The spaceship Enterprise 1 is moving directly away from earth at a velocity that an earth-based observer measures to be 0.62c. A sister ship, Enterprise 2, is ahead of Enterprise 1 and is also moving directly away from earth along the same line. The veolcity of Enterprise 2 relative to Enterprise 1 is 0.30c. What is the velocity of Enterprise 2

Answer:

[tex]291.67\ \text{m/s}[/tex]

Explanation:

[tex]m_1[/tex] = Mass of bullet = 4 g

[tex]m_2[/tex] = Mass of block = 1.3 kg

[tex]\mu[/tex] = Coefficient of friction = 0.17

[tex]s[/tex] = Displacement of block = 0.24 m

[tex]v_1[/tex] = Velocity of bullet

[tex]v[/tex] = Velocity of combined mass

[tex]g[/tex] = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

The energy balance of the system is given by

[tex]\dfrac{1}{2}(m_1+m_2)v^2=\mu(m_1+m_2)gs\\\Rightarrow v=\sqrt{2\mu gs}[/tex]

As the momentum is conserved in the system we have

[tex]m_1v_1=(m_1+m_2)v\\\Rightarrow m_1v_1=(m_1+m_2)\sqrt{2\mu gs}\\\Rightarrow v_1=\dfrac{(m_1+m_2)\sqrt{2\mu gs}}{m_1}\\\Rightarrow v_1=\dfrac{(4\times 10^{-3}+1.3)\times \sqrt{2\times 0.17\times 9.81\times 0.24}}{4\times 10^{-3}}\\\Rightarrow v_1=291.67\ \text{m/s}[/tex]

The initial speed of the bullet is [tex]291.67\ \text{m/s}[/tex].

Answer:

The photosphere is the visible "surface" of the sun. So your answer would be C.

Explanation: its right

Answer:

The equation of motion is [tex]x(t)=-[/tex][tex]\frac{1}{3} cos4\sqrt{6t}[/tex]

Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is [tex]32ft/s^2[/tex]

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet [tex]x=\frac{4}{12} =\frac{1}{3}feet[/tex]

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

[tex]W=kx[/tex] ⇒ [tex]k=\frac{W}{x}[/tex]

The spring constant , [tex]k=\frac{24}{1/3}[/tex]

                            = 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

    [tex]m\frac{d^2x}{dt} +kx=0[/tex]

    [tex]\frac{3}{4} \frac{d^2x}{dt} +72x=0[/tex]

  [tex]\frac{d^2x}{dt} +96x=0[/tex]

Auxiliary equation is, [tex]m^2+96=0[/tex]

                                 [tex]m=\sqrt{-96}[/tex]

                               =[tex]\frac{+}{} i4\sqrt{6}[/tex]

Thus , the solution is [tex]x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}[/tex]

                                 [tex]x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2[/tex]  [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6t}[/tex]

The mass is released from the rest x'(0) = 0

                    [tex]=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6(0)}[/tex] =0

                                                    [tex]c_2[/tex] [tex]4\sqrt{6} =0[/tex]

                                     [tex]c_2=0[/tex]

Therefore , [tex]x(t)=c_1[/tex] [tex]cos 4\sqrt{6t}[/tex]

Since , the mass is released from the rest from 4 inches

                    [tex]x(0)= -4[/tex] inches

[tex]c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}[/tex] feet

   [tex]c_1=-\frac{1}{3}[/tex] feet

Therefore , the equation of motion is  [tex]-\frac{1}{3} cos4\sqrt{6t}[/tex]

Answer:

Approximately [tex]261\; \rm K[/tex], if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.

Explanation:

Let [tex]P_1[/tex] and [tex]P_2[/tex] denote the pressure of this gas before and after the changes.

Let [tex]V_1[/tex] and [tex]V_2[/tex] denote the volume of this gas before and after the changes.

Let [tex]T_1[/tex] and [tex]T_2[/tex] denote the temperature (in degrees Kelvins) of this gas before and after the changes.

Let [tex]n_1[/tex] and [tex]n_2[/tex] denote the quantity (number of moles of gas particles) in this gas before and after the changes.

Assume that this gas is an ideal gas. By the ideal gas law, the ratios [tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex] and [tex]\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex] should both be equal to the ideal gas constant, [tex]R[/tex].

In other words:

[tex]R = \displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1}[/tex].

[tex]R =\displaystyle \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Combine the two equations (equate the right-hand side) to obtain:

[tex]\displaystyle \frac{P_1 \cdot V_1}{n_1 \cdot T_1} = \frac{P_2 \cdot V_2}{n_2 \cdot T_2}[/tex].

Rearrange this equation for an expression for [tex]T_2[/tex], the temperature of this gas after the changes:

[tex]\displaystyle T_2 = \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2} \cdot T_1[/tex].

Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: [tex]n_2 = n_1[/tex], [tex](n_2 / n_1) = 1[/tex].

[tex]\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}[/tex].

Answer:

kilogram

In the International System of Units (SI), the kilogram is the basic unit of mass, and the newton is the basic unit of force. The non-SI kilogram-force is also a unit of force typically used in the measure of weight.

Answer:

not enought info

Explanation:

tbh I just know it's not 5 10 or 1

Answer:

B. 5 A

Explanation:

10/2= 5

Educere

Answer:

Blank 1: Gasses

Blank 2: More

Blank 3: Solids

Blank 4: Fluids

Blank 5: Liquid

Blank 6: Gas

Blank 7: Higher

Blank 8: Lower

Blank 9: Sun

Blank 10: Radiation

Blank 11: Conductors

P.S. order of answers does not matter between Blank 5 and 6.

Answer:

Explanation:

Moment of inertia I = M k² , where M is mass and k is radius of gyration .

Putting the given values in the equation

5000 = 200 x k²

k² = 25

k = 5 cm .

Radius of gyration is 5 cm .

Answer:

como  ías

Explanation:

Answer:

a.

Explanation:

there would be a new planet is our solar system which could cause different gravitation pull on all the planets also there could be possible be new life form or other valuable metals that haven't been discovered on this planet. hope this helps somewhat

Answer:

harmful effects

1. that will cause air pollution

2. that will destroy our earth

Answer:

useful effects of volcano are :-

harmful effects of volcano are:-

hope it is helpful to you ☺️

Answer:

W = 1/2 K x^2

x^2 = 2 * W / K = 2 * 49 J / (N/m)  = .218 / m^2

x =  .467 m

Answer:

please give me brainlist and follow

Explanation:

Using locally available resources for art help in the preservation of environment. A significant and practical aspects of art is material significance. The items used by artists while making an art piece affects both the form and the material. Every material delivers something special in the creative process.

No, because some scientist are bad and they must have a scientific attitude like honesty, curiosity, open-mindedness, optimistic and more. And they should follow others sometimes because for their own good.

Answer:

No

Explanation:

Science operates in the context of national and international laws, agreements and conventions. It requires scientists to conduct and communicate scientific work for the benefit of society, with excellence, integrity, respect, fairness, trustworthiness, clarity, and transparency.

Answer:

speed: 35m/s

direction: left

Explanation:

Assuming the right side is the positive direction:

before explosion:

P = mv = 0

after explosion:

P' = 15P + 5P

(Set the velocity of the 15kg piece after explosion as v1' and the velocity of the 5kg piece after explosion as v2')

P' = 0.75mv1' + 0.25mv2'

P' = (15kg)v' + (5kg)(105m/s)

P' = 525kg/m/s + (15kg)v1'

P = P'

525kg/m/s + (15kg)v1' = 0

(15kg)v1' = -525kg/m/s

v1' = -35m/s

speed = |-35| = 35m/s

direction is to the left since the right side is the positive direction.

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

[tex]\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft[/tex]

Also, at any instant t

[tex]\Rightarrow l^2=x^2+y^2[/tex]

differentiate w.r.t.

[tex]\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s[/tex]

Answer:

13.6 N

Explanation:

Since one side of the wedge is vertical and the wedge makes and angle of 33 with the horizontal, the angle between the weight of the copper block on the incline and the incline is thus 90 - 33 = 57.

Let M be the mass of the block that hangs, m be the mass of the block on the incline and T be the tension in the weightless unstretchable cord.

We assume the motion is downwards in the direction of the hanging block, M.

We now write equations of motion for each block.

So

Mg - T = Ma    (1) and T - mgcos57 - F = ma where F is the frictional force on the block on the incline and a is their acceleration.

Now, since both blocks do not move, a = 0.

So, Mg - T = M(0) = 0     and T - mgcos57 - F = m(0) = 0

Mg - T = 0    (3) and T - mgcos57 - F = 0 (4)

From (3), T = Mg

Substituting T into (4), we have

T - mgcos57 - F = 0

Mg - mgcos57 - F = 0

So, Mg - mgcos57 = F  

F = Mg - mgcos57

F = (M - mcos57)g

Since g = acceleration due to gravity = 9.8 m/s², and M = 2.94 kg and m = 2.85 kg.

We find F, thus

F = (2.94 kg - 2.85 kgcos57)9.8 m/s²

F = (2.94 kg - 2.85 kg × 0.5446)9.8 m/s²

F = (2.94 kg - 1.552 kg)9.8 m/s²

F = (1.388 kg)9.8 m/s²

F = 13.6024 kgm/s²

F ≅ 13.6 N

Answer:

the total magnetic force on the 2 sides of the wire is 0.30 N

Option a) 0.30 N is the correct answer

Explanation:

Given the data in the question;

both the sides of the right angled triangle is the same and the magnetic is in the plane of the  triangle and is perpendicular to the hypotenuse.

hence, angle between the magnetic field and both the sides of the triangle is 45 degrees.

∅ = 45°

I = 2.0 A

L = 15 cm = 0.15 m

B = 0.7 T

Total magnetic force on the 2 sides of the wire will be;

F = 2BILsin∅

we substitute

F = 2 × 0.7 × 2.0 × 0.15 × sin(45°)

F = 0.42 × 0.70710678

F = 0.2970 ≈ 0.30 N

Therefore, the total magnetic force on the 2 sides of the wire is 0.30 N

Option a) 0.30 N is the correct answer

Answer:

#1 - True (touch) charging when electric conductors actually touch one another.

#2. True - electrical charges are conserved (not destroyed)

I believe the answer is a

Answer:

Answer: Oxygen

Explanation:

Oxygen would not be considered as a pollutant

Answer:

Hey there the answer is C. Oxygen

Explanation:

Oxygen is the chemical element with the symbol O and atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal, and an oxidizing agent that readily forms oxides with most elements as well as with other compounds. After hydrogen and helium, oxygen is the third-most abundant element in the universe by mass. At standard temperature and pressure, two atoms of the element bind to form dioxygen, a colorless and odorless diatomic gas with the formula O 2. Diatomic oxygen gas constitutes 20.95% of the Earth's atmosphere. Oxygen makes up almost half of the Earth's crust in the form of oxides. Hope this helps! Have a great day!

Answer:

im sorry i would help but thats too much

Answer:

The second bulb will have thicker filament

Explanation:

Given;

First electric bulb: Power, P₁ = 100 W and Voltage, V₁ = 220 V

Second electric bulb: Power, P₂ = 200 W and Voltage, V₂ = 200 V

Resistivity of tungsten, ρ = 4.9 x 10⁻⁸ ohm. m

Resistance of the first bulb:

[tex]P = IV = \frac{V}{R} .V = \frac{V^2}{R} \\\\R = \frac{V^2}{P} \\\\R_1 = \frac{V_1^2}{P_1} = \frac{(220)^2}{100} = 484 \ ohms[/tex]

Resistance of the second bulb:

[tex]R_2 = \frac{V_2^2}{P_2} = \frac{(200)^2}{200} = 200 \ ohms[/tex]

Resistivity of the tungsten filament is given by the following equation;

[tex]\rho = \frac{RA}{L}[/tex]

where;

L is the length of the filament

R is resistance of each filament

A is area of each filament

[tex]A = \pi r^2[/tex]

where;

r is the thickness of each filament

[tex]\rho = \frac{R (\pi r^2)}{L} \\\\\frac{\rho L}{\pi} = Rr^2 \\\\Recall ,\ \frac{\rho L}{\pi} \ is \ constant \ for \ both \ filaments\\\\R_1r_1^2 = R_2r_2^2\\\\(\frac{r_1}{r_2} )^2 = \frac{R_2}{R_1} \\\\\frac{r_1}{r_2} = \sqrt{\frac{R_2}{R_1} } \\\\\frac{r_1}{r_2} = \sqrt{\frac{200}{484} } \\\\\frac{r_1}{r_2} = 0.64\\\\r_1 = 0.64 \ r_2\\\\r_2 = 1.56 \ r_1[/tex]

Therefore, the second bulb will have thicker filament