How would an observer on train A, which is moving at nearly the speed of light, view a clock on train B, which is moving at the same speed and in the same direction?

A. The clock on train B would appear narrower and run more slowly.
B. The clock on train B would appear to be the same width and to run at the same rate. C. The clock on train B would appear narrower and run faster.
D. The clock on train B would appear wider and run more slowly.

Answer:

eight

Explanation:

it's atomic number is 8 which mean that an oxygen atom has eight protons in it's nucleus

Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

Answer:

a) the most probable speed of the molecules is 409.2 m/s

b) required temperature of xenon is 1322 K

Explanation:

Given the data in the question;

a)

Maximum probable speed of hydrogen molecule (H₂)

[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )

where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K

molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol

we substitute

[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = 409.2 m/s

Therefore, the most probable speed of the molecules is 409.2 m/s

b)

Temperature of xenon  = ?

Temperature of hydrogen = 20.3 K

we know that;

T = (Vxe² × Mxe) / 2R

molar mass of xenon; Mxe = 131.292 g/mol

so we substitute

T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314  )

T = 21984.14167 / 16.628

T = 1322 K

Therefore, required temperature of xenon is 1322 K

v1f = -0.16 ms

Explanation:

Use the conservation law of linear momentum:

m1v1i + m2v2i = m1v1f + m2v2f

where

v1i = v2i = 0

m1 = 160 kg

m2 = 0.50 kg

v2f = 50m/s

v1f = ?

So we have

0 = (160 kg)v1f + (0.5 kg)(50 m/s)

v1f = -(25 kg-m/s)/(160 kg)

= -0.16 m/s

Note: the negative sign means that its direction is opposite that of the arrow.

If a person wants to calculate the length of time it takes for a ball to fall from a height of 20m, the correct equation that they should use is:

D. Δt= √2Δd/a

The free fall formula should be used to obtain the length of time that it takes for a ball to fall from a given height. This formula also factors the height or distance from which the fall occurred and this is denoted by the letter d. The small letter 'a' is denotative of acceleration due to gravity and this is a constant pegged at -9.98 m/s².

So, the change in height is obtained and multiplied by two. This is further divided by the acceleration and the square root of the derived answer translates to the time taken for the ball to fall from the height of 20m. Of all the options listed, option D represents the correct equation.

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Answer:

Explanation:

Given that:

Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm

For objective lens of focal length f₁ = 20 mm

s₁' = 120 mm + 20 mm = 140 mm

∴

Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{140}{20}[/tex]

[tex]m_1 = 7 \ m[/tex]

For objective lens of focal length f₁ = 4 mm

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{124}{4}[/tex]

[tex]m_1 = 31 \ m[/tex]

For objective lens of focal length f₁ = 1.4 mm

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]

[tex]m_1 = \dfrac{121.4}{1.4}[/tex]

[tex]m_1 = 86.71 \ m[/tex]

The magnification of the eyepiece is given as:

[tex]m_e = 5X \ and \ m_e = 15X[/tex]

Thus, the largest angular magnification when  [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]

[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]

= 86.71 × 15

= 1300.65

The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]

[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]

= 7 × 5

= 35

The largest magnification will be 1300.65 and the smallest magnification will be 35.

Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.

Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.

It is given in the question:

Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm

now we will calculate the magnification for all three focal lengths of the objective lens.

Also, each objective forms an image 120 mm beyond its second focal point.

(1) For an objective lens of focal length   [tex]f_1=20 \ mm[/tex]

[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]

Magnification will be calculated as

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]

(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]

s₁' = 120 mm + 4 mm = 124 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]

(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]

s₁' = 120 mm + 1.4 mm = 121.4 mm

[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]

Now the magnification of the eyepiece is given as:

[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]

Thus, the largest angular magnification when  

[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]

[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]

[tex]m_{large}=86.71\times 15=1300.65[/tex]

The smallest angular magnification derived when

[tex]m_1=7\ \ \ \ m_e=5[/tex]

[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]

[tex]m_{small}=7\times 5=35[/tex]

Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.

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Answer:

a scale is used to measure the mass of the body

Answer:

Maria Corazon Sumulong Cojuangco Aquino, popularly known as Cory Aquino, was a Filipino politician who served as the 11th President of the Philippines, the first woman to hold that office.

Answer:

Former President of the Philippines

Explanation:

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Answer:

a) S =  63.2 km/h

b) V =  63.2 km/h*(-0.316 , 0.949)

Explanation:

Let's define:

North as the positive y-axis

East as the positive x-axis.

Also, remember the relation:

Distance = Time*Speed

Let's assume that she starts at the position (0km, 0km)

Then she travels due North at 90km/h for two hours, then the displacement is

90km/h*2h = 180km to the north

Then the new position is:

(0km, 180km)

Then she travels West at 60km/h for one hour.

Then the distance traveled to the West (negative x-axis) is:

60km/h*1h = 60km to the west

Then the new position is:

(-60km, 180km).

a) The average speed is defined as the quotient between the displacement and the time.

We know that the total time traveled is 3 hours.

And the displacement is the difference between the final position and the initial position.

this is:

D = √( -60km - 0km)^2 + (180km - 0km)^2)=

D = √( (60km)^2 + (180km)^2) = 189.7 km

Then the average speed is:

S = (189.7 km)/(3 h) = 63.2 km/h

b) Now we want to find the average velocity, this will be equal to the average speed times a versor that points from the origin to the direction of the final position.

So, if the final position is (-60km, 180km)

We need to find a vector that represents the same angle, but that is on the unit circle.

Then, if the module of the final position is 189.7 km (as we found above), then the versor is just given by:

(-60km/ 189.7 km, 180km/ 189.7 km)

(-60/189.7 , 180/189.7)

We can just check that the module of the above versor is 1.

[tex]module = \sqrt{(\frac{-60}{189.7} )^2 + (\frac{180}{189.7} )^2} = \frac{1}{189.7}* \sqrt{(-60 )^2 + (180 )^2} = 1[/tex]

Then the average velocity is:

V = 63.2 km/h*(-60/189.7 , 180/189.7)

We can simplify our versor so the velocity equation is easier to read:

V = 63.2 km/h*(-0.316 , 0.949)

Answer:

[tex]\triangle P=1.95*10^{-4}[/tex]

Explanation:

Mass [tex]m=0.001[/tex]

Diameter [tex]d=1.2m[/tex]

Length [tex]l=10m[/tex]

Generally the equation for Volume flow rate is mathematically given by

 [tex]Q=AV[/tex]

 [tex]V=\frac{Q}{\pi/4D^2}[/tex]

 [tex]V=\frac{0.001}{\pi/4(1.2)^2}[/tex]

 [tex]V=8.84*10^{-4}[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]F=\frac{64}{Re}[/tex]

Where Re

Re=Reynolds Number

 [tex]Re=\frac{pVD}{\mu}[/tex]

 [tex]Re=\frac{1000*8.84*10^{-4}*1.2}{1.002*10^{-3}}[/tex]

 [tex]Re=1040[/tex]

Therefore

 [tex]F=\frac{64}{Re}[/tex]

 [tex]F=\frac{64}{1040}[/tex]

 [tex]F=0.06[/tex]

Generally the equation for Friction factor is mathematically given by

 [tex]Head loss=\frac{fLv^2}{2dg}[/tex]

 [tex]H=\frac{0.06*10*(8.9*10^-4)^2}{2*1.2*9.81}[/tex]

 [tex]H=19.9*10^{-9}[/tex]

Where

[tex]H=\frac{\triangle P}{\rho g}[/tex]

[tex]\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}[/tex]

[tex]\triangle P=H*\rho g[/tex]

[tex]\triangle P=1.95*10^{-4}[/tex]

Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q

Answer:

To correct the defects of vision by measuring the radius of curvature and thus the power of the lenses.

Explanation:

A spherometer is an instrument used to measure the curvature of objects such as lenses and curved mirrors.

Generally it consists of a fine screw which is moving in a nut carried on the center of a 3 small legged table or frame. The feet forms the vertices of an equilateral triangle. The lower end of the screw and those of the table legs are finely tapered and terminate in hemispheres.

If the screw has two turns of the thread to the milli meter the head is generally divided into 50 equal parts, so that differences of 0.01 millimeter may be measured without using a vernier scale.

The spherometer is used to measure the radius of curvature of the lenses so that the opthalmologist find the focal length of the lens and then give the  power to the lens to correct the defects of vision.  

(a) When the spring is compressed 4.5 cm = 0.045 m, it exerts a restoring force on the block of magnitude

F = (1900 N/m) (0.045 m) = 85.5 N

so that at the moment the block is released, this force accelerates the block with magnitude a such that

85.5 N = (1.15 kg) a   ==>   a = (85.5 N) / (1.15 kg) ≈ 74.3 m/s²

The block reaches its maximum speed at the spring's equilbrium point, and this speed v is such that

v ² = 2 (74.3 m/s²) (0.045 m)   ==>   v = √(2 (74.3 m/s²) (0.045 m)) ≈ 2.59 m/s

(b) There is no friction between the block and plane, so the block maintains this speed as it slides over the edge. At that point, it's essentially in free fall, so if y is the height of the plane, then

(7 m/s)² - (2.59 m/s)² = 2gy   ==>   y = ((7 m/s)² - (2.59 m/s)²) / (2g) ≈ 2.16 m

Binary code is the answer

Answer:

binary code is the answer of blank

Answer: 31.89seconds

Explanation:

Based on the information given, we are meant to calculate deceleration which will be:

t = V/a

where, a = mg

Therefore, t = V/mg

t = 25/0.08 × 9.8

t = 25/0.784

t = 31.89seconds

Therefore, the time that it will take before the hockey puck slides to a stop is 31.89seconds.

Answer:

9 volts

Explanation:

[tex]v = i \times r \\ v = 2 .5 \times 3.6 \\ v = 9[/tex]

[tex] \huge \mathfrak{Answer.... }[/tex]

The Correct Answer is :

B. if there is even a slight imbalance in third law there will be a net force causing acceleration.

A slight difference in the forces can result in acceleration of an object.

[tex] \mathrm{✌TeeNForeveR✌}[/tex]

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

Answer:

2.8 MW

Explanation:

There are 7 wind turbines in the wind farm as shown in the diagram. Thus, the energy output by one turbine is 1/7 if the total energy output. So, 19.6/7=2.8MW

Answer:

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

Answer:

time of fall is 1.75 s and the velocity with which it strikes the water is 17.15 m/s.

Explanation:

Height, h =  15 m

Newton's second law

Force = mass x acceleration

The unit of gravitational force is Newton and the value is m x g.

where, m is the mas and g is the acceleration due to gravity.  

Let the time of fall is t.

Use second equation of motion

[tex]s= u t +0.5 at^2\\\\15 = 0 +0.5\times 9.8\times t^{2}\\\\t = 1.75 s[/tex]

Let the final speed is v.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s\\\\v^2 = 0 + 2 \times 9.8\times 15\\\\v =17.15 m/s[/tex]

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

Answer:

The velocity after 1.5 s is 22.7 m/s downwards.

Explanation:

Initial velocity = - 8 m/s

acceleration, a = - 9.8 m/s2

time, t = 1.5 s

Use first equation of motion

v = u + at

v = - 8 - 9.8 x 1.5

v = - 8 - 14.7

v = - 22.7 m/s  

Thus, the velocity after 1.5 s is 22.7 m/s downwards.

Answer:

The tub turns 37.520 revolutions during the 25-second interval.

Explanation:

The total number of revolutions done by the tub of the washer ([tex]\Delta n[/tex]), in revolutions, is the sum of the number of revolutions done in the acceleration ([tex]\Delta n_{1}[/tex]), in revolutions, and deceleration stages ([tex]\Delta n_{2}[/tex]), in revolutions:

[tex]\Delta n = \Delta n_{1} + \Delta n_{2}[/tex] (1)

Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2})[/tex] (1b)

Where:

[tex]\ddot n_{1}, \ddot n_{2}[/tex] - Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

[tex]t_{1}, t_{2}[/tex] - Acceleration and deceleration times, in seconds.

And each acceleration is determined by the following formulas:

Acceleration

[tex]\ddot n_{1} = \frac{\dot n}{t_{1}}[/tex] (2)

Deceleration

[tex]\ddot n_{2} = -\frac{\dot n}{t_{2} }[/tex] (3)

Where [tex]\dot n[/tex] is the maximum angular velocity of the tub of the washer, in revolutions per second.

If we know that [tex]\dot n = 3\,\frac{rev}{s}[/tex], [tex]t_{1} = 13\,s[/tex] and [tex]t_{2} = 12\,s[/tex], then the quantity of revolutions done by the tub is:

[tex]\ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}[/tex]

[tex]\ddot n_{1} = 0.231\,\frac{rev}{s^{2}}[/tex]

[tex]\ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}[/tex]

[tex]\ddot n_{2} = -0.25\,\frac{rev}{s^{2}}[/tex]

[tex]\Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})[/tex]

[tex]\Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right][/tex]

[tex]\Delta n = 37.520\,rev[/tex]

The tub turns 37.520 revolutions during the 25-second interval.

I don't know

because I don't know