Which simple machine is shown in the diagram?
a wedge
a screw
an inclined plane
a wheel and axle

Answer:

to the right

Explanation:

if the car turns to the lift, the body forces energy to the left side, so according to the first law of Newton, the body will move to the right side to resist the sudden motion.

Answer:

the points are closer to each other

Explanation:

The expression for the diffraction of a grating is

         d sin θ = m λ

         sin θ = m λ / d            (1)

where d is the distance between slits and m is the order of diffraction, the most general is to work in the order m = 1, the angle te is the angle of diffraction

When we immerse the apparatus in a medium with refractive index n = 1.33, the light emitted by the laser must comply

         v = λ f

where v is the speed of light in the medium, the frequency remains constant

velocity and refractive index are related

          n = c / v

          v = c / n

we substitute

          c / n = λf

          λ = [tex]\frac{c}{f} \ \frac{1}{n}[/tex]

          λ = λ₀ / m

where λ₀ is the wavelength in vacuum

we substitute is equation 1

         d sin θ = m λ₀ / n

         sin θ =  λ₀/ n d

         sin θ = [tex]\frac{1}{n}[/tex]  sin θ₀

we can see that the value of the sine is redueced since the refractive index is greater than 1,

consequently the points are closer to each other

Answer:

The Michelson-Morley was designed to detect the motion of the earth through the ether.

No such relation was found and the speed of light is assumed to be the same in all reference frames.

The Michelson-Morley experiment was designed to measure: A. the velocity of the Earth relative to the ether.

Michelson-Morley experiment is an experiment which was first performed in Germany by the American physicist named, Albert Abraham Michelson between 1880 to 1881.

However, the experiment was later modified and refined by Michelson and Edward W. in 1887.

The main purpose of the Michelson-Morley experiment was to measure the velocity of planet Earth relative to the luminiferous ether, which is a medium in space that is hypothetically said to carry light waves.

In conclusion, the Michelson-Morley experiment was designed to measure the velocity of the Earth relative to the hypothetical luminiferous ether.

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Answer:

Explanation:

a)     Fx = F cos (θ)

           = (200) cos(60)

           = 100 N

b)     FR = ma

       Fx + Ff = ma

      100 + Ff = (50)(1,5)

       Ff     = 75 - 100

               =  -25 N

c)    Fy = F sin θ

           = (200) sin(60)

           = 173,2 N

Answer:

x = 68.3 m

Explanation:

tan 32.8 = x / 106

Answer:

don't know what class are you you are using which mobile or laptop

Answer:

The SI unit of coefficient of linear expansion can be expressed as °C-1 or °K-1. ... The dimension of coefficient of linear expansion will be M0L0T0K−1.

Answer:

The speed at B is 16.18 ft/s .

Explanation:

Speed at A, u = 11 ft/s

Speed at C, v' = 18 ft/s

Time from A to C = 5 s

Time from B to C = 1.3 s

Let the speed of car at B is v.

Let the acceleration is a.

From A to B

Use first equation of motion

v = u + a t

18 = 11 + a x 5

a = 1.4 ft/s^2

Let the time from A to B is t' .

t' = 5 - 1.3 = 3.7 s

Use first equation of motion from A to B

v = 11 + 1.4 x 3.7 = 16.18 ft/s  

Answer:

Period is 86811.5 seconds.

Explanation:

[tex]{ \boxed{ \bf{T {}^{2} = (\frac{4 {\pi}^{2} }{GM}) {r}^{3} }}}[/tex]

[tex]{ \tt{T {}^{2} = \frac{4 {(3.14)}^{2} }{(6.6 \times {10}^{ - 11} ) \times (5.98 \times {10}^{24} )} \times {((35880\times {10}^{3}) } + (6370 \times {10}^{3} )) {}^{3} }} \\ \\ { \tt{T {}^{2} = 7.54 \times {10}^{9} }} \\ { \tt{T = \sqrt{7.54 \times {10}^{9} } }} \\ { \tt{T = 86811.5 \: seconds}}[/tex]

Answer:

a)  [tex]F_g=1.5*10^9Ibf[/tex]

b)  [tex]F_t=12490Ibf/ft^2[/tex]

     [tex]F_b=0[/tex]

Explanation:

From the question we are told that:

Height [tex]h=200ft[/tex]

Width [tex]w=1200ft[/tex]

a)

Generally the equation for Dam's Hydro static force is mathematically given by

[tex]F_g=\rho*g*\frac{h}{2}(w*h)[/tex]

Where

[tex]\rho=Density\ of\ water[/tex]

[tex]\rho=62.4Ibm/ft^3[/tex]

Therefore

[tex]F_g=62.4*32.2*\frac{200}{2}(1200*200)[/tex]

[tex]F_g=1.5*10^9Ibf[/tex]

b)

Generally the equation for Dam's Force per unit area is mathematically given by

[tex]F=\rho*g*h[/tex]

For Top

[tex]F_t=\rho*g*h[/tex]

[tex]F_t=62.4*32.2*200[/tex]

[tex]F_t=12490Ibf/ft^2[/tex]

For bottom

[tex]Here \\H=0 zero[/tex]

Therefore

[tex]F_b=0[/tex]

The hydrostatic force on the dam is [tex]2.995 \times 10^9 \ lbF[/tex].

The force per unit area near the top is 86.74 psi.

The force per unit area near the bottom is zero.

The hydrostatic force on the dam is the force exerted on the dam by the column of the water.

[tex]F = PA\\\\F = (\rho gh) \times (wh)\\\\F = (62.4 \times 32.17 \times 200) \times (1200 \times 200)\\\\F = 9.636 \times 10^{10} \ lb-ft/s^2\\\\1 \ lbF = 32.17\ lb-ft/s^2\\\\F = 2.995 \times 10^9 \ lbF[/tex]

The force per unit area is the pressure exerted near the top of the dam.

[tex]P = \rho gh\\\\P = 0.052 \times \rho h[/tex]

where;

P is pressure in PSI

ρ is density of water in lb/gal

h is the vertical height in ft

[tex]P = 0.052 \times 8.34 \times 200\\\\P = 86.74 \ Psi[/tex]

The pressure near the bottom is zero, become the vertical height is zero.

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Answer:

1.25

Explanation:

(C)

Explanation:

The capacitance C of a parallel plate capacitor is given by

[tex]C = \epsilon_0 \dfrac{A}{d}[/tex]

Let C' be the new capacitance where the area and the plate separation distance are doubled. This gives us

[tex]C' = \epsilon_0\dfrac{A'}{d'} = \epsilon_0\left(\dfrac{2A}{2d}\right) = \epsilon_0 \dfrac{A}{d} = C[/tex]

B will have the greater force

Fc=MV2 /R=Fm

The A particle has less centipetal force and larger radius so larger curve

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D = D_A = D_B[/tex]

where;

length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]

[tex]\mathbf{Q_1 = 2Q}[/tex]

The flow rate in pipe B is 2Q of the flow rate of the pipe A

The flow rate is defined as the flow of the fluid across the cross section in per unit time.

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D=D_A=D_B[/tex]

where;

length of the first pipe A  [tex]L_A=L[/tex] and the length of the second pipe B  

[tex]L_B=2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]

[tex]Q_1=2Q[/tex]

Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A

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Answer:

80

Explanation:

the mechanical advantage is the ratio of the load to the effort so it doesn't have units.to calculate it you use the formula

mechanical advantage=load/effort

in this case the load is 6000N and the effort is 75N

Ma=6000/75

=80

I hope this helps

Answer:

100 Joules

Explanation:

Applying,

W = mgh................... Equation 1

Where W = workdone to hold the box above the ground, mg = weight of the box, h = height of the box.

From the question,

Given: mg = 10 newtons, h = 10 meters.

Substitute these values into equation 1

W = 10×10

W = 100 Joules.

Hence the amount of workdone is 100 Joules

Answer:

the study of the past

Explanation:

dogs and cats

Answer:

The answer is 312.5j

Explanation:

The kinetic energy (KE):

KE=1/2*m*v^2

M= mass of the object

v= velocity of the object

We have;

m=25g

v=5m/s

KE=1/2*25g*5^2m/s

KE =312.5j

Answer:

5.4uC

Explanation:

Answer:

down below

Explanation:

Image 1- wheels of train showing both translatory motion as well as rotatory motion.

Image 2- rotation of ball shows both rotatory motion as well as translatory motion.

Image 3- the earth rotates about its axis, same time it revolves around the sun thus showing both rotatory motion and curvilinear motion in a fixed time. (perodic motion)

Image 4- while cutting wood, the

carpenter's saw has both

translatory motion and oscillatory

motion, as it moves down while

oscillating.

Answer:

[tex]R=0.023m[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=1.16*10^{-26}[/tex]

Potential difference [tex]V=523V[/tex]

Magnitude [tex]m=0.370 T[/tex]

Generally the equation for Velocity is mathematically given by

[tex]\frac{1}{2}mv^2=ev[/tex]

[tex]v=\frac{2ev}{m}[/tex]

[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]

[tex]v=12.22*10^4m/s[/tex]

Generally the equation for Force is mathematically given by

[tex]F=qvBsin \theta[/tex]

Where

[tex]qVB=m\frac{v^2}{R}[/tex]

[tex]F=m\frac{v^2}{R}sin\theta[/tex]

Therefore

[tex]R=\frac{mv}{qB sin \theta}[/tex]

[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]

[tex]R=0.023m[/tex]

Answer:

a) The proton's speed is 5.75x10⁵ m/s.

b) The kinetic energy of the proton is 1723 eV.  

Explanation:

a) The proton's speed can be calculated with the Lorentz force equation:

[tex] F = qv \times B = qvBsin(\theta) [/tex]     (1)          

Where:

F: is the force = 9.14x10⁻¹⁷ N

q: is the charge of the particle (proton) = 1.602x10⁻¹⁹ C

v: is the proton's speed =?

B: is the magnetic field = 3.28 mT

θ: is the angle between the proton's speed and the magnetic field = 17.6°

By solving equation (1) for v we have:

[tex]v = \frac{F}{qBsin(\theta)} = \frac{9.14 \cdot 10^{-17} N}{1.602\cdot 10^{-19} C*3.28 \cdot 10^{-3} T*sin(17.6)} = 5.75 \cdot 10^{5} m/s[/tex]

Hence, the proton's speed is 5.75x10⁵ m/s.

b) Its kinetic energy (K) is given by:

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

Where:

m: is the mass of the proton = 1.67x10⁻²⁷ kg

[tex] K = \frac{1}{2}mv^{2} = \frac{1}{2}1.67 \cdot 10^{-27} kg*(5.75 \cdot 10^{5} m/s)^{2} = 2.76 \cdot 10^{-16} J*\frac{1 eV}{1.602 \cdot 10^{-19} J} = 1723 eV [/tex]  

Therefore, the kinetic energy of the proton is 1723 eV.

I hope it helps you!        

Answer:

D.

Explanation:

The crest of a wave refers to the highest point of a wave. This is illustrated by D.

Answer:

to prevent friction on the surfaces

Answer:

Explanation:

Oiling reduces friction between parts with relative motion between them.

Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.

Oil also becomes oxidized which reduces the oil's ability to decrease friction.

Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.

Hydrogen bomb is more powerful than atom bomb

Answer:

a)  [tex]v=206.896m/s[/tex]

b)  [tex]L=6.749m[/tex]

Explanation:

From the question we are told that:

Frequency [tex]F=46Hz[/tex]

Length [tex]l=579m[/tex]

Total Mass [tex]T=4.3kg[/tex]

Tension [tex]T=3423[/tex]

a)

Generally the equation for velocity is mathematically given by

[tex]v=\sqrt{\frac{T}{\rho}}[/tex]

Where

[tex]\pho=m*l\\\\\pho=46*579\\\\\pho=0.0799kg/m[/tex]

Therefore

[tex]v=\sqrt{\frac{3423}{0.0799}}[/tex]

[tex]v=206.896m/s[/tex]

b)

Generally the equation for length of string is mathematically given by

[tex]L=\frac{3\lambda}{2}[/tex]

Where

[tex]\lambda=\frac{v}{f}[/tex]

[tex]\lambda=\frac{206.89}{46}[/tex]

[tex]\lambda=4.498[/tex]

Therefore

[tex]L=\frac{3*4.498}{2}[/tex]

[tex]L=6.749m[/tex]

Answer:

Explanation:

Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

11.54 minutes

Explanation:

The decay rate equation is given by

[tex]N = N_0e^{-\frac{t}{\lambda}}[/tex]

where [tex]\lambda[/tex] is the half-life. We can rewrite this as

[tex]\dfrac{N}{N_0} = e^{-\frac{t}{\lambda}}[/tex]

Taking the natural logarithm of both sides, we get

[tex]\ln \left(\dfrac{N}{N_0}\right) = -\left(\dfrac{t}{\lambda}\right)[/tex]

Solving for [tex]\lambda[/tex],

[tex]\lambda = -\dfrac{t}{\ln \left(\frac{N}{N_0}\right)}[/tex]

[tex]\:\:\:\:= -\dfrac{(24\:\text{minutes})}{\ln \left(\frac{1000\:\text{counts/min}}{8000\:\text{counts/min}}\right)}[/tex]

[tex]\:\:\:\:=11.54\:\text{minutes}[/tex]

Answer:

A. Equals to that of the smaller sphere

B. 3 times less than that of the smaller sphere

Explanation:

(a) Equals to that of the smaller sphere

The potential of an isolated metal sphere, with charge Q and radius R, is kQ=R, so a sphere with charge 3Q and radius 3R has the same potential

b) 3 times less than that of the smaller sphere

However, the electric field at the surface of the smaller sphere is ?=? 0 = kQ=R2 , so tripling Q and R reduces the surface field by a factor of 1/3