I NEED HELP THIS QUESTION IS SO HARDDD!!

Answer:

yong may check po ayon yong sagot

Answer:

the wavelength of the sound in seawater is 90.1 m.

Explanation:

Given;

frequency of the sound, f = 17 Hz

speed of the sound in seawater, v = 1531 m/s

The wavelength of the wave is calculated as follows;

v = fλ

λ = v / f

where;

λ is the wavelength of the sound

λ = 1531 / 17

λ = 90.1 m

Therefore, the wavelength of the sound in seawater is 90.1 m.

Answer:

ook soooooo

Explanation:

Answer:

Height of the door = 2m = 2000 cm

1 in = 2.54 cm

So 1 cm = 1/2.54 in

2000 cm = 200000/ 254

=

787.401574803

So no.c is correct

The door is  78.7 inch tall. Hence, option (A) is correct.

Any arbitrarily selected and widely used reference standard for length measurement is referred to as a unit of length. The metric system, which is adopted by every nation on earth, is the most widely utilized in modern times.

The American customary units are also in use in the United States. In the UK and several other nations, British Imperial units are still used sometimes. There are SI units and non-SI units in the metric system.

Given that: the height of the door is = 2 meter

= 2*100 centimeter

= 200 centimeter.

There are 2.54 centimeter in 1 inch.

Hence,  the height of the door is = 2 meter  = 200 centimeter

= (200/2.54) inch

= 78.7 inch.

The door is 78.7 inch tall.

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Answer:

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m  

Period:

The period is given as:

[tex]Period = \frac{time}{no.\ of\ cycles} \\\\Period = \frac{10\ s}{1}\\\\[/tex]

Period = 10 s

Frequency:

The frequency is given as:

[tex]f = \frac{1}{time\ period}\\\\f = \frac{1}{10\ s}\\\\[/tex]

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m

Answer:

[tex]f_{U}= 1019.72hz[/tex]

Explanation:

From the equation we are told that:

Velocity of French sub [tex]V_{U}= 43.00km/h[/tex]

Velocity of U.S. sub at [tex]V_{F}|=64.00 km//h[/tex]

French Wave Frequency  [tex]F_{F}=1000Hz[/tex]

Velocity of wave  [tex]V_{s}=5470 km/h[/tex]

Generally the equation for Signal's frequency as detected by the U.S. is mathematically given by

Doppler effect

 [tex]f_{U} = \frac{ f_F (vs + v_{U})}{(v_s - v_F) }[/tex]

 [tex]f_{U}= \frac{ 1000 x ( 5470+64)}{(5470-43)}[/tex]

 [tex]f_{U}= 1019.72hz[/tex]

The relative error in this weight measurement as percentage is

is the ratio of the absolute error of a measurement to the measurement being taken. In other words, this type of error is relative to the size of the item being measured.

Therefore,

[tex]\% relative error = \frac{smallest division}{mass of apples}*100\\\\\% relative error = \frac{2.1g}{1900}*100\\\\\% relative error = 0.1105\%[/tex]

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Answer:

The relative error in this weight measurement is approximately 0.1105 %.

Explanation:

We know that,
Relative Error is the ratio of the absolute error of a measurement to the actual measurement . It can be mathematically represented as,

Relative Error  = (Measured Value - Actual Value)  / Actual Value * 100

In this case, the actual weight is 1.9 kg, and the smallest division of the spring scale is 2.1 g.

Actual Weight = 1.9  * 1000  = 1900  (As 1 kg = 1000g)

We know that the absolute error will be equal to the smallest division hence,

Measured Value - Actual Value = 2.1 g

By replacing this in the general formula we get,

Relative Error = (Measured Value - Actual Value) / Actual Value * 100

= [tex]\frac{2.1}{1900} *100[/tex] %

= [tex]\frac{2.1}{19}[/tex] %

≈ 0.1105 %

Therefore, the relative error in this weight measurement is approximately 0.1105 %.

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Answer: [tex]1.102\ J[/tex]

Explanation:

Given

Mass [tex]m=1.3\ kg[/tex]

Spring constant [tex]k=58\ N/m[/tex]

Compression in the spring [tex]x=19.5\ cm\ or\ 0.195\ m[/tex]

When the mass leaves the spring, the elastic potential energy of spring is being converted into kinetic energy of mass i.e.

[tex]\Rightarrow \dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}\cdot 58\cdot (0.195)^2=\dfrac{1}{2}mv^2\\\\\Rightarrow \dfrac{1}{2}mv^2=1.102\ J[/tex]

The kinetic energy of the mass is 1.102 J.

Answer:

check out of phase

Explanation:

this is my answer

Answer:

     λ = 5.75 10⁻⁷ mm

Explanation:

This is a slit interference exercise, we analyze each wavelength separately                    

            λ = 657 nm                                     indicate that the third dark pattern

          a sin θ = (m + ½) lam

          a sin θ = (3 + ½) 657 10⁻⁹

          a sin θ = 2299.5 10⁻⁹ nm

for the other wavelength in the same place we have m = 4 bright

          a sin θ = m lam

we substitute

           2299.5 10⁻⁹ = 4 λ

           λ = [tex]\frac{2299.5 \ 10^9 }{ 4}[/tex]

           λ = 5.75 10⁻⁷ mm

Answer:

1. F = 3400 N = 3.4 KN

2. [tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3. v = 14.9 m/s

Explanation:

1.

First, we will calculate the acceleration of Pam by using the third equation of motion:

[tex]2as = v_f^2-v_i^2[/tex]

where,

a = acceleration = ?

s = distance = 27.3 m

vf = final speed = 62 m/s

vi = initial speed = 0 m/s

Therefore,

[tex]2a(27.3\ m) = (62\ m/s)^2-(0\ m/s)^2\\\\a = 70.4\ m/s^2[/tex]

Now, we will calculate the force by using Newton's Second Law of Motion:

F = ma

F = (48.3 kg)(70.4 m/s²)

F = 3400 N = 3.4 KN

2.

Final kinetic energy is given as:

[tex]K.E_f = \frac{1}{2}mv_f^2\\\\K.E_f = \frac{1}{2} (48.3\ kg)(62\ m/s)^2[/tex]

[tex]K.E_f=92832.6\ J = 92.83\ KJ[/tex]

3.

According to the law of conservation of energy:

[tex]Potential\ Energy\ at\ top = Kinetic\ Energy\ at\ bottom\\mgh = \frac{1}{2}mv_2 \\\\v = \sqrt{2gh}[/tex]

where,

v = speed at bottom = ?

g = acceleration due to gravity = 9.81 m/s²

h = height at top = 11.3 m

Therefore,

[tex]v = \sqrt{(2)(9.81\ m/s^2)(11.3\ m)}[/tex]

v = 14.9 m/s

Answer:

Explanation:

Let after time t , Tina catches up David .

Distance travelled by them are equal ,

Distance travelled by Tina

s = ut + 1/2 a t²

= .5 x 2.10 t²

= 1.05 t²

Distance travelled by David

= 30 t ( because of uniform velocity )

1.05 t² = 30t

t = 28.57 s

Distance travelled by Tina

= 1/2 a t²

= .5 x 2.10 x 28.57²

= 857 m approx.

Answer: [tex]857\ m[/tex]

Explanation:

Given

Speed of David car [tex]v=30\ m/s[/tex]

Tina begins to accelerate [tex]2.1\ m/s^2[/tex] after David pass the tina

Suppose it took t time for tina to catch David

Distance traveled by David in t time

[tex]\Rightarrow s_d=30\times t[/tex]

Using the equation of motion to get the distance of Tina is

[tex]s_t=ut+\dfrac{1}{2}at^2\\\\s_t=0+\dfrac{1}{2}\times 2.1t^2[/tex]

now, [tex]s_d=s_t[/tex]

[tex]30t=\dfrac{2.1}{2}t^2\\\\\Rightarrow 2.1t^2-60t=0\\\Rightarrow t(2.1t-60)=0\\\Rightarrow t=0,28.57\ s[/tex]

Neglecting [tex]t=0[/tex]

Distance traveled by tina in [tex]28.57\ s[/tex] is

[tex]s_t=\dfrac{1}{2}\times 2.1\times (28.57)^2\\\\s_t=857.057\approx 857\ m[/tex]

[tex]\huge{ \mathfrak{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Copper wires are used as connecting wires in circuits because they offer very less resistance.

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Answer:

Option C. 4 Hz

Explanation:

To know the correct answer to the question given above, it is important we know the definition of frequency.

Frequency can simply be defined as the number of complete oscillations or circles made in one second.

Considering the diagram given above, the wave passes through the medium over a period of one second.

Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.

From the diagram given above,

The number of circles = 4

Thus,

The frequency is 4 Hz

Answer:

Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit

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Answer:

3.11ev

Explanation:

[tex]E = hf[/tex]

[tex]E = 6.626 * 10^-^3^4 * 7.5 * 10^1^4\\E = 49.695 * 10^-^2^0[/tex]

[tex]1ev = 1.60 * 10^-^1^9\\xev = 49.695 * 10^-^2^0 / 1.60 * 10^-^1^9\\xev = 3.1059\\xev = 3.11eV[/tex]

Answer:

optimism

Explanation:

just an idea

....

Complete question is;

A small car of mass m and a large car of mass 2m drive around a highway curve of radius R. Both cars travel at the same speed (v). The

centripetal acceleration (Grad) of the large car is the centripetal acceleration of the small car. How does the Force of the small car FS compare to the force of the large car FL as they round the curve.

four times

twice

half

equal to

Answer:

Half

Explanation:

Formula for centripetal force is given as;

F = mv²/R

Where;

v is velocity

R is radius

Now, centripetal acceleration is given by;

a = v²/R

Since they both travel with the same velocity V and radius remains the same, we can say that;

F = ma

For the small car;

FS = ma

For the big car;

FL = 2ma

This means the force of the small car is half of that of the Large car

Thus;

FS = ½FL

Answer:

[tex](a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]

[tex](b)\ F_{No} = 771.125N[/tex]

Explanation:

Given

[tex]d_D = 6000ft[/tex] ---- Altitude of container in Denver

[tex]A = 0.0155m^2[/tex] -- Surface Area of the container lid

[tex]P_D = 79000Pa[/tex] --- Air pressure in Denver

[tex]P_{No} = 100250Pa[/tex] --- Air pressure in New Orleans

See comment for complete question

Solving (a): The expression for [tex]F_{No[/tex]

Force is calculated as:

[tex]F = \triangle P * A[/tex]

The force in New Orleans is:

[tex]F_{No} = \triangle P * A[/tex]

Since the inside pressure is half the pressure at sea level, then:

[tex]\triangle P = P_{No} - \frac{P_{area}}{2}[/tex]

Where

[tex]P_{area} = 101000Pa[/tex] --- Standard Pressure

Recall that:

[tex]F_{No} = \triangle P * A[/tex]

This gives:

[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]

Solving (b): The value of [tex]F_{No[/tex]

In (a), we have:

[tex]F_{No} = [P_{No} - \frac{P_{area}}{2}]* A[/tex]

Where

[tex]A = 0.0155m^2[/tex]

[tex]P_{No} = 100250Pa[/tex]

[tex]P_{area} = 101000Pa[/tex]

So, we have:

[tex]F_{No} = [100250 - \frac{101000}{2}] * 0.0155[/tex]

[tex]F_{No} = [100250 - 50500] * 0.0155[/tex]

[tex]F_{No} = 49750* 0.0155[/tex]

[tex]F_{No} = 771.125N[/tex]

Answer:

S = Vo t * 1/2 a t^2     equation for distance traveled

w = w0 t + 1/2 a t^2    equivalent circular equation where w equals omega

w = 5.98 * 27.5  + a (27.5)^2/ 2

a = (w2 - w1) / t = -5.98 / 27.5 = -.217

w = 5.98 * 27.5 - 1/2 * .217 * 27.5^2 = 82.4

Since 1 Rev = 2 pi  radians      

Rev = 82.4 / 2 * pi = 13.1 Rev

Answer:

Cellular respiration takes place in the cells of all organisms. It occurs in autotrophs such as plants as well as heterotrophs such as animals. Cellular respiration begins in the cytoplasm of cells. It is completed in mitochondria

Explanation:

Cellular respiration takes place in the cells of all organisms. It happening in autotrophs such as plantas as well as heterotrophs such as animals. Cellular respiration starts in the cytoplasm of cells.

It is finished in mitochondria.

Answer:

four glorps

Explanation:

We know :

[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]

[tex]$\Rightarrow -1 \text{glorp} = 0 - \frac{g}{2} \times (1 g\text{ gleep})^2$[/tex]

[tex]$\Rightarrow 1 \text{ glorp}= \frac{g}{2} (1 \text{ gleep})^2$[/tex]   .............(i)

Now, t' = 2 gleep

[tex]$y=v_{0y}t + \frac{1}{2}a_yt^2$[/tex]

  [tex]$=0+ \frac{-g}{2} (2 \text{ gleep})^2$[/tex]

  [tex]$=-\frac{4g}{2}(2 \text{ gleep})^2$[/tex]

 [tex]$=4\left[\frac{-g}{2} (\text{gleep})^2\right]$[/tex]

 = 4 (-1 gleep)     (From (i))

So, |y| = 4 glorp

You really don't need any help on the question.  It's all right there in the picture.  What you need help with is the answer.

The number of times the same thing happens each second is called its "frequency".  The frequency of the dragonfly's flaps is 477 Hz.  (If you're close enough to the dragonfly, you can hear the wings flapping.  It sounds like a raspy tone with a frequency of 477 Hz.)

The "period" is just the length of time it takes to happen once.  That length of time is just  (1 / frequency) .

The dragonfly flaps its wings once every  (1 / 477 Hz) = 0.0021 second (C)

Answer:

The answer is "[tex]4659.2 \times 10^{-24} \ N[/tex]"

Explanation:

The magnetic field at ehe mid point of the coils is,

[tex]\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\[/tex]

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.

[tex]\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\[/tex]

       [tex]=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\[/tex]

Calculating the force experienced through the protons:

[tex]F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N[/tex]

Rt = 3 ohms

Explanation:

Let R1 = 4-ohm resistor

R2 = 12-ohm resistor

For 2 resistors connected in parallel, the total resistance Rt is given by

1/Rt = 1/R1 + 1/R2

or

Rt = R1R2/(R1 + R2)

= (4 ohms)(12 ohms)/(4 ohms + 12 ohms)

= 48 ohms^2/16 ohms

= 3 ohms

If the output work of a simple machine equals the input work, the machine is said to have 100% efficiency.

[tex]\circ \: \: { \underline{ \boxed{ \sf{ \color{green}{Happy\:learning.}}}}}∘[/tex]

Explanation:

x-component:

Vx = Vcos(theta)

= (92.5 m)cos(32.0)

= 78.4 m

Answer:

-78.4

Explanation:

For acellus students