Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that Careful measurements reveal that a star maintains a steady apparent brightness at most times except that at precise intervals of 127 hours the star becomes dimmer for about 4 hours. The most likely explanation is that:________

a. the star is a white dwarf.
b. the star is periodically ejecting gas into space, every 127 hours.
c. the star is a Cepheid variable.
d. the star is a member of an eclipsing binary star system.

Answer:

hope you can understand

This question is incomplete, the complete question is;

When light with a wavelength of 221 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 3.28 × 10⁻¹⁹ J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answer:

the required wavelength of light is 161.9 nm

Explanation:

Given the data in the question;

Let us represent work function of the metal by W.

Now, using Einstein photoelectric effect equation;

[tex]E_{proton[/tex] = W + [tex]K_{max[/tex]

hc/λ = W + [tex]K_{max[/tex] ------- let this be equation 1

we solve for W

W = hc/λ - [tex]K_{max[/tex]

given that; λ = 221 nm = 2.21 × 10⁻⁷ m, [tex]K_{max[/tex]= 3.28 × 10⁻¹⁹ J

we know that speed of light c = 3 × 10⁸ m/s and Planck's constant h = 6.626 × 10⁻³⁴ Js.

so we substitute

W = [( (6.626 × 10⁻³⁴)(3 × 10⁸) )/2.21 × 10⁻⁷ ] -  3.28 × 10⁻¹⁹

W =  8.99457 × 10⁻¹⁹ -   3.28 × 10⁻¹⁹

W = 5.71457 × 10⁻¹⁹ J

Now, to determine λ for which maximum kinetic energy is double

so;

[tex]K'_{max[/tex] = double = 2( 3.28 × 10⁻¹⁹ J ) = 6.56 × 10⁻¹⁹ J

from from equation 1

we solve for λ'

λ' = hc / W + [tex]K'_{max[/tex]

we substitute

λ' = ( (6.626 × 10⁻³⁴)(3 × 10⁸) ) / ( (5.71457 × 10⁻¹⁹ J) + ( 6.56 × 10⁻¹⁹ J ))

λ' = 1.9878 × 10⁻²⁴ /  1.227457 × 10⁻¹⁸

λ' =  1.619 × 10⁻⁷ m

λ' =  161.9 nm

Therefore, the required wavelength of light is 161.9 nm

Answer:

the force of attraction between the two charges is 4.2 x 10⁹ N.

Explanation:

Given;

the magnitude of first charge, q₁ = 0.06 C

the magnitude of the second charge, q₂ = 0.07 C

distance between the two charges, r = 3 m

The force of attraction between the two charges is calculated as ;

[tex]F = \frac{Kq_1q_2}{r^2}[/tex]

where;

k is Coulomb's constant = 9 x 10⁹ Nm²/C²

[tex]F = \frac{Kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(0.06)(0.07)}{3^2} \\\\F = 4.2 \times 10^{6} \ N[/tex]

Therefore, the force of attraction between the two charges is 4.2 x 10⁹ N.

Answer:

R = 7.915 10⁶ m,  M = 1.04 10³⁵ kg

Explanation:

Let's start by finding the acceleration of the planet's gravity, let's use the kinematic relations

           v = v₀ - g t

the velocity of the body when it falls is the same for equal height, but it is positive when it rises and negative when it falls

          v = -v₀

         -v₀ = v₀ - g t

          g = 2v₀ / t

          g = 2 15 / 2.7

          g = 11.11 m / s²

I now write the law of universal gravitation and Newton's second law

          F = m a

          G m M / R² = m a

         a = g

          g = G M / R²

Now let's work with the cruiser in orbit

         F = ma

acceleration is centripetal

         a = v² / r

         G m M / r² = m v² / r                        (1)

the distance from the center of the planet is

        r = R + h

        r = R + R = 2R

we substitute in 1

        G M / 4R² = v² / 2R

        G M / 2R = v²

The modulus of the velocity in a circular orbit is

         v = d / T

the distance is that of the circle

          d = 2π r

          v = 2π 2R / T

          v = 4π R / T

          G M / 2R = 16pi² R² / T²

          T² = 32 pi² R³ / GM

let's write the equations

             g = G M / R² (2)

             T² = 32 pi² R³ / GM

we have two equations and two unknowns, so it can be solved

let's clear the most on the planet and equalize

             g R² / G = 32 pi² R³ / GT²

              g T² = 32 pi² R

             R = g T² / 32 pi²

let's reduce the period to SI units

           T = 250 min (60 s / 1 min) = 1.5 104 s

let's calculate

             R = 11.11 (1.5 10⁴) ² / 32 π²

           R = 7.915 10⁶ m

from equation 2 we can find the mass of the planet

             M = g R² / G

             M = 11.11 (7.915 10⁶) ² / 6.67 10⁻¹¹

             M = 1.04 10³⁵ kg

Answer:

. always start on the north pole and terminate (end) on the South Pole

Explanation:

Explanation:

I think you forgot to add the other part!

Answer:

uhm what plant...

Explanation:

Answer:

a)  P = 807.85 N,  b)  P = 392.15 N,  c)  P = 444.12 N

Explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

         sin θ = Wₓ / W

         cos θ = W_y / W

         Wₓ = W sin θ

         W_y = W cos θ

         Wₓ = 1200 sin 30 = 600 N

          W_y = 1200 cos 30 = 1039.23 N

Y axis  

      N- W_y = 0

      N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

       P -Wₓ -fr = 0

       P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

      fr = μ N

      fr = 0.20 1039.23

      fr = 207.85 N

we substitute

       P = 600+ 207.85

       P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

        P + fr -Wx = 0

       fr = μ N

       fr = 0.20 1039.23

        fr = 207.85 N

we substitute

        P =  Wₓ -fr

        P = 600 - 207,846

        P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

         P - Wₓ + fr = 0

         P = Wₓ - fr

         fr = 0.15 1039.23

         fr = 155.88 N

         P = 600 - 155.88

         P = 444.12 N

Answer:

charge = electrons + protons

=92+92

=184

Answer:

You get Day and Night

It takes 24 hour

Answer:

Explanation:

The Earth's orbit makes a circle around the sun. At the same time the Earth orbits around the sun, it also spins.Since the Earth orbits the sun and rotates on its axis at the same time we experience seasons, day and night, and changing shadows throughout the day.It only takes 23 hours, 56 minutes and 4.0916 seconds for the Earth to turn once on its axis.

Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!  

bit.[tex]^{}[/tex]ly/3a8Nt8n

Answer:

C. Slippery feel

Explanation:

Answer:

Solid:

- has definite shape and volume.

- change to liquid on heating.

Liquid:

- has definite volume but does not have definite shape .

- changes to solid on cooling.

Gas :

- does not have definite shape or volume.

- change to liquid on cooling

Answer:

The car will stop moving

Explanation:

If a balanced force is applied to the car in motion, the car will stop accelerating and remain stationary because all the forces acting on the car are equal.

Also, you can say, since a balanced force is applied, the net force or resultant force on the car is zero. According to Newton's second law of motion, an object will move in the direction of the applied force. When the resultant force on the object is zero, it means the object will not move.

Answer:  

a) 1m

b) 2μs

c) 3mm

Explanation:

Answer:

I think it is C, I'm not for sure.

Answer:

Attached below is the required diagram related to the question

answer :

q'3 = WбT^4

engineer's proposal has merit

Explanation:

Let : A'3 represent the deep space

      A'1 represent the surface area , F13 and  F23  represent the view factors

      T1 , T2, T3 ; represent temperatures

      q'3 represent net rate of heat radiation

 Derive the expression for the rate per unit length at which radiation is transferred from the surfaces to deep space

derived expression ; q'3 = WбT^4

attached below is a detailed solution

Given that The emission is proportional to the area of the opening and the surfaces ( 1 and 2 ) have the same temperature hence this problem can be treated as a two surface enclosure. hence the engineer's proposal have merit .

attached below is a prove ( b )

Answer:

Half life is 3.23 hours

Explanation:

Given

Decay rate at starting = 1160 decays per minute

Decay rate after 4 hours = 170 decays per minute

As we know know

[tex]N = N_0 *e ^{\Lambda *T}[/tex]

Substituting the given values, we get -

[tex]170 = 1160 *e ^{-4*\Lambda}\\0.1465 = e ^{-4*\Lambda}\\-0.834 = -4 * \Lambda\\\Lambda = 0.834/4\\\Lambda = 0.2085[/tex]

Also

[tex]t_{1/2} = \frac{ln2}{\Lambda}[/tex]

Substituting the given values we get -

[tex]t_{1/2} = =0.693/0.2085\\= 3.23[/tex]hours

Answer:

Electron

Explanation:

An object can become electrically charged when it gains or loses an electron. Because an electron is negatively charged, when an object gains an electron it becomes negatively charged. Also, when it gives up an electron, it becomes positively charged. This positive charge is because the atom has one proton more than electron. In a neutral atom, the number of the proton is equal to the number of the electron. An electron is negatively charged, and a proton is positively charged.

Answer:

soluble soluble soluble soluble

Explanation:

solublesolublesolublesolublesolublesolublesoluble dguhjjewugbcsbdc csyuhjci

Answer:

Work input = Work output * Work against friction is your answer so C

Explanation:

I hope this helps you :)

Answer:

A) Work input = Work output + Work against friction

Explanation:

Answer:

the red pointer on the magnet ( grey region) : points towards north

red pointer outside the magnet ( white region) is pointing towards south

Explanation:

please see the attached image

Answer:

A. 3.9 V B. 1.9 fA

Explanation:

Part A

What emf is induced in the ring as the field changes?

Express your answer to two significant figures and include the appropriate units.

The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time

ε = ΔΦ/Δt

ε = ΔABcos0°/Δt

ε = AΔB/Δt

A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T.  ΔB = B₁ - B₀  = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.

So, ε = AΔB/Δt

ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s

ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s

ε = 3.926 V

ε ≅ 3.9 V

Part B

If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.

Express your answer to two significant figures and include the appropriate units.

Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².

So, i =  ε/R

=  ε/ρl/A

= εA/ρl

= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)

= ‭15.704‬ × 10⁻⁶ V-m²/(82.9356‬ × 10⁸ Ωm²

= 0.1894 × 10⁻¹⁴ A

= 1.894 × 10⁻¹⁵ A

≅ 1.9 fA

Answer:

[tex]12\; \rm N[/tex].

[tex]3\; \rm m\cdot s^{-2}[/tex].

Explanation:

By Newton's Second Law, the acceleration of an object is proportional to the size of the resultant force on it, and inversely proportional to the mass of this object.

[tex]\displaystyle \text{acceleration} = \frac{\text{resultant force}}{\text{mass}}[/tex].

Rearrange this equation for the resultant force on the object:

[tex]\text{resultant force} = \text{acceleration} \cdot \text{mass}[/tex].

For the [tex]6\; \rm kg[/tex] object in this question:

[tex]\begin{aligned} F &= m \cdot a \\ &= 6\; \rm kg \times 2\; \rm m \cdot s^{-2} \\ &=12\; \rm N\end{aligned}[/tex].

When the resultant force on the [tex]4\; \rm kg[/tex] object is also [tex]12\; \rm N[/tex], the acceleration of that object would be:

[tex]\begin{aligned} a &= \frac{F}{m} \\ &= \frac{12\; \rm N}{4\; \rm kg} = 3\; \rm m \cdot s^{-2}\end{aligned}[/tex].

Answer:

The answer is "1.26".

Explanation:

[tex]D=18^{\circ}[/tex]

The refractive index is:

[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]

       [tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]

Answer:

The acceleration of 2.0 kg block is 2.7 m/s²

Explanation:

Since, µ₁ < µ₂ acceleration of 2 kg block down the plane will be more than the acceleration of 4 kg block, if allowed to move separately. But, as the 2.0 kg block is behind the 4.0 kg block both of them will move with same acceleration say a. Taking both the blocks as a single system:

Force down the plane on the system

= (4 + 2) g sin30°

= (6)(10)(½)

= 30N

Force up the plane on the system

= µ₁ (2)(g)cos30° + µ₂ (4)(g)cos30°

= (2µ₁ + 4µ₂) g cos30°

= (2 × 0.2 + 4 × 0.3)(10)(√3/2)

≈ 13.76 N

∴ Net force down the plane is F

F = 30 - 13.76

F = 16.24 N

∴Acceleration of both the blocks down the

plane will b a

a = F ÷ (4 + 2)

a = 16.24 ÷ 6

a = 2.7 m/s²

Thus, The acceleration of 2.0 kg block is

2.7 m/s²

-TheUnknownScientist

Explanation:

2.7m/s2

I hope its helpful

[tex]\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}[/tex]

Actually Welcome to the concept of Logic.

Since in the above statement it is given that,

All S are P ==> True,

then obviously Some S are also P always, hence it is true.

Answer is True.

Answer:

Explanation:

Resultant is a single force that can replace the effect of a number of forces. "Equilibrant" is a force that is exactly opposite to a resultant. Equilibrant and resultant have equal magnitudes but opposite directions.

Answer:

use the formula to calculate acceleration and you'll get the answers

Solution :

In the question, it is given that the collision is inelastic and the blocks stick together.

In an inelastic collision, the linear momentum is conserved but the kinetic energy is not conserved.

The linear momentum is given by :

[tex]$\vec p = m \vec v$[/tex] (mass x velocity)

So according to the conservation of linear momentum,

[tex]$\vec p_{(\text{before collision})}=\vec p_{(\text{after collision})}$[/tex]

Let the velocity after the collision is [tex]$v_F$[/tex]

[tex]$m_1v_0+m_2 \times 0 = m_1v_F+m_2v_F$[/tex]

Putting the values of [tex]$m_1 \text{ and}\ m_2$[/tex]

[tex]$m_1=2M \text{ and}\ m_2=M$[/tex]

∴ [tex]$2Mv_0=2Mv_F+Mv_F$[/tex], as the blocks stick together after the collision.

and  [tex]$2MV_0=3Mv_F$[/tex], as the blocks stick together after the collision.