. A ball weighs 120g on the earth surface,

i) What is its mass on the surface of the moon? 1mk

​

Answer:

a. 200 m

b. 100 m

Explanation:

Solution:-

- We will first draw three points marked A,B and Q from left most to right most.

- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).

- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:

                             AQ - BQ = n*λ/2

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ/2

- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ/2

                           λ = 200 m  .... Answer

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:

                             AQ - BQ = n*λ

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ

- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ

                           λ = 100 m  .... Answer

Answer:

Explanation:

Initial kinetic energy of the system = 1/2 mA v0²

If Vf be the final velocity of both the carts

applying conservation of momentum

final velocity

Vf = mAvo / ( mA +mB)

kinetic energy ( final ) =  1/2 (mA +mB)mA²vo² /  ( mA +mB)²

= mA²vo²  / 2( mA +mB)

Given 1/2 mA v0²  / mA²vo²  / 2( mA +mB) = 6

mA v0² x ( mA +mB) / mA²vo² = 6

( mA +mB) / mA = 6

mA + mB = 6 mA

5 mA = mB

mB / mA = 5 .

Answer:

The youngest person

Explanation:

Hearing worsens with age

Please mark brainliest

Answer:

A

Explanation:

The person closest to the origin of the sound will most likely hear the loudest sound. ^^

Answer:

Explanation:

There's a formula for this:

[tex]F = k*displacement[/tex]

F being force, k being the spring constant, and displacement being the change in x

We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters

Answer:

2500 N/m²

Explanation:

Pressure: This can be defined as the force acting normally on a surface per unit area.

The expression for pressure is give as

P = F/A...................... Equation 1

Where P = pressure (N/m²), F = force (N), A = Contact area (m²)

Given: F = 500 N, A = 2000 cm² = (2000/10000) m = 0.2 m.

Substitute into equation  1

P = 500/0.2

P = 2500 N/m²

Hence the pressure exerted to the surface is 2500 N/m²

Answer:

St Lawrence Sea way

Explanation:

The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.

Answer:

3985 s or 66.42 mins

Explanation:

Given:-

- The length of the rod, L = 83.0 cm

- The cross sectional diameter of rod , d = 2.4 cm

- The temperature of reservoir, Tr = 540°C

- The amount of ice in chamber, m = 1.43 kg

- The temperature of ice, Ti = 0°C

- Thermal conductivity of silver, k = 406 W / m.K

- The latent heat of fusion of water, Lf = 3.33 * 10^5 J / kg

Find:-

How much time is required for the ice to melt completely

Solution:-

- We will first determine the amount of heat ( Q ) required to melt 1.43 kg of ice.

- The heat required would be used as latent heat for which we require the latent heat of fusion of ice ( Lf ). We will employ the first law of thermodynamics assuming no heat is lost from the chamber ( perfectly insulated ):

                              [tex]Q = m*L_f\\\\Q = ( 1.43 ) * ( 3.33 * 10 ^5 )\\\\Q = 476190 J[/tex]

- The heat is supplied from the hot reservoir at the temperature of 540°C by conduction through the silver rod.

- We will assume that the heat transfer through the silver rod is one dimensional i.e along the length ( L ) of the rod.

- We will employ the ( heat equation ) to determine the rate of heat transfer through the rod as follows:

                             [tex]\frac{dQ}{dt} = \frac{k.A.dT}{dx}[/tex]

Where,

                           A: the cross sectional area of the rod

                           dT: The temperature difference at the two ends of the rod

                           dx: The differential element along the length of rod ( 1 - D )

                           t: Time ( s )

- The integrated form of the heat equation is expressed as:

                            [tex]Q = \frac{k*A*( T_r - T_i)}{L}*t[/tex]

- Plug in the respective parameters in the equation above and solve for time ( t ):

                           [tex]476190 = \frac{406*\pi*0.024^2 * ( 540 - 0 ) }{0.83*4}*t \\\\t = \frac{476190}{119.49619} \\\\t = 3985 s = 66.42 mins[/tex]

Answer: It would take 66.42 minutes to completely melt the ice

Answer:

[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]

Explanation:

In order to calculate the angular momentum of the particle you use the following formula:

[tex]\vec{L}=\vec{r}\ X\ \vec{p}[/tex]       (1)

r is the position vector respect to the point (0 , 5.0), that is:

r = 0m i + 5.0m j    (2)

p is the linear momentum vector and it is given by:

[tex]\vec{p}=m\vec{v}=(2.0kg)(3.0m/s)(\hat{i+\hat{j}})=6\frac{kgm}{s}(\hat{i}+\hat{j})[/tex]   (3)

the direction of p comes from the fat that the particle is moving along the i + j direction.

Then, you use the results of (2) and (3) in the equation (1) and solve for L:

[tex]\vec{L}=-30\frac{kgm^2}{s}\hat{k}[/tex]

The angular momentum is -30 kgm^2/s ^k

Answer:

decreases, decreases

Explanation:

Answer:

Explanation:

From Newton's 2nd Law,

F = m×a

Where F is Force

m is mass

a is acceleration

Hence a= F/m

a= 16/8= 2m/s2

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

Answer: 1.00c

Explanation: I got it correct on the homework

Answer:

a) a = 5.03x10¹³ m/s²

b) [tex]V_{f} = 4.4 \cdot 10^{5} m/s [/tex]

Explanation:    

a) The acceleration of the positron can be found as follows:

[tex] F = q*E [/tex]    (1)

Also,

[tex] F = ma [/tex]    (2)

By entering equation (1) into (2), we have:

[tex] a = \frac{F}{m} = \frac{qE}{m} [/tex]

Where:

F: is the electric force

m: is the particle's mass = 9.1x10⁻³¹ kg

q: is the charge of the positron = 1.6x10⁻¹⁹ C    

E: is the electric field = 286 N/C

[tex] a = \frac{qE}{m} = \frac{1.6 \cdot 10^{-19} C*286 N/C}{9.1 \cdot 10^{-31} kg} = 5.03 \cdot 10^{13} m/s^{2} [/tex]

b) The positron's speed can be calculated using the following equation:

[tex] V_{f} = V_{0} + at [/tex]

Where:

[tex]V_{f}[/tex]: is the final speed =?

[tex]V_{0}[/tex]: is the initial speed =0

t: is the time = 8.70x10⁻⁹ s

[tex] V_{f} = V_{0} + at = 0 + 5.03 \cdot 10^{13} m/s^{2}*8.70 \cdot 10^{-9} s = 4.4 \cdot 10^{5} m/s [/tex]

I hope it helps you!

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

Answer:

I =1.8 kgm^2

Explanation:

In order to calculate the moment of inertia of the door you use the following formula, which relates the torque applied to the door with its moment of inertia and angular acceleration:

[tex]\tau=I\alpha[/tex]        (1)

τ: torque applied to the door

I: moment of inertia of the door

α: angular acceleration = 5 rad/s^2

The torque is also given by τ = Fd, where F is the force applied at a distance of d to the pivot of the door (hinge axis).

F = 10 N

d = 0.9 m

You replace the expression for τ, and solve for I:

[tex]Fd=I\alpha\\\\I=\frac{Fd}{\alpha}\\\\I=\frac{(10N)(0.9m)}{5rad/s^2}=1.8kgm^2[/tex]

The moment of inertia of the door is 1.8 kgm^2

Answer:

The moment of inertia is [tex]I = 1.8 \ kg m^2[/tex]

Explanation:

From the question we are told that

   The force applied is  [tex]F = 10 \ N[/tex]

    The distance of the knob to the hinge is  [tex]d = 0.9 \ m[/tex]

     The angular acceleration is  [tex]a = 5 \ rad/s[/tex]

The moment is mathematically represented as

        [tex]I = \frac{d Fsin(\theta)}{a}[/tex]

Here [tex]\theta = 90^o[/tex] This is because the force direction is perpendicular to the plane of the door

substituting values

          [tex]I = \frac{0.9 * 10 * sin (90)}{5}[/tex]

          [tex]I = 1.8 \ kg m^2[/tex]

Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?

Answer:

Radius of the handle = 3 cm = 0.03 m

Explanation:

Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)

Diameter of the rod = 1 cm

Radius of the rod = Diameter/2

Radius of the rod = 1/2

Radius of the rod = 0.5 cm

Mechanical Accuracy of the machine, MA = 6

Substitute the values into equation (1)

6 =  (Radius of the handle)/0.5

Radius of the handle = 6 * 0.5

Radius of the handle = 3 cm

The Radius of the handle is = 3 cm = 0.03 m

Since

Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)

Here,

Diameter of the rod = 1 cm

We know that

The radius of the rod = Diameter/2

So,

Radius of the rod = 1/2

So,

Radius of the rod = 0.5 cm

Now

Mechanical Accuracy of the machine, MA = 6

Now

6 =  (Radius of the handle)/0.5

Radius of the handle = 6 * 0.5

Radius of the handle = 3 cm

Learn more about radius here: https://brainly.com/question/18648019

Answer:

Δm Δt> h ’/ 2c²

Explanation:

Heisenberg uncertainty principle, stable uncertainty of energy and time, with the expressions

     ΔE Δt> h ’/ 2

     h’= h / 2π

to relate this to the masses let's use Einstein's relationship

      E = m c²

let's replace

     Δ (mc²) Δt> h '/ 2

the speed of light is a constant that we can condense exact, so

      Δm Δt> h ’/ 2c²

Answer:

correct answer is c

v = -0.8 m / s

Explanation:

This is a problem of quantity of movement, for this we must define a system formed by the two cars, so that the forces during the collision are internal and therefore the quantity of movement is conserved

initial

    p₀ = m₁ v₁ - m₂ v₂

final

    = (m₁ + m₂) v

We have taken the direction to the right as positive

    p₀ =p_{f}

    m₁ v₁ - m₂ v₂ = (m₁ + m₂) v

    v = (m₁ v₁ - m₂ v₂) / (m₁ + m₂)

we calculate

    v = (2  4 - 8  2) / (2 + 8)

    v = (8 -16) / 10

     v = -0.8 m / s

the negative sign indicates that the set is moving to the left

correct answer is c

Answer:

E = 3.45*10^-19 N/C

Explanation:

a) The electric field between two parallel plates id given by the following formula:

[tex]E=\frac{\sigma}{\epsilon_o}[/tex]           (1)

where:

σ: surface charge density of the plates = 39.0nC/m^2

εo: dielectric permittivity of vacuum = 8.85*10^-12 C/Nm^2

You replace these values in the equation (1):

[tex]E=\frac{39.0*10^{-9}C/m^2}{8.85*10^{-12}C^2/Nm^2}\\\\E=3.45*10^{-19}\frac{N}{C}[/tex]

The electric field in between the parallel plates is 3.45*10^-19 N/C

Answer:

Explanation:

Distance between fringe or fringe width =  xλ /  d

where x is location of screen and d is slit separation

Given x = 4 m

λ = 694 nm

d = .085 x 10⁻³ m

distance between fringes

= 4 x 694 x 10⁻⁹ / .085 x 10⁻³

= 4 x 694 x 10⁻⁹ / 85 x 10⁻⁶

= 32.66 x 10⁻³ m

= 32.66 mm .

3.267 cm

b )

when submerged in water , wavelength in water becomes as follows

wavelength in water = wave length / refractive index

= 694 / 1.333 nm

= 520.63 nm

new distance between fringes

3.267 / 1.333

= 2.45 cm .

Answer:

91.5 miles

Explanation:

61 miles per hour so 61(x amount of hours)

so 61 x 1.5 hours is 91.5 miles

Answer:

λ1 = 0.0129m = 1.29cm

λ2 = 0.00923m = 0.92 cm

Explanation:

To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:

[tex]y_m=\frac{m\lambda D}{d}[/tex]    (1)

m: order of the bright fringe = 1

λ: wavelength of the light = 660 nm, 470 nm

D: distance from the screen = 5.50 m

d: distance between slits = 0.280mm = 0.280 *10^⁻3 m

ym: height of the m-th fringe

You replace the values of the variables in the equation (1) for each wavelength:

For λ = 660 nm = 660*10^-9 m

[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]

For λ = 470 nm = 470*10^-9 m

[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]

Answer:

Explanation:

The velocity at the inlet and exit of the control volume are same [tex]V_i=V_e=V[/tex]

Calculate the inlet and exit velocity of water jet

[tex]V=V_j+V_e\\\\V=30+14\\\\V=44m/s[/tex]

The conservation of mass equation of steady flow

[tex]\sum ^e_i\bar V. \bar A=0\\\\(-V_iA_i+V_eA_e)=0[/tex]

[tex]A_i\ \texttt {is the inlet area of the jet}[/tex]

[tex]A_e\ \texttt {is the exit area of the jet}[/tex]

since inlet and exit velocity of water jet are equal so the inlet and exit cross section area of the jet is equal

The expression for thickness of the jet

[tex]A_i=A_e\\\\\frac{\pi}{4} D_j^2=2\pi Rt\\\\t=\frac{D^2_j}{8R}[/tex]

R is the radius

t is the thickness of the jet

D_j is the diameter of the inlet jet

[tex]t=\frac{(100\times10^{-3})^2}{8(230\times10^{-3}} \\\\=5.434mm[/tex]

(b)

[tex]R-x=\rho(AV_r)[-(V_i)+(V_c)\cos 60^o]\\\\=\rho(V_j+V_c)A[-(V_i+V_c)+(V_i+V_c)\cos 60^o]\\\\=\rho(V_j+V_c)(\frac{\pi}{4}D_j^2 )[V_i+V_c](\cos60^o-1)][/tex]

[tex]1000kg/m^3=\rho\\\\44m/s=(V_j+V+c)\\\\100\times10^{-3}m=D_j[/tex]

[tex]R_x=[1000\times(44)\frac{\pi}{4} (10\times10^{-3})^2[(44)(\cos60^o-1)]]\\\\=-7603N[/tex]

The negative sign indicate that the direction of the force will be in opposite direction of our assumption

Therefore, the horizontal force is -7603N

Answer:

a)188.65m

b)154.35m

c)243.7m

Explanation:

Given data:

[tex]t_A=0.55s[/tex]

[tex]t_B=0.45s[/tex]

(a) The distance from the kicker to each of the 2 spectators is given by:

[tex]d_A=v \times t_A[/tex]

where,

v= speed of sound

[tex]t_A[/tex]=time taken for the sound waves to reach the ears

[tex]d_A=343\times 0.55=188.65[/tex]m

(b)[tex]d_B=v \times t_B[/tex]

where,

v= speed of sound

[tex]t_B[/tex]=time taken for the sound waves to reach the ears

[tex]d_B=343\times 0.45=154.35m[/tex]

(c)As the angle b/w slight lines  from the two spectators to the player is right angle,

hypotenuse=the distance b/w 2 spectators

and, the slight lines are the other 2 lines

[tex]D^2=d_A^2+d_B^2\\D=\sqrt{188.65^2+154.35^2} \\D= 243.7m[/tex]

Answer:

A. Final Floor = 15.5 = 15 (Considering downward portion of elevator)

B. T = 14859.5 N = 14.89 KN

Explanation:

A.

First we calculate distance covered by the elevator during downward motion. Downward motion consists of two parts. First one is uniformly accelerated. For that part we use 2nd equation of motion:

s₁ = Vi t + (0.5)at²

where,

s₁ = distance covered during accelerated downward motion = ?

Vi = initial speed = 0 m/s (since elevator is initially at rest)

t = time taken = 6 s

a = acceleration = 1.5 m/s²

Therefore,

s₁ = (0 m/s)(6 s) + (0.5)(1.5 m/s²)(6 s)²

s₁ = 4.5 m

also we find the final velocity using 1st equation of motion:

Vf = Vi + at

Vf = 0 m/s + (1.5 m/s²)(6 s)

Vf = 9 m/s

Now, the second part of downward motion is with constant velocity. So:

s₂ = vt

where,

s₂ = distance covered during constant speed downward motion = ?

v = Vf = 9 m/s

t = 6 s

Therefore,

s₂ = (9 m/s)(6 s)

s₂ = 54 m

Now for distance covered during upward motion is given by the 2nd equation of motion. Since the values of acceleration and time are same. Therefore, it will be equal in magnitude to s₁:

s₃ = s₁ = 4.5 m

Therefore, the total distance covered by elevator is given by following equation:

s = s₁ + s₂ - s₃      (Downward motion taken positive)

s = 4.5 m + 54 m - 4.5 m

s = 54 m

Therefore, net motion of the elevator was 54 m downwards.

So the final floor will be:

Final Floor = Initial Floor - Distance Covered/Length of a floor

Final Floor = 29 - 54 m/4m

Final Floor = 15.5 = 15 (Considering the downward portion or floor of elevator)

B.

The maximum tension will occur during the upward accelerated motion. It is given by the formula:

T = m(g + a)

where,

T = Max. Tension in Cable = ?

m = total mass of person and elevator = 115 kg + 1200 kg = 1315 kg

g = 9.8 m/s²

a = acceleration = 1.5 m/s²

Therefore,

T = (1315 kg)(9.8 m/s² + 1.5 m/s²)

T = 14859.5 N = 14.89 KN

Answer:2000

Explanation:

Answer:

49.725× 10^-24J

Explanation:

The Energy associated with a Photon us defined as;

E = hf

Where h is Planck's constant = 6.63× 10^-34m2kg/s

f is the frequency= 7.5 x 10^14 Hz

Hence

E = 6.63× 10^-34 × 7.5 x 10^14 =49.725× 10^-24J

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex]        ............1

put here value and we will get value

normal force = [tex]\frac{0.400}{0.04}[/tex]  

solve it we get

normal force = 10 N

Answer:

11,700Newton

Explanation:

According to Newton's second law, Force = mass × acceleration

Given mass = 65kg.

Acceleration if the car can be gotten using one of the equation of motion as shown.

v² = u²+2as

v is the final velocity = 18m/s

u is the initial velocity = 0m/s

a is the acceleration

s is the distance travelled = 0.9m

On substitution;

18² = 0²+2a(0.9)

18² = 1.8a

a = 324/1.8

a = 180m/²

Net force acting on the body = 65×180

Net force acting on the body = 11,700Newton