Which conclusion can be made based on the information in the table? Wave speed and wavelengths can vary inversely to produce the same frequency. Frequency and wave speed can vary directly to produce the same wavelength. Wavelengths and frequency can vary inversely to produce the same wave speed. Frequency and wavelengths can vary directly to produce the same wave speed.

Answer:  YES

Explanation: The main difference between flood and tsunami  is that the flood is overflow of water that submerges land and tsunami is a series of water waves caused by the displacement by large volume of a volume of water

Answer:

WEEEEEEEEEEEEEEEEEEEEEEEEE

Explanation:

Answer:

The correct answer is "164.06 nm".

Explanation:

Refractive index,

n = 1.75

Wavelength,

λ = 525 nm

Now,

The maximum thickness of the film will be:

⇒ [tex]t=\frac{\lambda}{2n}[/tex]

On substituting the values, we get

⇒    [tex]=\frac{525}{2\times 1.75}[/tex]

⇒    [tex]=\frac{525}{3.5}[/tex]

⇒    [tex]=164.06 \ nm[/tex]

Answer:

The total resistance is 106 Ω and the current in the circuit is 0.11 A.

Explanation:

Given that,

Voltage of the battery, V = 12 V

Resistors 34 Ω, 42Ω, and 30Ω are connected in series.

The total resistance is given by :}

R = 34 + 42 +30

= 106Ω

Let I is the total current in the circuit. Using ohm's law to find it such that,

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{12}{106}\\\\I=0.11\ A[/tex]

Hence, the total resistance is 106 Ω and the current in the circuit is 0.11 A.

Answer:

Gy = 3.14x10⁻⁴ Gy

Explanation:

To get the dose in Gy we need to use the following expression:

Gy = E / m  (1)

Where:

Gy: dose

E: energy absorbed per atom

m: mass of the human body.

We don't have the energy per atom, but we can calculate that by following the next procedure.

First, let's determine the number of atoms of potassium in our body. For that we need to determine the moles in the 140 g of potassium, with the molecular mass and then, use the avogadro's number:

moles = m/MM

moles = 140 / 39.1 = 3.58 moles

N° atoms = 3.58 * 6.02x10²³ atoms = 2.16x10²⁴ atoms of K.

The abundance of the ⁴⁰K is 0.012% so the atoms of this isotope would be:

N = 2.16x10²⁴ * (0.012/100) = 2.59x10²⁰ atoms of ⁴⁰K.

With this number, and the half life rate, we can determine the number of decay atoms in a year (λ) using the following expression:

λ = ln2 / t(1/2)

λ = ln2 / 1.3x10⁹ = 5.33x10⁻¹⁰ year⁻¹

This number, multiplied by the number of atoms:

R = 5.33x10⁻¹⁰ * 2.59x10²⁰ = 1.38x10¹¹ atoms/year

Now, each atom of K gives an average energy of 1 MeV, so with the atoms we have:

E = 1.38x10¹¹ * 1x10⁶ eV = 1.38x10¹⁷ eV

This value can be expressed in Joules so:

E = 1.38x10¹⁷ eV * (1 J / 6.24x10¹⁸ eV) = 0.022 J

Finally, we can use (1) to get the dose in Gy:

Gy = 0.022 / 70

Hope this helps

Answer:

B

I think it is false.

It does not move to lower energy state.

Answer:

Explanation:

Geostationary satellites orbit around the Earth at the same rate as the Earth rotates so that the satellites are over the same spot on Earth all the time.

They are called geostationary due to their movement.

Geostationary satellites are a key tool for scientists to monitor and observe the Earth's atmosphere.  

Answer:

the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Explanation:

Given the data in the question and as illustrated in the diagram below;

Now, Lets assume line AB represent the movement of the car,

AC is the movement of the ball been thrown back and forth in the back seat

Ax is the motion of the ball it flies off the window

so from the diagram, We can see triangle ABC

where AB is 45 mph and AC = 7 mph

and angle ∠CAB = 90°

using SOH CAH TOA

TOA; tanθ = Opposite / Adjacent

tanθ = Opposite / Adjacent

tan( ∠ ABC ) = AC /  AB

we substitute

tan( ∠ ABC ) = 7 /  45

tan( ∠ ABC ) = 0.15555

( ∠ ABC ) = tan⁻¹ 0.15555

( ∠ ABC ) = 8.84°

Therefor, angle ( ∠ ABC )  is 8.84°

Meaning angle ( ∠ XAA' ) is also 8.84°

Therefore, the ball will fly in AX direction, making angle of 8.84° from the motion of the car

Answer:

the magnitude of the average contact force exerted on the leg is  3466.98 N

Explanation:

Given the data in the question;

Initial velocity of hand v₀ = 5.25 m/s

final velocity of hand v = 0 m/s

time interval t = 2.65 ms = 0.00265 s

mass of hand m = 1.75 kg

We calculate force on the hand F[tex]_{hand[/tex]

using equation for impulse in momentum

F[tex]_{hand[/tex] × t = m( v - v₀ )

we substitute

F[tex]_{hand[/tex] × 0.00265  = 1.75( 0 - 5.25 )

F[tex]_{hand[/tex] × 0.00265  = 1.75( - 5.25 )

F[tex]_{hand[/tex] × 0.00265  = -9.1875

F[tex]_{hand[/tex] = -9.1875 / 0.00265  

F[tex]_{hand[/tex] = -3466.98 N

Next we determine force on the leg F[tex]_{leg[/tex]

Using Newton's third law of motion

for every action, there is an equal opposite reaction;

so, F[tex]_{leg[/tex] = - F[tex]_{hand[/tex]

we substitute

F[tex]_{leg[/tex] = - ( -3466.98 N )

F[tex]_{leg[/tex] = 3466.98 N

Therefore, the magnitude of the average contact force exerted on the leg is  3466.98 N

Answer:

B. Frequency

Explanation:

Answer:

Magnitude of the average force = 7.85 N

Explanation:

Data given:

Mass of bullets, m = 6.4 g

Rate of bullets/min, r = 92 bullets/min

Speed of each bullet, v = 400 m/s

Change in momentum here, Δ B = Bf - Bi

where f is the final and i is the initial

Note that change in momentum = force X time

So, Δ B = m(vf - vi)

              = 2mv

             = 2 X 0.0064 kg X 400 m/s        (convert g to kg)

            = 5.12 kg.m/s (for one bullet)

        so for the 92 bullets = 92 X 5.12 kg.m/s

           = 471.04 kg.m/s

         The force = Δ B ÷ Δt

where t = time measured in 60 seconds

        =  471.04 kg.m/s ÷ 60 seconds

       = 7.85 N

Answer:

DragonFly or a dog

Explanation:

Answer:

19

Explanation:

I legit did this and it took 19.

Ship handling is both a science and an art. Science because it requires knowledge of various forces acting on the ship. Art because it requires the skills of an experienced navigator to use these forces in his favour. We may learn the science part from the various ship handling courses.

please give brainliest

Answer:

1700 kg

Explanation:

Let’s use conservation of momentum

32.5 * 388 = 7.42 * mc

mc = 1699.46

mc = 1700 kg

Answer:

[tex]I=0.0987kg.m^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=1.80kg[/tex]

Deviation [tex]d=0.250[/tex]

Time [tex]t=0.940s[/tex]

Generally the equation for moment of inertia is mathematically given by

 [tex]I=\frac{T}{2\pi}^2(mgd)[/tex]

 [tex]I=\frac{0.94}{2.3.142}^2(1.80*9.8*0.250)[/tex]

 [tex]I=0.0987kgm^2[/tex]

Answer:

100N in the opposite direction

Explanation:

as the block is not moving, we can apply the formula f=ma where the total force is equal to the mass times accelaration, since the block is not moving, the net force is zero meaning that the other force is completely cancelling out the force of 100N, or to put it in a maths equation, 100-force=0 as the block is stationary. we can therefore assume that the force cancelling the 100N out is equal to it and therefore 100N in the opposite direction

Answer:

5525 N/C

Explanation:

Magnitude of electric field ( E ) = 3500 N/c

Direction of electric field : positive X axis

point charge ( q ) = -9.0 * 10^-9

Calculate the Magnitude of the net electric field  at (a) x = -0.20 m

Magnitude =  5525 N/C

Electric field due to q = ( 9 * 10^9 * 9 * 10^-9 ) / ( -0.2 )^2  

                                    = 81 / 0.04 = 2025 N/c

Therefore the magnitude of the net electric field

= 2025 + 3500

= 5525 N/C

Answer:

the flywheel's average angular acceleration is -2.05 rad/s²

Explanation:

Note: counterclockwise is positive

         clockwise is negative

Given;

initial angular velocity, [tex]\omega _i[/tex] = 5.03 rev/s = [tex]5.03\frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 31.61 \ rad/s[/tex]

final angular velocity, [tex]\omega_f[/tex]= -2.63 rev/s = [tex]-2.63 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = -16.53 \ rad/s[/tex]

duration of the flywheel rotation, Δt = 23.5 s

The average acceleration of the flywheel is calculated as;

[tex]a_r = \frac{\Delta \omega}{\Delta t} = \frac{\omega_f - \omega _i}{t_2-t_1} = \frac{-16.53 \ - \ 31.61}{23.5} = -2.05 \ rad/s^2[/tex]

Therefore, the flywheel's average angular acceleration is -2.05 rad/s²

Answer:

the velocity of the mass is 8.44 m/s

Explanation:

Given;

mass of the object, m = 2 kg

spring constant, k = 180 N/m

extension of the spring, x = 0.89 m

The maximum velocity of the mass is calculated as follows;

By the principle of conservation of energy;

Elastic potential energy = kinetic potential energy

¹/₂kx² = ¹/₂mv²

kx² = mv²

[tex]v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{180 \times 0.89^2}{2} }\\\\v = 8.44 \ m/s[/tex]

Therefore, the velocity of the mass is 8.44 m/s

Answer:

A)    q₂ = 75.98 cm, B)     q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

         [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

          [tex]\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}[/tex]

         [tex]\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}[/tex]  

          1 / q₁ = 0.04756

           q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

          p₂ = 39.4 -q₁

          p₂ = 39.4 -21.0227

          p₂ = 18.38 cm

let's use the constructor equation

            1 / q₂ = 1 / f - 1 / p2

             [tex]\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}[/tex]

            [tex]\frac{1}{q_2}[/tex] = 0.01316

            q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

            q₂’= q₂ + 39.4

             q₂'= 75.98 +39.4

              q₂' = 115.38 cm

C) the magnification of a lens is

              m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

          m = - 75.98 / 50

          m = -1.5 X

the negative sign indicates that the image is inverted

Answer:

9.2 V

Explanation:

The RMS value of an AC is the effective value of a varying voltage or current in DC, that is the equivalent value of the AC which produces the same effect as an DC. For example if a motor is supplied by a 9V RMS voltage, it will rotate as if the voltage applied was 9V DC.

The RMS value is given by:

RMS voltage = Peak voltage * 1/√2

Given that the maximum voltage should not exceed 13 V, this means that the peak voltage is 13 V. The maximum RMS voltage is:

RMS voltage = Peak voltage * 1/√2 = 13 * 1/√2 = 9.2 V

Answer:

P = 756.84 Watts

Explanation:

As the tension is stationary or innitial, T₀ = 31 N, the mean would be:

T₁ + T₂ / 2 = T₀    (1)

T₁ + T₂ = 2 * 31 = 62 N

Now, with the following expression we can determine the linear speed:

V = πWD  (2)

W: angular speed of the wheel  (rev/s)

D: diameter of the wheel (in meters)

W = 1547 rev/min * (1 min/60 s) = 25.78 rev/s

V = π * 25.78 * 0.379 = 30.695 m/s

We also know that:

T₁ / T₂ = exp (μθ)

T₁ = T₂ exp(μθ)   (3)

We already have those values so replacing:

T₁ = T₂ exp(0.3 * 161 * π/180)

T₁ = 2.32T₂   (4)

We can now replace (4) in (1) like this:

T₁ + T₂ = 62 N

2.32T₂ + T₂ = 62

3.32T₂ = 62

T₂ = 18.67 N

Which means that T₁:

T₁ = 2.32(18.67)

T₁ = 43.33 N

Finally, the power can be determined using the following expression:

P = (T₁ - T₂)V  (5)       Replacing we have:

P = (43.33 - 18.67)*30.695

Hope this helps

Answer:

Solution given:

Radius of small sphere[r]=5cm=0.05m

Radius of large sphere[R]=10cm=0.1m

capacitance of small sphere[c]=4πε0r

=[tex]4π*8.85×10^{-12}×0.05=5.56*10^{-12}F[/tex]

Charge for small sphere[Q1]=100C

Charge for small sphere[Q2]=50C

Potential difference [V1]=[tex] \frac{charge}{capacitance}=\frac{100}{5.56*10^{-12}}=1.8×10^{13}[/tex]V

.

again

capacitance of small sphere[C]=4πε0R

=[tex]4π*8.85×10^{-12}×0.1=1.11*10^{-11}F[/tex]

Potential difference [V2]=[tex] \frac{charge}{capacitance}=\frac{50}{1.11*10^{-11}}=4.5×10^{12}[/tex]V

Now

Loss of energy:

[tex] \frac{cC(V1-V2)^{2}}{2(c+C)}[/tex]

=[tex] \frac{5.56*10^{-12}*1.11*10^{-11}(1.8*10^{13}-4.5*10^{12})^{2}}{2(5.56*10^{-12}+1.11*10^{-11})}[/tex]

=25Joule

Answer:

the ratio Kf : Ki is  1 / 4 or 1 : 4

Explanation:

Given the data in the question;

Since this is a perfectly inelastic collision, momentum is conserved;

[tex]P_{initial[/tex] = [tex]P_{final[/tex]

Now for BLOCK 1

mass = M₁ = M

KE = K[tex]_i[/tex]

[tex]\frac{1}{2}[/tex]mv[tex]_i[/tex]₁² = K[tex]_i[/tex]

we solve for v[tex]_i[/tex]₁

mv[tex]_i[/tex]₁² = 2K[tex]_i[/tex]

v[tex]_i[/tex]₁ = √( 2K[tex]_i[/tex] / m )

for BLOCK 2

mass = M₂ = 3m and since its at rest v[tex]_i[/tex]₂ = 0

Now after the collision; Total mass = m + 3m = 4m

KE = K[tex]_f[/tex]

[tex]\frac{1}{2}[/tex]( 4m )v[tex]_f[/tex]² = K[tex]_f[/tex]

(2m)v[tex]_f[/tex]² = K[tex]_f[/tex]

v[tex]_f[/tex] = √(K[tex]_f[/tex] / 2m)

so since [tex]P_{initial[/tex] = [tex]P_{final[/tex]

[m₁ × v[tex]_i[/tex]₁] + [m₂ × v[tex]_i[/tex]₂] = ( m + 3m ) × v[tex]_f[/tex]

so

[ m₁ × √( 2K[tex]_i[/tex] / m ) ] + [ m₂ × 0 ] = ( m + 3m ) × [ √(K[tex]_f[/tex] / 2m) ]

[ m × √( 2K[tex]_i[/tex] / m ) ] = 4m × [ √(K[tex]_f[/tex] / 2m) ]

square both side

m² ×  2K[tex]_i[/tex] / m  = (4m)² × K[tex]_f[/tex] / 2m

m² ×  2K[tex]_i[/tex] / m  = 16m² × K[tex]_f[/tex] / 2m

m × 2K[tex]_i[/tex]  = 8m × K[tex]_f[/tex]

2K[tex]_i[/tex]  =  8K[tex]_f[/tex]

K[tex]_f[/tex] = 2K[tex]_i[/tex] / 8

K[tex]_f[/tex] / K[tex]_i[/tex] = 2 / 8

K[tex]_f[/tex] / K[tex]_i[/tex] = 1 / 4

Therefore,  the ratio Kf : Ki is  1 / 4 or 1 : 4

The ratio of the final kinetic energy to initial kinetic energy of the blocks is determined as 1:4.

The speed of the two blocks after collision is determined by applying the principle of conservation of linear momentum as follows;

m1u1 + m2u2 = v(m1 + m2)

mu1 + m2(0) = v(m + 3m)

mu1 = v(4m)

u1 = 4v

[tex]\frac{K.E_f}{K.E_i} = \frac{0.5(4m)v_f^2}{0.5(m)u_1^2} \\\\\frac{K.E_f}{K.E_i} = \frac{4(v_f)^2}{u_1^2} \\\\\frac{K.E_f}{K.E_i} = \frac{4v_f^2}{(4v_f)^2} = \frac{4v_f^2}{16v_f^2} = \frac{1}{4} = 1:4[/tex]

Learn more about kinetic energy here: https://brainly.com/question/25959744

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Answer:

The correct answer is -

B. The velocity would double (v = 2v).

C. The wavelength would be half (λ = λ/2).

Explanation:

A wave has a speed or velocity that is related to the wavelength of the wave and the frequency of the wave and this relationship can be represented by the following equation-

Wave velocity V = Wavelength (λ) * Frequency (f)

Frequency (f) = Velocity (V) / Wavelength(λ).

The frequency and wavelength are inversely proportional and frequency and velocity are directly proportional to each other.

So, if f = 2f then,

putting value in the formula,

2f = 2v/λ, which means, f = 2v and f = λ/2

when the frequency is doubled, the wavelength will be halved and velocity will be doubled.

Answer:

Explanation:

a= (v-u)/t where v = the final velocity at 50m/s, u= the inital velocity at 30m/s and t= the time taken at 30 seconds

a=(50-30)/30

a= 2/3m/s^-2 or 0.67m/s^-2  (m/s^-2 is the units for accelaration)

Answer:

Un'auto si muove lungo un percorso rettilineo con velocità variabile come mostrato in figura. Quando l'auto è in possesso di A, la sua velocità è 10 ms-1 e quando è in posizione B, la sua velocità è 20 ms-1. Se l'auto impiega 5 secondi per spostarsi da A a B, trova l'accelerazione dell'auto.

Explanation: