Question 3 of 10
The Moon has much less gravitational force than Earth. What would happen
if you went to the Moon?
OA. Your mass would decrease.
O
B. Your weight would increase.
O
C. Your mass would increase.
O
D. Your weight would decrease.
SUBMIT

The frequency of the photon emitted by a hydrogen atom is   [tex]26.2*10^-^4 hz[/tex]

A particle representing a quantum of light or other electromagnetic radiation. A photon carries energy proportional to the radiation frequency but has zero rest mass.

In the case of light, the frequency, symbolized by the Greek letter nu (ν), of any wave equals the speed of light, c, divided by the wavelength λ:

v = c/λ

Since the wavelength λ is in the bottom of the fraction, the frequency is inversely proportional to the wavelength.

v = c/λ

v= [tex]\frac{3*10^8}{434*10^-^9} = 6.91*10^1^4 Hz[/tex]

By using Rydberg's formula,

1/λ [tex]= Rz^2(\frac{1}{n_1^2} - \frac{1}{n_2^2})[/tex]

1/λ = [tex]109677 *(\frac{1}{4^2} - \frac{1}{6^2}) = 3808.22 cm^-^1[/tex]

λ  = [tex]26.2*10^-^4 hz[/tex]

[tex]26.2*10^-^4 hz[/tex] is  frequency of the photon emitted by a hydrogen atom.

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Jupiter is the largest. the closest Jovian planet to the sun is Jupiter and most farthest is  Neptune. moons of Jupiter are Europa, lo, Ganymede, etc. Saturn has an orbital period equal to 29 earth years. Uranus's rotational period is equal to 17h 14m.

In terms of size-

In terms of location-

Jovian planets are generally farther from the sun while terrestrial plants are closer to the sun. therefore Jupiter, Saturn, Uranus, and Neptune are away from the sun whole earth is closer to the sun.

In terms of moons-

Jovian planets have more moons as compared to earth. Jupiter has 53 moon, Uranus has 27 Miranda, ariel ,Oberon) while the earth has one moon.

In terms of the orbital period-

Jovian planets have a longer orbital period than earth as they are further from the sun.

In terms of the rotational period-

The rotational period of the Jovian planet is approximately 10 to 17 hours.

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Answer:

500 g

Explanation:

Specific heat of water = 1 j/g-c

Heat given up by hot water = 200(75-25)(1) = 10 000 j

this is the heat GAINED by the cold water

10000 j =  x ( 25 -5)(1)  

          x =500g

The magnitude of the vector is 34.092 N

The resultant is the square root of sum of square of the vector component of x and y direction.

R =√ [(Nx)² +(Ny)²]

Given is the vector with horizontal length of the vector = -28.7 m and the vertical length of the vector = 18.4 m

The resultant will be

R =√ [(-28.7)² +(18.4)²]

R = 34.092 N

Thus, the magnitude of the vector is  34.092 N.

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The average force exerted on the ball by the glove will be 129 N.

Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body. Force is defined as the product of mass and acceleration. Its unit is Newton.

Given data;

Mass,(m)=0.15-kilogram

Initial speed,u=43 meters per second

Time period,t= 0.050 seconds

Average force exerted,F=?

The average force exerted on the ball by the glove is;

[tex]\rm F = m\frac{v-u}{t} \\\\ F=0.15( \frac{43-0}{0.050} )\\\\ F=129 N[/tex]

Hence, the average force exerted on the ball by the glove will be 129 N.

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The magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

u is the initial velocity of canonball= 47.4 m/sec

g is the acceleration of free fall = 9.81 m/sec²

v is the velocity after 1.23 s

According to Newton's first equation of motion,

[tex]\rm v=u +gt \\\\ v= 47.4 \ m/sec + 9.81 \ m/sec^2 \times 1.23 sec \\\\\ v=59.46 \ m/sec[/tex]

Hence, the magnitude of the cannonball's velocity after 1.23 s will be 59.46 m/sec.

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The minimum stopping distance when the car is moving at 32.0 m/s is 348.3 m.

The acceleration of the car before stopping at the given distance is calculated as follows;

v² = u² + 2as

when the car stops, v = 0

0 = u² + 2as

0 = 15² + 2(76.5)a

0 = 225 + 153a

-a = 225/153

a = - 1.47 m/s²

If the same force is applied, then acceleration is constant.

v² = u² + 2as

0 = 32² + 2(-1.47)s

2.94s = 1024

s = 348.3 m

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d. is wrong...the temperature cannot be measured due to physical changes

Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.

So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

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Answer:

cost=$15

Explanation:

given;

P= 2100 W

t= 12'= 720 sec=0.2h

V= 120 V

required;

Energy?

cost?

solution;

Energy(kWh) = power*time(in hour)

E=2100*0.2= 420 kWh

cost= 420 kWh* $0.25/kWh

cost= $15

since blow dryer use 120-V

Answer:

inertia of motion

Explanation:

it's because when a passenger is jumping from a bus his/her body is in motion after falling in a road he/she remains or tends to remain in the state of motion that is the reason

Answer:(note that values will be on top in small)

cobalt :- 60Co

potassium :- 40K

neon :- 24Ne

lead :- 208Pb

The angle (θcornea)  when light passes from air to cornea is 16.86°

It states that the ratio of sine of angle of incidence and angle of refraction is equal to the refractive index of second medium to the  first medium.

sini/sinr =n₂ / n₁

Most of the bending of light in the eye is done at the air-cornea interface. The beam of light deviated as it passes from air to the cornea if the incident angle is θi = 23.6°.

Given the angle of incidence i = 23.6°, refractive index of air n₁ =1, refractive index of cornea n₂ = 1.38, then the angle of refraction at cornea is

sinr = sini x (n₁/n₂)

Plug the values, we get

sinr = sin23.6 x (1/1.38)

sinr = 16.86°

The angle of refraction is less than angle of incidence due to refraction.

Thus, the angle (θcornea)  when light passes from air to cornea is 16.86°

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An object, which weighs 98 N on the surface of Earth, weigh on this point is 2.23 x 10²⁴ kg.

The force of attraction felt by a person at the center of a planet or Earth is called as the gravity.

Given, the Earth has the acceleration due to gravity, g =  9.81 m/s².

Force of gravity W = mass x acceleration due to gravity

98N = m x 9.8m/s²

m = 10 kg

The value of acceleration due to gravity (g) on a point 10,000 km above sea level is about 1.49 m/s².

The acceleration due to gravity and mass is related as

g = GM/R²

where G = gravitational constant = 6.67 x 10⁻¹¹ N.m²/kg² and R is the distance between two masses.

Substituting the values, we get

1.49 m/s² =  6.67 x 10⁻¹¹ N.m²/kg² x M/ (10000 x 10³ m)²

M = 2.23 x 10²⁴ kg

Therefore, an object will weigh on this point is 2.23 x 10²⁴ kg

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Answer:

212N/m

Explanation:

First of all, what is spring constant
As per Hooke's law, The restoring force applied by the spring is proportional to the change in length, and restoring force is directed towards the equilibrium position. The proportionality constant is called the spring constant and it is a measure of the stiffness of the spring. So as Hooke's law

F = -Kx

Where

F -> Restoring force

K-> Spring constant

x -> change in length

the negative sign indicates it is acting toward the center

Rearranging this for the Spring constant, Spring constant formula is given by

k = Fx

Unit of Spring Constant is Newton/meter

(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (4ti + j) m/s and a = a = 4i m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 3.98 m/s².

x = at²i + btj

x = 2t²i + tj

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

v = dx/dt

v =  (4ti + j) m/s

a = dv/dt

a = 4i m/s²

v(2) = 4(2)i + j

v(2) = 8i + j

|v(2)| = 8.06 m/s

v(3) = 4(3)i + j

v(3) = 12i + j

|v(3)| = 12.04 m/s

a = (12.04 - 8.06)/(3 - 2)

a = 3.98 m/s²

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Answer:

Explanation:

Half-life problems are modeled as exponential equations.  The half-life formula is [tex]P=P_o\left (\dfrac{1}{2} \right)^{\frac{t}{k}}[/tex] where [tex]P_o[/tex] is the initial amount, [tex]k[/tex] is the length of the half-life, [tex]t[/tex] is the amount of time that has elapsed since the initial measurement was taken, and [tex]P[/tex] is the amount that remains at time [tex]t[/tex].

[tex]P=14.2\left (\dfrac{1}{2} \right)^{\frac{t}{8.0252}}[/tex]

Deriving the half-life formula

If one forgets the half-life formula, one can derive an equivalent equation by recalling the basic an exponential equation, [tex]y=a b^{t}[/tex], where [tex]t[/tex] is still the amount of time, and [tex]y[/tex] is the amount remaining at time [tex]t[/tex].  The constants a and b can be solved for as follows:

Knowing that amount initially is 14.2g, we let this be time zero:

[tex]y=a b^{t}[/tex]

[tex](14.2)=ab^{(0)}[/tex]

[tex]14.2=a *1[/tex]

[tex]14.2=a[/tex]

So, [tex]a=14.2[/tex], which represents out initial amount of the substance, and our equation becomes: [tex]y=14.2 b^{t}[/tex]

Knowing that the "half-life" is 8.0252 days (note that the unit here is "days", so times for all future uses of this equation must be in "days"), we know that the amount remaining after that time will be one-half of what we started with:

[tex]\left(\frac{1}{2} *14.2 \right)=14.2 b^{(8.0252)}[/tex]

[tex]\dfrac{7.1}{14.2}=\dfrac{14.2 b^{8.0252}}{14.2}[/tex]

[tex]0.5=b^{8.0252}[/tex]

[tex]\sqrt[8.0252]{\frac{1}{2}}=\sqrt[8.0252]{b^{8.0252}}[/tex]

[tex]\sqrt[8.0252]{\frac{1}{2}}=b[/tex]

Recalling exponent properties, one could find that  [tex]\left ( \frac{1}{2} \right )^{\frac{1}{8.0252}}=b[/tex], which will give the equation identical to the half-life formula.  However, recalling this trivia about exponent properties is not necessary to solve this problem.  One can just evaluate the radical in a calculator:

[tex]b=0.9172535661...[/tex]

Using this decimal approximation has advantages (don't have to remember the half-life formula & don't have to remember as many exponent properties), but one minor disadvantage (need to keep more decimal places to reduce rounding error).

So, our general equation derived from the basic exponential function is:

[tex]y=14.2* (0.9172535661)^t[/tex]  or [tex]y=14.2*(0.5)^{\frac{t}{8.0252}}[/tex] where y represents the amount remaining at time t.

Solving for the amount remaining

With the equation set up, substitute the amount of time it takes to cross the Pacific to solve for the amount remaining:

[tex]y=14.2* (0.9172535661)^{(31.8)}[/tex]          [tex]y=14.2*(0.5)^{\frac{(31.8)}{8.0252}}[/tex]

[tex]y=14.2* 0.0641450581[/tex]                    [tex]y=14.2*(0.5)^{3.962518068}[/tex]

[tex]y=0.9108598257[/tex]                              [tex]y=14.2* 0.0641450581[/tex]

                                                        [tex]y=0.9108598257[/tex]

Since both the initial amount of Iodine, and the amount of time were given to 3 significant figures, the amount remaining after 31.8days is 0.911g.

The work done by the electric field on the proton when it reaches the positive plate willl be 9.1 ×10⁻¹⁵ J.

A physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is referred to as an electric field,

The electric field formula is as follows:

E = F /Q

Where,

The electric field is denoted by the letter E.

F is a powerful letter.

The charge is Q.

The potential difference between the plate;

[tex]\rm V = E d \\\\ V = 3.40 \times 10^6 \times 0.13 \ m \\\\ V= 442000 \\\\ V = 4.42 \times 10^5 \ v[/tex]

Force on the charged particle ;

[tex]\rm F = qV \\\\ F= 1.6 \times 10^{-19} \times 4.42 \times 10^ 5 \\\\\ F = 7.0 \times 10^{-14} \ N[/tex]

Work done in  order to move the charge;

[tex]\rm W = F \times d \\\\ W= 7.0 \times 10^{-14} \ N \times 0.13 m \\\\ W=9.1 \times 10^{-15} \ J[/tex]

Hence, the work done by the electric field on the proton when it reaches the positive plate willl be 9.1 ×10⁻¹⁵ J.

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Answer:

35 m/s^2

Explanation:

Decceleration =   change in velocity / change in time

                        =  ( 70 m/s ) / 2 seconds = 35 m/s^2

   DEcceleration = 35 m/s^2

       Jason's speed changes by  - 35 m/s^2

The car's acceleration will be 0.575 m/s².The unit of acceleration is m/sec².

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u= 0 m/s

Final velocity, v= 4.2 m/s

Time elapsed, t = 7.3 seconds.

To find ;

Acceleration, a

The acceleration when the change in velocity is observed by the formula as:

a= (v-u)/(t)

Substitute the given values:

a= (4.2-0)/(7.3)

a=(4.2)/(7.3)

a= 0.575 m/s²

Hence, the car's acceleration will be 0.575 m/s².

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Answer:

Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.

Explanation:

Hard water is formed when water percolates through deposits of limestone, chalk or gypsum which are largely made up of calcium and magnesium carbonates, bicarbonates and sulfates.

Answer:

Hard water indicates a high amount of minerals.

Explanation:

This most often occurs when natural water sources pass through mineral deposits such as limestone or chalk. The process leads to a relatively high level of calcium, iron, and magnesium.

Answer:

E = 5.52 x 10⁻¹⁹  J

Explanation:

The formula for energy of a photon, E, is:

[tex]E = hf[/tex]

where h is Plank's constant = 6.62 x 10 ⁻³⁴ .

Using the formula:

E = 6.62 x 10⁻³⁴   x   8.34 x 10¹⁴

  = 5.52 x 10⁻¹⁹  J

Answer:

CABD

.....................

The image is cut and jumbled and placed into different places, so the correct order of the image that it can be back together is CABD.

A specific piece of something is an image. It can provide information to the optical system in two dimensions, three components, or in another way. An item that mimics a subject, such as a picture or other two-dimensional image, might be considered an image. An image in the context of signal analysis is a dispersed color amplitude.

A graphical image need not make use of the complete visual system. A common illustration of this is a grayscale image, which does not employ color, but instead relies on the visual game's sensitivity to brightness across all wavelengths. Even if it doesn't make proper use of the visual system, a black-and-white and white visual depiction of something is nonetheless an image.

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The pressure of a gas inside a rigid container at 0 °C will not be zero, but will somewhat be less than it would be at 25 °C.

According to Charle's law of gases, the pressure of a gas is directly proportional to its temperature. This is based on the condition that the volume of the gas does not change.

Thus, for a gas inside a rigid container, the pressure at 0 °C may not be zero. However, the pressure will be less than it would be if the temperature were to be 25 °C.

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The high salt concentration make sea water significantly less resistive than pure water due to presence of charged ions in the sea water.

The resistivity of a substance is the opposition to the flow of charges offered by the substance.

The greater the resistivity of a substance, the lesser its conductivity.

A high salt concentration make sea water highly conductive due to presence of charged ions in the water. The greater conductivity reduces the resistivity of the sea water.

A pure water has no charged ions in the water, thereby decreasing its conductivity and increasing its resistivity.

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Answer:

Something which occupies space and have mass.

Let's see

[tex]\\ \rm\Rrightarrow mgh=\dfrac{1}{2}mv^2[/tex]

[tex]\\ \rm\Rrightarrow 2gh=v^2[/tex]

[tex]\\ \rm\Rrightarrow 2(10)(247)=v^2[/tex]

[tex]\\ \rm\Rrightarrow v²=4940[/tex]

[tex]\\ \rm\Rrightarrow v=70.3ms^{-1}[/tex]