The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/m. The tension in the string is __________.

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)

Answer:

a) 0 V

b) 10 turns

c) 4000 turns

d) 12.5 A

e) 400 W

f) 0.5 A

g) 95.4%

Explanation:

A

0

B

To solve this, we would be using the simple relationship between voltage and number of turns

V1/V2 = N1/N2

500/25 = 200/N2

20 = 200/N2

N2 = 200/20

N2 = 10 turns

C

Here also, we would be using the relationship between current and the number of turns

I1/I2 = N2/N1

500/25 = N2/20

20 = N2/20

N2 = 20 * 20

N2 = 4000 turns

D

Like in the previous question, current and the number of turn relationship is used

N1/N2 = I2/I1

400/80 = I2/2.5

5 = I2/2.5

I2 = 5 * 2.5

I2 = 12.5 A

E

The power remains unchanged at 400 W

F

Power = Voltage * Current

P = VI

I = P/V

I = 60/120

I = 0.5 A

G

95.4%

The transformer is a device used to step up or step down voltage.

Part A;

Given that;

Es/Ep = Ns/Np

Es = voltage in the secondary coil

Ep = voltage in primary coil

Ns = Number of turns in secondary coil

Np = Number of coils in primary coil

Es = Ns/Np ×  Ep

Es = 200/100  × 1.5 V

Es = 3 V

Part B

Ns = Es/Ep × Np

Ns = 25/500 × 200

Ns = 10 turns

Part C

Ns/Np = Ip/Is

Ns = Ip/Is × Np

Ns = 500/25 × 200

Ns = 4000 turns

Part D

Ns/Np = Ip/Is

NsIs = NpIp

Is = NpIp/Ns

Is = 400 × 2.5/80

Is =12.5 A

Part E

The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.

Part F

Power supplied = 60 watts

Voltage of primary coil = 120 V

Since;

P = IV

I = P/V = 60/120 = 0.5 A

Part G

Since;

E = 100Pout/Pin

Pin = 120 V × 2 A = 240 W

Pout =  19.4 V ×  11.8 A = 228.92 W

E = 100(228.92/240)

E = 95.4%

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Answer:

The value is  [tex]\mu_k = 0.102[/tex]

Explanation:

From the question we are told that

   The initial speed of the pluck is  [tex]u = 10 \ m/s[/tex]

    The  distance it slides on the horizontal ice is  [tex]s = 50 \ m[/tex]

Generally from kinematic equation we have that

       [tex]v^2 = u^2 + 2as[/tex]      

Here v is  is the final velocity and the value is 0 m/s given that the pluck came to rest, so

      [tex]0^2 = 10 ^2 + 2* a * 50[/tex]      

=>   [tex]a = - 1 \ m/s^2[/tex]

Here the negative sign show that the pluck is decelerating  

 Generally the force applied on the pluck is  equal to the frictional force experienced by the pluck

      So  

                [tex]F = F_f[/tex]

=>            [tex]m * a = m* g * \mu_k[/tex]

=>             [tex]1 = 9.8 * \mu_k[/tex]

=>              [tex]\mu_k = 0.102[/tex]

Answer:

Angular speed = 27.78 rad/s (Approx)

Explanation:

Given:

Diameter = 21.6 cm

Speed = 3 m/s

Find:

Angular speed

Computation:

Radius = 21.6 / 2 = 10.8 cm = 0.108 m

Angular speed = v / r

Angular speed = 3 / 0.108

Angular speed = 27.78 rad/s (Approx)

its A or D but im not sure which one ik it moves fast

Answer:

816

Explanation:

We must remember that torque is defined as the product of a force by a distance.

This distance is measured from the pivot point of the Bolt to the point where the force is applied. in this way, we have the following equation to be able to determine the torque.

[tex]T=F*d[/tex]

Now if we double the turning distance we have the torque should also be double.

[tex]T =F*2*d[/tex]

Answer:

The maximum angular velocity is 20 rad/s

Explanation:

Given;

torque, τ = 10 N

maximum mechanical power, P = 200 J/s

The output power of the pmdc is given as;

P = τω

where;

P is the maximum mechanical power

ω is the maximum angular velocity

ω = P / τ

ω = (200) / (10)

ω = 20 rad/s

Therefore, the maximum angular velocity is 20 rad/s

Answer: 54.88 meters/s

Explanation:

The final velocity will be calculated by using the formula:

v = u + at

where,

v = final velocity

u = initial velocity = 0

a = 9.8

t = 5.6

Therefore, we slot the value back into the formula. This will be:

v = u + at

v = 0 + (9.8 × 5.6)

v = 0 + 54.88

v = 54.88 meters per second

Therefore, the final velocity is 54.88m/s

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

Answer:

12445 J

Explanation:

Given that

Power output, P = 5 kW

efficiency of the engine, e = 27% = 0.27

Thermal energy expelled, Q(c) = 9085 J

Heat absorbed, Q(h) = ?

Using the formula

e = W/Q(h)

e = [Q(h) - Q(c)] / Q(h)

e = 1 - Q(c)/Q(h)

Now, substituting the values into the formula, we have

0.27 = 1 - 9085/Q(h)

9085/Q(h) = 1 - 0.27

9085/Q(h) = 0.73

Q(h) = 9085 / 0.73

Q(h) = 12445 J

Thus, the heat absorbed is 12445 J

Answer:

Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.

Explanation:

Hope this helps

Answer:

Δx = 39.1 m

Explanation:

        [tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

        [tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]        

The car will travel a distance of 39.1 m before its stops.

To solve the problem above, use the equations of motion below.

Equation:

Where:

From the question,

Given:

Substitute these values into equation 1

Solve for s

Hence, The car will travel a distance of 39.1 m before its stops.

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Answer:

the placement of the ammeter in the circuit

Explanation:

Given :

The decibel scale intensity for busy traffic is 80 dB.

Two people having a loud conversation have a decibel intensity of 70 dB.

To Find :

The approximate combined sound intensity.

Solution :

We know, intensity in decibel can be converted to W/m² by :

[tex]\beta(dB) = 10\ log_{10}( \dfrac{I}{I_o})[/tex]

Putting intensity in decibel scale, we get :

[tex]I(80\ dB ) =10^{-4}\ W/m^2\\\\I(70\ dB ) = 10^{-5}\ W/m^2[/tex]

Let, combine intensity is I .

I = I(80 dB) + I(70 dB)

[tex]I = 10^{-4} + 10^{-5} \ W/m^2\\\\I = 1.1 \times 10^{-4} \ W/m^2[/tex]

Therefore, the combined sound intensity is [tex]1.1 \times 10^{-4} \ W/m^2[/tex] .

The exercise involves filling in the gaps with the possible wave

properties that can be obtained from a spring.

The correctly completed exercise is presented as follows;

A wave is a disturbance that moves through a medium. There are two

types of waves. A mechanical wave requires matter to travel. List some

examples of this type: sound wave, water wave, spring waves.

A electromagnetic wave does not require a medium. Examples include: Light waves

In order to start and transmit a mechanical wave, a source of

disturbance and a physical medium are required. A single disturbance is

referred to as a pulse, and a series of disturbance is a wave train.

This type of wave is called transverse wave. Its pulses are called crest

and troughs.

Now send a pulse by quickly pushing the spring forward and pulling it

back, as shown. This type of wave is called longitudinal wave. Watch the

motion of the ribbon. In this type, the particles move parallel to the

direction the wave travels. Its pulses are called compression and

rarefactions. Note that all waves transfer energy without transferring

matter. In mechanical waves, particle of the medium vibrate back and

forth in simple harmonic motion while the disturbance (or energy)

moves from one place to another.

B. Wave speed

Does the pulse catch up to the other? yes. The size of the pulse is called

the amplitude of the wave.

Did the size of the pulse affect the speed? No.

The average time wave it takes the wave to travel

Without changing your positions therefore keeping the distance the

pulse travels the same), pull the slinky tighter using only about 3/4 of

coils. This makes a completely different medium through which the

pulse will travel. Time the journey as before time record. Does the kind

of medium affect the speed of the pulse? Yes

C. Wavelength and Frequency

High frequency waves have short wavelengths and low frequency waves

have long wavelengths.

The speed of a wave in any medium is equal to the frequency of the wave × the wavelength. This wave equation [tex]\underline{f = \dfrac{v}{\lambda } }[/tex] shows that f and λ are

inversely proportional. The units of the variables are;

Learn more about waves here:

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Answer:

[tex]MA = 3[/tex]

Explanation:

Given

[tex]Box = 10kg[/tex]

[tex]Ramp\ Height = 5ft[/tex]

[tex]Ramp\ Length = 15ft[/tex]

Required

Determine the mechanical advantage

This is calculated as follows:

[tex]MA = \frac{Ramp\ Length}{Ramp\ Height}[/tex]

[tex]MA = \frac{15ft}{5ft}[/tex]

[tex]MA = 3[/tex]

Hence, the mechanical advantage is 3

Answer:

C) 600 Hz

Explanation:

The fundamental frequency can be related to the driving frequency by the expression below;

f(n) = n * f(1)

Where f(1)= fundamental frequency

f(n) = driving frequency

There are four equal segments in the standing wave , then our n= 4 and our f(n)=4, then we can get the fundamental frequency here

f(4) = 4× f(1)

480 = 4× f(1)

f(1) = 480/4

f(1)=120Hz

Hence, fundamental frequency is 120Hz

To calculate the driving frequency that will set up a standing wave with five equal segments?

n=5

f(n) = n× 120Hz

f(5) = 5×120Hz

= 600Hz.

Hence, the driving frequency that will set up a standing wave with five equal segments is 600Hz

Answer:

The answer is "1557 meters".

Explanation:

speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]

[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]

Given :

Brandon hits a golf ball with an initial velocity of 30 m/s at an angle of 30 above the horizontal.

To Find :

How long is it in the air.

Solution :

We know, the formula of time of flight is :

[tex]T = \dfrac{2usin\ \theta}{g}\\\\T = \dfrac{2\times 30\times sin\ 30^o}{9.8}\\\\T = 3.06\ seconds[/tex]

Therefore, the ball is in air for 3.06 seconds.

Answer: The correct answer is C

Explanation:

it would be the 3rd one. so C

The awnser is sulfur dioxide.

The gas that is responsible for lowering the temperature is sulfur dioxide gas.

A volcanic eruption refers to the sudden escape of lava and gases from the earth's core. It usually occurs at a very high temperature. The solidification of the volcano leads to rock formation.

During a volcanic eruption, the gas that is responsible for lowering the temperature is sulfur dioxide gas.

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Answer:

[tex]\theta = 16.70 ^{\circ}[/tex]

Explanation:

The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.  

Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.

Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.

We are given the mass of the block (kg) and we know that g = 9.8 m/s².

Now we can use this force in the equation:

Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.

Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].

Evaluate this equation by taking the inverse tangent of both sides of the equation.

The minimum angle at which the block will begin to slide is about 16.70 degrees.

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]

[tex]\\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]

[tex]\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]

[tex]\\[/tex]

☯ For left penetration (s₂)

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]

[tex]\\[/tex]

[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]

Answer:

-240

Explanation:

A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.

The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.

Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.

When the body slows down from its present level to a condition of rest, the same quantity of energy is used.

Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.

As per the given information in the question,

Distance, d = 2.0 m

Friction, f = 120 N

The angle between displacement and friction force, θ = 180°

Now, the change in kinetic energy for the motorcycle = Work done by the friction.

K.E = f × d(cos θ)

= 120 (2.0 m)(cos 180°)

Δ K.E = -240 J

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Answer:

the gravitational force decreases