrating both sides, we have

∫dV/V = ∫kdt

㏑(V/V₀) = kt

V/V₀ = [tex]e^{kt}[/tex]

Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V

Since dV/dt = kV

-0.01V = kV

k = -0.01

So, V/V₀ = [tex]e^{-0.01t}[/tex]

V = V₀[tex]e^{-0.01t}[/tex]

Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀

So, V = 0.1V₀

Thus

V = V₀[tex]e^{-0.01t}[/tex]

0.1V₀ = V₀[tex]e^{-0.01t}[/tex]

0.1V₀/V₀ = [tex]e^{-0.01t}[/tex]

0.1 = [tex]e^{-0.01t}[/tex]

to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have

㏑(0.01) = -0.01t

So, t = ㏑(0.01)/-0.01

t = -4.6052/-0.01

t = 460.52 s