A geologist finds that a Moon rock whose mass is 8.94 kg has an apparent mass of 6.18 kg when submerged in water. What is the density of the rock

188m - 556m is the wavelength range for AM radio.

2.77 - 3.40m is the wavelength range for FM radio.

1) For AM radio

f(max) = 1600 × 10³ Hz

f(min) = 540 ×10³Hz

c = fλ

λ = c/f where,

c = speed of light

f = frequency

λ= wavelength

So,

λ(min) = 3 × [tex]10^{8}[/tex] / 1600 × 10³ = 188 m

λ(max) = 3 × 10⁸/540 ×10³ = 556 m

So wavelength range of AM radio is 188 m - 556m

2) For FM radio

f(max) = 108 × 10⁶ Hz

f(min) = 88 × 10⁶ Hz

λ(min)  = 3 × 10⁸ / 108 × 10⁶ = 2.77 m

λ(max) = 3 × 10⁸ / 88 × 10⁶ = 3.40 m

So wavelength range of FM radio is 2.77m - 3.40m

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The work done by the force is 140 J.

Calculation:

It is given that, an 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of [tex]30^\circ[/tex] with the horizontal. The force pushing the crate is parallel to the slope and the acceleration is 1.5 [tex]\text{ m/s}^2[/tex].

It is required to find the work done by the force.

The free-body diagram is shown below,

Here, W=80 N, a=-1.5 [tex]\text{ m/s}^2[/tex]

We can write the net force equation as,

[tex]\begin{gathered}F-W \sin 30^{\circ}=m a \\F-m g \sin 30^{\circ}=\left(\frac{W}{g}\right) a\end{gathered}[/tex]

Here W is the weight of the object, m is the mass of the body, and F is the applied force.

Thus,

[tex]\begin{aligned}F-(80 \mathrm{~N}) \sin 30^{\circ} &=\left(\frac{80 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\right)\left(-1.5 \mathrm{~m} / \mathrm{s}^{2}\right) \\F &=27.755 \mathrm{~N}\end{aligned}[/tex]

It is known that the work done by the force is calculated as the multiplication of the applied force and the displacement.

Thus the work one is,

[tex]\begin{aligned}W &=(27.755 \mathrm{~N})(5 \mathrm{~m}) \\& \approx 140 \mathrm{~J}\end{aligned}[/tex]

Thus the last option is the correct one.

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It takes four times the work and four times the stopping distance.

Given:

original speed of car = x

brakes applied so final speed = 0

moving speed of another car = 2x

To Find:

work done to stop the car

Solution: Work is done whenever a force moves something over a distance. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

Work done is given by

W = KE(final) - KE(initial)

W = 1/2m(v)^2 - 1/2m(u)^2

W = 1/2m(0 - (x)^2)

W = 1/2m(4x^2)

W = 4(1/2mx^2)

W = 4W

So, work done will be 4 times initial work

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Answer: d) income only includes salary

Explanation:

i just took a test for economics and personal finance and got it correct

The statement that "Income only includes salary " is not true.

Income is the money that an individual or organization receives in return for their services or goods. 

Depending on the context—such as taxation, financial accounting, or economic analysis—income may have a variety of definitions.

For the majority of people, their total earnings include their wages and salaries, investment returns, pension payments, and other receipts. 

For businesses, income refers to the money made from selling goods and services as well as any interest or dividends paid on the company's cash holdings and reserves.

So, the amount of money, property, and other value transfers received over a predetermined period in return for goods or services is often referred to as "income."

That's why income does not include only salary.

Hence Statement (D) about the income is not true.

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The solution is (v, u) = (- 585.709, 593,034). Please notice that the value of v only means that the direction of the real vector is antiparallel to the supposed one.

According to the definition of vector sum and vectors in rectangular form, we must solve the following vector equation:

(0, 0) = 205 · (cos 23°, - sin 23°) + v · (- cos 75°, sin 75°) + u · (- cos 55°, - sin 55°)

(0, 0) = (188.703, 80.100) + v · (- 0.259, 0.966) + u · (- 0.574, 0.819)

(- 188.703, - 80.100) =  v · (- 0.259, 0.966) + u · (- 0.574, 0.819)

Then, we must solve the following system of linear equations:

- 0.259 · v - 0.574 · u = - 188.703

0.966 · v + 0.819 · u = - 80.100

The solution is (v, u) = (- 585.709, 593,034). Please notice that the value of v only means that the direction of the real vector is antiparallel to the supposed one.

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2. 0 mc charge in an external field of 20n/c, north (direction) will experience a force of 0.4 newtons, the direction of the force is north

Force in an electric field  = charge * electric field

given

charge =  2* [tex]10^{-3}[/tex] C

electric field  = 20 N/C

Force  in an electric field   = 2* [tex]10^{-3}[/tex] * 20

                                            = 0.04 N

since , external field is in north direction then the force must be in north direction because the direction of an electric field at a point is the same as the direction of the electric force acting on a positive test charge

hence,  2. 0 mc charge in an external field of 20n/c, north (direction) will experience a force of 0.4 newtons, the direction of the force is north

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A radar receiver indicates that a pulse return as an echo in 20 μs after it was sent. The reflecting object would be 3000 m away .

Phenomenon of hearing back our own sound is called an echo. It is due to successive reflection of sound waves from the surfaces or obstacles of large size. To hear an echo, there must be a time gap of 0.1 second in original sound and the reflected sound.

Given

time =  20 μs = 20 * [tex]10^{-6}[/tex] s

let distance to the reflecting surface be = x

total distance travelled by pulse will be  = 2x

speed = 3.0 × [tex]10^{8}[/tex] m/s

distance = speed * time

2x = 3.0 × [tex]10^{8}[/tex] * 20 * [tex]10^{-6}[/tex]

   x = 3000 m

The reflecting object would be 3000 m away

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Time for which box will remain in contact with spring is 1.1 sec

Given:

mass of box = 8 kg

initial speed of box = 1.5 m/s

spring constant = 69 N/m

To Find:

time for which box will remain in contact with spring

Solution: The time taken to complete one oscillation is called the time period. where m is the mass, and k is the spring constant.

Time period is given as

t = π √m/k

t = 3.14 √8/69

t = 1.1 sec

So time for which box will remain in contact with spring is 1.1 sec

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When the north-pole end of a bar magnet is held near a stationary positively charged piece of plastic, it remains unaffected.

What is a bar magnet?

A bar magnet is made of ferromagnetic substance that produces

magnetic field.

What are the properties of a bar magnet?

Since plastic is not a ferromagnetic material or any other material like gold and silver ,they remain unaffected in the presence of magnetic field.

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Answer:

a whole new glacier will start because the other one is melting

Answer: the radial distance between the 500-v equipotential surface and the 1000 v surface will be 8.91*106 times the charge Q.

Explanation: To find the answer, we have to know more about the equipotential surfaces.

                            [tex]V=\frac{Qk}{r}[/tex]  , where k is equal to 1/(4π∈₀) =[tex]8.99*10^9[/tex] .

                          [tex]r_d=r_1-r_2[/tex]

                   [tex]500V=\frac{Qk}{r_1} \\thus, r_1=\frac{Qk}{500V}= (1.79*10^7 *Q) m[/tex]

                  [tex]1000V=\frac{Qk}{r_2}\\ r_2= \frac{Qk}{1000V} =(8.99*10^6*Q) m[/tex]

                  [tex]r_d=1.79*10^7Q-8.99*10^6Q=(8.91*10^6*Q)m[/tex]

Thus, we can conclude that, the radial distance between the equipotential surface of 500V and 1000V will be,8.91*106 times the charge Q.

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Answer:

5.1 The DNA double helix is the most recognizable nucleic acid structure, but these are ribozymes.

Two 1-d waves of amplitude a and period t and constant phase 0 are superposed. If wave 1 travels 13. 5 wavelengths farther than wave 2,  the interference is 3/2π.

Interference of light is the phenomena of multiple light waves interfering with each other under certain circumstances, causing the mixed amplitudes of the waves to both boom or lower.one of the exceptional examples of interference is demonstrated by the light contemplated from a film of oil floating on water.

The phenomenon of addition or superposition of mild waves which produces an increase in intensity at some factors and a decrease in depth at a few other points are known as interference of light.

The intensities of light at exceptional factors of the medium are distinctive. At a few factors of the medium, depth is most and at a few different points intensity is maximum ( about 0) This phenomenon is called the interference of mild waves, the interference is known as negative interference.

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Relation can you state about the acceleration of the charge and it mass is they are inversely proportional .

According to the question

Newton's second law :

Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

F = m * a

a = [tex]\frac{F}{m}[/tex]  

Relation can you state about the acceleration of the charge and it mass is

According to Newton's second law of motion, acceleration is inversely proportional to the mass when the force is a constant.

The acceleration is directly proportional to the force when the mass remains constant.

Hence, Relation can you state about the acceleration of the charge and it mass is they are inversely proportional .

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Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero.  Thus, neither object is accelerating.

Recall Newton's second law: [tex]\sum {\vec {F}}=m \vec {a}[/tex]

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

Since the only forces acting on the objects are gravity and the force of air friction, in order to zero out, the force of air friction must be equal in magnitude and opposite in direction to the force of gravity.

Recall that near the surface of the earth, [tex]F_{gravity}=mg[/tex], so the Force of Gravity acting on an object is directly proportional to the object's mass.  (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.  

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

The wavelength of the ultrasound wave in 20 °c air is 1.715 * 10⁻⁵ m.

The frequency of sound is the number of complete oscillations per second.

The wavelength is the distance between successive positions in the wave.

Velocity of sound is the distance covered per second.

The frequency and wavelength of a wave are related by the formula given below:

The velocity of sound in 20°C air = 343 m/s

frequency of the sound wave = 2.00 × 10⁷ hz or 2.00 × 10⁷ s⁻¹

wavelength = velocity/frequency

wavelength = 343 /  2.00 × 10⁷

wavelength = 1.715 * 10⁻⁵ m

In conclusion, the wavelength of the ultrasound wave is determined from the ratio of the velocity and the frequency of the sound.

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If the separation between the openings in a laser is increased, then the distance between the interference fringes decreases

Interference fringe refers to bands caused by different lights which can be found in phase or not each other.

In conclusion, if the separation between the openings in a laser is increased, then the distance between the interference fringe decreases

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Answer:

As the plane moves toward the listener, the apparent frequency of the plane would be [tex]4250\; {\rm Hz}[/tex] ([tex]2250\; {\rm Hz}[/tex] higher than the frequency at the source.)

As the plane moves away from the listener, the apparent frequency of the plane would be approximately [tex]1308\; {\rm Hz}[/tex] (approximately [tex]692\; {\rm Hz}[/tex] lower than the frequency at the source.)

Assumption: the speed of sound in the air is [tex]340\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Crests of this sound wave travel toward the listener at a constant [tex]v = 340\; {\rm m\cdot s^{-1}}[/tex]. Since there is a pause of [tex]t = 1 / f = (1/2000)\; {\rm s}[/tex] between every two consecutive crests of this sound wave, the distance between each pair of consecutive crests would be:

[tex]\begin{aligned}\lambda &= \frac{v}{f} \\ &= \frac{340\; {\rm m \cdot s^{-1}}}{2000\; {\rm s^{-1}}} \\ &= 0.17\; {\rm m} \end{aligned}[/tex].

Hence, if the aircraft wasn't moving, the first crest would have a head start of [tex]\lambda = 0.17\; {\rm m}[/tex] relative to the second one. This head start would ensure that the first crest arrive [tex]t = \lambda / v = 0.17\; {\rm m} / (340\; {\rm m \cdot s^{-1}}) = (1/2000)\; {\rm s}[/tex] earlier than the second crest.  

However, at a speed of [tex]v_{\text{s}} = 180\; {\rm m\cdot s^{-1}}[/tex], the aircraft would have travelled an additional [tex]v_\text{s}\, t = 180\; {\rm m\cdot s^{-1}} \times (1/2000)\; {\rm s} = 0.09\; {\rm m}[/tex] within that [tex]t = (1 / 2000)\; {\rm s}[/tex].

In other words, if the aircraft was moving towards the listener, the period of the sound would appear to the listener to be approximately [tex]0.000235\; {\rm s}[/tex]. in contrast, if the aircraft was moving away from the listener, the period of the sound would appear to the listener as approximately [tex]0.000765\; {\rm s}[/tex].

Therefore:

The field strength needed to produce a 24.0 V peak emf is 0.73T.

To find the answer, we need to know about the expression of emf.

What's the expression of peak emf produced in a rotating rectangular loops?

B= 24/(300×0.00261×42) = 0.73T

Thus, we can conclude that the magnetic field is 0.73T.

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The path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

Based on the law of inertia, which is the reluctance of an object to stop moving once in motion or start moving when it is at rest.

The bowling ball will maintain the path of the airline in the first few seconds of fall, after which it will change its path to vertical direction.

Thus, the path the bowling ball would most closely follow after leaving the airplane is horizontal direction.

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The ray will not emerge into the air medium from glass medium.

To find the answer, we need to know about the critical angle.

Thus, we can conclude that the light will not emerge into air medium.

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The magnitude is 5.69 × 10^4  N/C and the direction is  opposite to the acceleration of electron.

Equation of force on a charge is -

                F = ma

→             qE = ma

The acceleration of an electron is 1.201016 m/s².

→         E =  ma/q

  plug in the values

          E = 9.11 × 10^-31 kg × 1×10^16 m/s² i / -1.6 × 10^-19 C

          E = 5.69 × 10^4  N/C   (-i)

so, magnitude is 5.69 × 10^4  N/C  and field is along negative x-axis

Thus, it has opposite direction to the acceleration of electron.

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   After impact velocity = 14.968 ft/s

Weight and mass of Bullet and wooden block:

Bullet: w = 1oz = 1/16 lb m = 0.001941 lb

wooden block : W = 5lb M = 0.15528 lb

velocity of block and bullet immediately after impact:

Σmv1 + ΣImp = mv2

Resolving vertical component

( m× v₀cos30⁰) + 0 = ( m+M) v'

v' = ( m× v₀cos30⁰)/ (m+M)

v' = 14.968 ft/s

Horizontal and vertical component of the impulse exerted by block on the bullet:

   Here we will apply the principle of impulse and momentum.

 Horizontal component:

          -mv₀ cos30⁰ + RxΔt  =0

                                 RxΔt = mv₀sin30⁰

                                          = 0.001941 × 1400sin30⁰

                                   RxΔt = 1.3587 lb.s

         Vertical component:

                           -mv₀cos30⁰  + RyΔt =  -mv'

                                     RyΔt = m( v₀cos30⁰-v')

                                     RyΔt = 0.001941(1400cos30⁰ - 14.968)

                                              = 2.32 lb.s

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Power require in Watts is 50 watts

Given:

Force = 30 N

Distance = 100m

Time for which force applied = 1 min = 60 sec

To Find:

Power

Solution: A watt is a unit of power. It is the amount of energy an item needs to function; the rate at which energy is consumed. One watt is equivalent to electricity flowing at a rate of one joule (unit of energy) per second

P = W/t

where W is work done by force and t is the time for which force is applied

W = F x d

W = 30 x 100 = 3000

P = 3000/60 = 50 watts

So Power require in Watts is 50 watts

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Answer:

[tex]\huge\boxed{\sf v = 448\ m/s}[/tex]

Explanation:

Frequency = f = 16 Hz

Wavelength = λ = 28 m

Speed = v = ?

v = fλ

Put the givens.

v = (16)(28)

v = 448 m/s

[tex]\rule[225]{225}{2}[/tex]

The buoyant force acting on the block is 58.8 N.

The dimensions of the block are given as 0.10 m, 0.20 m and 0.30 m.

The volume of the block will be = 0.10 × 0.20 × 0.30 m³

                                                     = 0.006 m³

We know that,

W = mg                                 .....(1)

where,

W = weight of the block

m = mass of the block

g = acceleration due to gravity = 9.8 m/s²

We also know that mass is proportional to density and volume

m = ρ.V                                  .....(2)

Substituting the equation (2) in (1)

W = ρ.V.g                               .....(3)

Density of water, ρ = 1000 kg/m³

Force is the net weight acting on the block

F = W = ρ.V.g

F = 1000 × 0.006 × 9.8

F = 58.8 N

Therefore, the buoyant force acting on the block is 58.8 N.

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a metaphor for generosity and joy and joy and joy and joy organised a large wedding guests had to choose their meal from beef chicken tikka and a kind young women in the streets of St

The number of excess electrons needed to be distributed uniformly in the sphere to produce the given electric field is 1.69*10^(10).

What is an electric field?

The electric force exerted on a unit charge is called the electric field.

For a sphere of uniform charge density, the magnitude of the electric field on its surface is given by,

E=k*q/r^2

where E is the magnitude of the electric field, k is the coulomb's constant which has a value of 9*10^9 N m^(2) C^(-2), and q is the total charge of the sphere, and r is the radius of the sphere.

The radius of the sphere is half of the diameter. Given the diameter of the sphere is 26.0 m, the radius of the sphere r=26.0/2 cm or r=13.0 cm.

It is given that E=1440 N/C, r=13.0 cm, and the value of k is 9*10^9 N m^(2) C^(-2). Substitute these values in the formula of the electric field and solve it to get the value of the total charge on the sphere.

Note: 1 cm = 0.01 m.

1440 (N/C)= (9*10^9 N m^(2) C^(-2))*q/( 13 cm)^2

1440 (N/C)= (9*10^9 N m^(2) C^(-2))*q/( 13*0.01 m)^2

1440 ( N/C)= (532.54*10^(9) N C^(-2))*q

q=1440/(532.54*10^(9)) C

q=2.704*10^(-9) C

Since the charge is quantized, the total charge inside the sphere is the integral multiple of an elementary charge. In the given case, the value of that elementary charge is 1.6*10^(-19) C. The total charge is then given by,

q=ne

where q is the total charge, n is the number of elementary charges and e is the value of the elementary charge.

The value of q is 2.704*10^(-9) C and the value of e is 1.6*10^(-19) C.

Substitute these values in the formula of total charge and solve it to get the number of elementary charges.

2.704*10^(-9) C =n*1.6*10^(-19) C

n=2.704*10^(-9)/ 1.6*10^(-19)

n=1.69*10^(10)

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OPTION (C) IS CORRECT

ANSWER - UNIFORM MOTION

If there is no change in the velocity of the object then it is known to be in UNIFORM MOTION.

EXPLORE MORE:-

EXAMPLE - A CAR COVERS A DISTANCE OF 15 KM WE'D EVERY 2 HOURS

FOR AN UNIFORM MOTION, ACCELERATION IS ZERO , AS THERE IS NO CHANGE IN VELOCITY....

-THANKS.!!

Dead space is the name of the air in the conducting zone that is not available for gas exchange.

What is conducting zone?

The conducting zone consists of all of the structures that provide passageways for air to travel into and out of the lungs.

The volume of air that is inhaled that does not take part in the gas exchange is dead space. Because it either remains in the conducting airways or reaches alveoli that are not perfused or poorly perfused.

Dead space volume is air in the conducting airways that does not participate in gas exchange between the alveoli and blood.

Hence,

Dead space is the name of the air in the conducting zone that is not available for gas exchange.

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