A bat emits a sonar sound wave
(343 m/s) that bounces off a
mosquito 8.42 m away. How
much time elapses between
when the bat emits the sound
and when it hears the echo?
(Unit = s)

Please help I am very confused with this topic

Answer:

Could I get brainest answer

I think it is B and e

Answer:

The answer is D hope it helps

Explanation:

Hi there!

[tex]\large\boxed{5N}[/tex]

From the given diagram, 20 N is going left, and 15 N is going right.

To find the net force, we must subtract the two since the two forces go in opposite directions:

20 N - 15 N = 5N.

Answer:

19 protons, 20 neutrons and 18 electrons.

Explanation:

The atomic number gives the number of protons 19

p

=

19

The atomic mass is the sum of the protons and neutrons

p

+

n

=

39

p

=

19

put p into the equation and solve for n the neutrons.

19

+

n

=

39

Subtract 19 from both sides

19

−

19

+

n

=

39

−

19

n

=

20

The number of electrons equals the number of protons in a neutral atom. The positive charge equals the negative charge. The negative charge is the number of electrons. This ion has a charge of +1. So solve for the negative charge.

−

19

+

1

=

−

18

The negative charge is -18 so

e

=

18

Answer:

her acceleration is 1 m/sec

Explanation:

The following information is given in the question

The initial velocity is 5 m/s

After 10 seconds, she would be moved at 15 m/s

We need to find the acceleration

As we know that

Acceleration = Change in speed ÷ time

Acceleration = (15 - 5) ÷ (10)

= 1 m/sec

Hence, her acceleration is 1 m/sec

The same would be considered  

The most common type of hydroelectric power plant is an impoundment facility. An impoundment facility, typically a large hydropower system, uses a dam to store river water in a reservoir. Water released from the reservoir flows through a turbine, spinning it, which in turn activates a generator to produce electricity.

"A turbine is a rotary mechanical device that extracts energy from a fluid flow and converts it into useful work." The work produced by a turbine can be used for generating electrical power when combined with a generator.

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its wave length

its wave lenght because how its measure

Answer:

Answer D. Inertial mass constant, weight decreases.

Explanation:

Khan Academy

[tex]\huge \fbox \pink {A}\huge \fbox \green {n}\huge \fbox \blue {s}\huge \fbox \red {w}\huge \fbox \purple {e}\huge \fbox \orange {r}[/tex]

A black hole is a tremendous amount of matter crammed into a very small — in fact, zero — amount of space. The result is a powerful gravitational pull, from which not even light can escape — and, therefore, we have no information or insight as to what life is like inside. A black hole is not empty, It's actually a lot of matter condensed into a single point. This point is known as a singularity.

Answer:

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Explanation:

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Answer:

The mass of the fish is 6.493 kilograms.

Explanation:

The spring-mass system experiments a Simple Harmonic Motion, then the angular frequency ([tex]\omega[/tex]), in radians per second, is determined by the following equation:

[tex]\omega = 2\pi\cdot f[/tex] (1)

Where [tex]f[/tex] is the frequency, in hertz.

In addition, the mass of the fish ([tex]m[/tex]), in kilograms, is:

[tex]m = \frac{k}{\omega^{2}}[/tex] (2)

Where [tex]k[/tex] is the spring constant, in newtons per meter.

And the spring constant is determined by Hooke's Law:

[tex]k = \frac{F}{\Delta x}[/tex] (3)

Where:

[tex]F[/tex] - Elastic force, in newtons.

[tex]\Delta x[/tex] - Spring elongation, in meters.

If we know that [tex]F = 225\,N[/tex], [tex]\Delta x = 0.125\,m[/tex] and [tex]f = 2.65\,hz[/tex], then the mass of the fish is:

[tex]\omega = 2\pi\cdot f[/tex]

[tex]\omega \approx 16.650\,\frac{rad}{s}[/tex]

[tex]k = \frac{F}{\Delta x}[/tex]

[tex]k = 1800\,N[/tex]

[tex]m = \frac{k}{\omega^{2}}[/tex]

[tex]m = 6.493\,kg[/tex]

The mass of the fish is 6.493 kilograms.

This question involves the concepts of Hooke's Law, and the frequency of spring-mass system.

The mass of the fish is "6.5 kg".

First, we will calculate the angular speed of the system:

[tex]\omega=2\pi f[/tex]

where,

ω = angular speed = ?

f = frequency = 2.65 Hz

Therefore,

[tex]\omega = 2\pi(2.65\ Hz)\\\omega = 16.65\ rad/s\\[/tex]

Now, we will use the Hooke's Law to find out the spring constant of the spring:

[tex]K =\frac{F}{\Delta x}[/tex]

where,

K = spring constant = ?

F = force applied = maximum weight = 225 N

Δx = extension = 12.5 cm = 0.125 m

Therefore,

[tex]k=\frac{225\ N}{0.125\ m}\\\\k = 1800\ N/m\\[/tex]

Now, we will use the formula for the angular speed of a spring-mass system to find out the mass of the fish:

[tex]\omega = \sqrt{\frac{k}{m}}\\\\m = \frac{k}{\omega^2}\\\\m=\frac{1800\ N/m}{(16.65\ rad/s)^2}[/tex]

m = 6.5 kg

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The attached picture illustrates Hooke's Law.

Answer:

I am calculating the total area of a solar panel for a particular load demand by ... designing according to energy demand for a day then how will it affect total solar area? ... Total Power Output=Total Area x Solar Irradiance x Conversion Efficiency ... would need is a 1 m2 solar panel to produce 1000 Watts of electrical energy.

Explanation:

Answer:

The balloon prohibits the flow of air through the air capacitor.

Explanation:

Just like an electric capacitor has an insulator between the plates, the air capacitor has a balloon between the chambers.

The balloon analogy is frequently used in electrical capacitors to assist visualise the notion of the capacitor's behaviour.

The balloon illustrates the capacitor's physical structure, specifically the two conducting plates and the dielectric material between them.

The capacitor's stored energy is equivalent to the air pressure within the balloon. When the voltage is withdrawn, the stored charge is released, and the capacitor returns to its uncharged condition, exactly as the balloon deflates when the air exits.

The balloon analogy aids in understanding how a capacitor stores electrical energy in its electric field and how that energy may be released when linked to a circuit.

Thus, it depicts the relationship between a capacitor's voltage, charge, and capacitance.

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Explanation:

Using the ideal gas equation, which I presume you are since you didn't specify using any other EOS, we have PV=nRT. Solving for what changes, i.e. pressure(P) and temperature(T), we have P/T=nR/V. Now, we can set up a relationship between the two pressures and temperatures and solve for what's necessary.

So, we have:

P1/T1=P2/T2

Solving for P2, we have:

P2=(P1*T2)/T1

NOTE: We MUST convert our temperatures to kelvin, otherwise you will end up with a NEGATIVE AND INCORRECT pressure!

Plugging in our values of P1=3.00x10^5 N/m^2, T1 of 308.15K, and T2 of 235.15K. Now we are free to evaluate:

P2=[(3.00x10*5 N/m^2)(235.15K)]/[308.15K]

P2=228930.7156 N/m^2

Or, to the appropriate amount of significant figures: 2.29x10^5 N/m^2

Which makes sense intuitively, as things tend to deflate slightly when the temperature drops!

Hope this helps!

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Answer:

Explanation:

The light waves in the reflected sunlight are horizontally polarized, which illustrates that they oscillate at a [tex]90^o[/tex] angle related to a vertical line.

Depending on the condition of the height of the light, the glare can be almost entirely horizontally polarized. Furthermore, all reflections from over-water surfaces are partially polarized. The water becomes more translucent when using polarized sunglasses.

If polarized sunglasses are to be efficient against solar glare, the transmission axis should be positioned at an angle of [tex]\theta = 45^{o}[/tex]

Answer:

Explained below.

Explanation:

Electromagnetism is defined as the study of the electromagnetic force that takes place between two or more electrically charged particles.

Now, in electromagnetic interaction, the charged particles tend to interact in an electromagnetic manner through the process of exchanging photons. Therefore, it was concluded from various experiments that the magnetic fields lines of forces were always closed, thereby eliminating the possibility of any magnetic monopole in such a manner that within a particular magnetic field, there will be two poles and in the solar system, the electromagnetic attraction between earth magnetic field and solar wind will be by way of a field.

Answer:

v = √{2(GmE(1/RE - 1/(RE + h)]}

Explanation:

From the law of conservation of energy, the total kinetic + potential energy at h = total kinetic  + potential energy at the surface of the earth(h = 0)

So, K + U = K' + U'

where K = kinetic energy of hammer at h = 0

U = gravitational potential energy of hammer at h = -GmEm/(RE + h)

K' = kinetic energy of hammer at the earth's surface (h = 0) = 1/2mv²

U = gravitational potential energy of hammer at earth's surface (h = 0) = -GmEm/RE

So, K + U = K' + U'

0 + [-GmEm/(RE + h)] = 1/2mv² +  [-GmEm/RE]

0 - GmEm/(RE + h) = 1/2mv² - GmEm/RE

collecting like terms,we have

1/2mv² = GmEm/RE - GmEm/(RE + h)

Factorizing GmE and multiplying both sides by 2, we have

v² = 2(GmE(1/RE - 1/(RE + h)]

taking square-root of both sides, we have

v = √{2(GmE(1/RE - 1/(RE + h)]}

The expression for the speed of the hammer will be:

[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]

The law of conservation says that the total energy at the height h and the total energy at the surface of the earth will remain constant.

From the law of conservation of energy

[tex]KE+PE=KE'+PE'[/tex]

where

KE = kinetic energy of hammer at h = 0

PE = gravitational potential energy of hammer at h height

[tex]PE=\dfrac{-Gm_em}{R_e+h}[/tex]

KE' = kinetic energy of hammer at the earth's surface (h = 0)

[tex]KE'=\dfrac{1}{2} mv^2[/tex]  

PE' = gravitational potential energy of hammer at earth's surface (h = 0)

[tex]PE'=\dfrac{Gm_em}{R_e}[/tex]

Now by putting the values

[tex]KE+PE=KE'+PE'[/tex]

[tex]0+ \dfrac{-Gm_em}{R_e+h}[/tex]   [tex]=\dfrac{1}{2} mv^2[/tex] [tex]-\dfrac{Gm_em}{R_e}[/tex]

[tex]\dfrac{1}{2} mv^2=\dfrac{Gm_em}{R_e} -\dfrac{Gm_em}{R_e+h}[/tex]

[tex]v^2=2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )[/tex]

[tex]v=\sqrt{2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )}[/tex]

Thus the expression for the speed of the hammer will be:

[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]

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Explanation:

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Explanation:

1200 is your answer for this question

Answer:

4m/s

Explanation:

Speed=distance/time

So the speed will be 16m/4s which gives 4m/s

Answer:

B

Explanation:

the answer as said is picture 2

Answer:

C.  10

Explanation:

The half-life of argon-44 is 12 minutes, 10 atoms of argon-44 will be left. The correct option is C.

A radioactive substance's half-life is the amount of time it takes for half of the atoms in a sample to decay.

The half-life of argon-44 in this situation is 12 minutes. Starting with 20 argon-44 atoms, half of them will have disintegrated within 12 minutes, leaving 10 atoms.

This happens because radioactive decay is an exponential process with a constant rate of decay. Every half-life cuts the number of atoms in half.

So, if we waited another 12 minutes, half of the remaining 10 atoms would decay, leaving us with 5 atoms. This process is repeated, with half of the remaining atoms dying with each half-life.

Thus, the correct option is C.

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Answer:

(a) 6.283 Wb (b) 69.11 Wb (c) I = 0.628 A

Explanation:

Given that,

The diameter of the loop, d = 40 cm

Radius, r = 20 cm

Initial magnetic field, B = 5 mT

Final magnetic field, B' = 55 mT

Initial magnetic flux,

[tex]\phi_i=BA\\\\=5\times 10^{-3}\times \pi \times 20^2\\\\=6.283\ Wb[/tex]

Final magnetic flux,

[tex]\phi_f=B'A\\\\=55\times 10^{-3}\times \pi \times 20^2\\\\=69.11\ Wb[/tex]

Due to change in magnetic field an emf will be generated in the loop. It is given by :

[tex]\epsilon=-\dfrac{d\phi}{dt}\\\\=\phi_f-\phi_i\\\\=69.11-6.283\\\\=62.827\ V[/tex]

Let I be the current in the loop. We can find it using Ohm's law such that,

[tex]\epsilon=IR\\\\I=\dfrac{\epsilon}{R}\\\\I=\dfrac{62.827}{100}\\\\=0.628\ A[/tex]

Hence, this is the required solution.

Answer:

a)   λ = 4.862 10⁻⁷ m,  b)  λ = 4.341 10⁻⁷ m

Explanation:

The spectrum of hydrogen can be described by the expression

         [tex]\frac{1}{\lambda} = R_H ( \frac{1}{n_o^2} - \frac{1}{n^2} ) \ \ \ \ n>n_o[/tex]

in the case of the initial state n = 2 this series is the Balmer series

a) Find the wavelength for n = 4

let's calculate

          [tex]\frac{1}{ \lambda}[/tex] = 1,097 10⁷ ([tex]\frac{1}{2^2} - \frac{1}{4^2}[/tex])

          \frac{1}{ \lambda} = 1.097 10⁷ 0.1875 = 0.2056 10⁷

           λ = 4.862 10⁻⁷ m

b) n = 5

     \frac{1}{ \lambda} = 1,097 10⁷ ([tex]\frac{1}{2^2} - \frac{1}{5^2}[/tex])

     \frac{1}{ \lambda} = 1.097 10⁷ 0.21 = 0.23037 10⁷

      λ = 4.341 10⁻⁷ m

Answer:

Here we only need to look at the vertical problem, so first, let's look at the forces acting vertically on the wallet.

When the wallet starts to fall, the only force acting on it will be the gravitational force (where we are ignoring the effects of air friction).

Then the acceleration of the wallet will be equal to the gravitational acceleration, g = 9.8m/s^2

Then we can write:

a(t) = (-9.8m/s^2)

Where the negative sign is because this acceleration is downwards.

To get the vertical velocity equation of the wallet we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

Where v0 is the constant of integration, and in this case is the initial velocity of the wallet, which we know is equal to 6m/s, then the velocity equation is:

v(t) = (-9.8 m/s^2)*t + 6m/s

To get the position equation we need to integrate over time again, we get:

p(t) = (1/2)*(-9.8 m/s^2)*t^2 + (6m/s)*t + p0

Where p0 is the initial vertical position, in this case, is the height at which the wallet is dropped, which is also the altitude of your jet pack when the wallet falls.

Now we want to know two things:

Determine the speed of your wallet when it hits the ground at t = 8s

Here we just need to evaluate the velocity equation in t = 8s.

v(8s) =  (-9.8 m/s^2)*8s + 6m/s = -72.4 m/s

We also want to determine the altitude of the jet pack (when the wallet drops).

To find this, we can use the fact that the wall hits the ground at t = 8s.

The wallet hits the ground when it's vertical position is equal to zero, then:

p(8s) = 0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0

Now we can solve this for p0.

0m = (1/2)*(-9.8 m/s^2)*(8s)^2 + (6m/s)*8s + p0

(1/2)*(9.8 m/s^2)*(8s)^2 -  (6m/s)*8s = p0

265.6m = p0

This means that the altitude of the jet pack when the wallet drops is 265.6m

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

         p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

         p_f = (m + M) v

the moment is preserved

          p₀ = p_f

          m v₀ = (m + M) v

          v = [tex]\frac{m}{m+M} \ v_o[/tex]

for when n balls have collided

          v = [tex]\frac{m}{n \ m + M}[/tex]  v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

        p₀ = m v₀

final moment

        p_f = m v_{1f} + M v_{2f}

        p₀ = p_f

        m v₀ = m v_{1f} + M v_{2f}

       m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

        K₀ = K_f

        ½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

        v₁ = v_{1f}            v₂ = v_{2f}

        m (v₀² - v₁²) = M v₂²

let's use the identity

         (a² - b²) = (a + b) (a-b)

we write our equations

         m (v₀ - v₁) = M v₂                       (1)

         m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

         v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

          m (v₀ - v₁) = M (v₀ + v₁)

          v₀ (m -M) = (m + M) v₁

          v₁ = [tex]\frac{m-M}{m + M}[/tex]   v₀

we substitute in equation 1 to find v₂

            [tex]\frac{M}{m}[/tex]  v₂ = v₀ -  [tex]\frac{m-M}{m+M}[/tex]   v₀

            v₂ = [tex]\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o[/tex]

            v₂ = [tex]\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o[/tex]

            v₂ = [tex]\frac{2m}{m +M} \ v_o[/tex]  

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.