Certain neutron stars (extremely dense stars) are believed to be rotating at about 500 rev/s. If such a star has a radius of 17 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation

Answer: the at which the bar conducts now is 5 Js⁻¹

Explanation:

Given the data in the question;

we know that; Heat transfer by conduction is given by;

Q ∝ [tex]A / l[/tex]

such that,

[tex]Q_{1}/Q_{2}[/tex] = [tex]A_{1}l_{2} / A_{2}l_{1}[/tex]

[tex]Q_{2} = Q_{1} A_{2}l_{1}/A_{1}l_{2}[/tex]

so

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r}{2}[/tex])² × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])  

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × π([tex]\frac{r^{2} }{4}[/tex]) × [tex]l[/tex] ) / (πr² × [tex]\frac{l}{2}[/tex])

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × πr² × 1/4 × [tex]l[/tex] ) / (πr² × 1/2 × [tex]l[/tex] )

[tex]Q_{2} =[/tex] ( 10 Js⁻¹ × 1/4) / ( 1/2)

[tex]Q_{2} =[/tex]  ( 10 Js⁻¹ × 1/4) / ( 1/2)

[tex]Q_{2} =[/tex] 5 Js⁻¹

Therefore, the at which the bar conducts now is 5 Js⁻¹

Explanation:

There are two answers

m/(s/s)=m

or

(m/s)/s=m/s²

Answer: 30 to 40 s

Explanation:

Answer:

hello your question is incomplete attached below is the complete question

answer :

20.16 v

Explanation:

The reading of the voltmeter at the instant the switch returns  to position a

L = 5H

i ( current through inductor ) = 1/L ∫ V(t) d(t) + Vo

                                               = 1/5 ∫ 3*10^-3  d(t)  + 0 = 0.6 * 10^-3 t

iL ( 1.6 s ) = 0.6 * 10^-3 * 1.6 = 0.96 mA

Rm ( resistance ) = 21 * 1000 = 21 kΩ

 The reading of the voltmeter ( V )

V = IR

   = 0.96 mA * 21 k Ω  = 20.16 v

Answer:

the force exerted on the foot by the tibia would be 2975 N

Explanation:

Given the data in the question;

To maintain equilibrium between the foot and the ball vertically, the addition normal normal force [tex]N^>[/tex] (750 N)  and the tension in the Achilles tendon [tex]F^>_{Achilles}[/tex] (2225 N) must be equal to the force exerted on the foot by the tibia;

so

| [tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] | = | [tex]F^>_{Tibia}[/tex] |

so force exerted on the foot by the tibia will be;

| [tex]F^>_{Tibia}[/tex] | = |[tex]N^>[/tex] | + |[tex]F^>_{Achilles}[/tex] |

so we substitute IN OUR VALUES

| [tex]F^>_{Tibia}[/tex] | = 750 N + 2225 N

| [tex]F^>_{Tibia}[/tex] |  = 2975 N

Therefore, the force exerted on the foot by the tibia would be 2975 N

Answer:

v = 10.46 m/s

x =  30.134 m from house

Explanation:

given data

speed = 3.05 m/s

time = 7 s

height = 40.8 mm

solution

we get here first time required to fall that is t

t = [tex]\sqrt{\frac{2\times 40.8}{9.8}}[/tex]       ..................1

t = 2.88 s

now we take here initial speed that is v

so v × t = 3.05 × ( t+ 7)

v = [tex]\frac{3.05\times (2.88 + 7)}{2.88}[/tex]

v = 10.46 m/s

and

when she catch the bagel henrietta was at

x = 3.05 × ( 2.88 + 7)

x =  30.134 m from house

Answer:

The acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Explanation:

Given;

first net force, F₁ = 100 N

second net force, F₂ = 50 N

If we consider equal mass for the two net forces, and apply Newton's second law of motion, the acceleration produced by the 100 N net force will be two times greater than the acceleration produced by 50 N net force.

Let a₁ be the acceleration produced by the first net force

then, a₂ be the acceleration produced by the second net force

Thus, a₁ = 2a₂

Answer:

The speed of the cyclist is 2.75 km/min.

Explanation:

Given

To determine

We need to find the speed of a cyclist.

In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.

Using the formula involving speed, time, and distance

[tex]s=\frac{d}{t}[/tex]

where

substitute d = 88, and t = 32 in the formula

[tex]s=\frac{d}{t}[/tex]

[tex]s=\frac{88}{32}[/tex]

Cancel the common factor 8

[tex]s=\frac{11}{4}[/tex]

[tex]s=2.75[/tex] km/min

Therefore, the speed of the cyclist is 2.75 km/min.

Answer:

58.469 kg.m/s are moving in the xy plane

Answer: [tex]26.460\times 10^3\ J[/tex]

Explanation:

Given the mass of skier m=72 kg

distance traveled d=75 m

constant speed v=3.4 m/s

If speed is constant  then there must no force acting in the direction of motion

i.e. tension force must be equal to the component of weight

[tex]T=mg\sin 30^{\circ}[/tex]

Work done is given by

[tex]\Rightarrow W=F\cdot d=Td\cos 0^{\circ}\\\Rightarrow W=mg\sin 30^{\circ} d=72\times 9.8\times 0.5\times 75\\\Rightarrow W=26.460\times 10^3\ J[/tex]

Answer:

the force your friend applied on the box is 40 N.

Explanation:

Given;

speed of the box, v₁ = 1 m/s

force applied to the box, F₁ = 20 N

the speed of the box when your friend pushes it, v₂ = 2 m/s

then your friends applied force, F₂ = ?

Assuming the time, t, through which both forces were applied and mass of the box, m, to be constant;

[tex]F_1 = \frac{mv_1}{t} \\\\\frac{m}{t} = \frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]

[tex]F_2 = \frac{F_1v_2}{v_1} \\\\F_2 = \frac{20\times 2}{1} \\\\F_2 = 40 \ N[/tex]

Therefore, the force your friend applied on the box is 40 N.

Answer:

Explanation:

Answer: IT B TRUST ME MANN OK

Explanation: OK BYE TRUST

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = [tex]\frac{1}{2}[/tex] mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = [tex]\frac{1}{2} \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }[/tex]

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             [tex]K_e = K_p[/tex]

              K_p = ½ m v_e²

              K_p = [tex]\frac{1}{2}[/tex]  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = [tex]\frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }[/tex]

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = [tex]\frac{7 \ 10^4 }{ 3 \ 10^8}[/tex]

        v/c= 2.33 10⁻⁴

Answer:

a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Explanation:

Given the data in the question;

as the equation of standing wave on a string is fixed at both ends

y = 2AsinKx cosωt

but k = 2π/λ and ω = 2πf

λ = 4 × 0.150 = 0.6 m

and f =  v/λ = 260 / 0.6 = 433.33 Hz

ω = 2πf = 2π × 433.33 = 2722.69

given that A = 2.20 mm = 2.2×10⁻³

so [tex]V_{max1}[/tex] = A × ω

[tex]V_{max1}[/tex] = 2.2×10⁻³ × 2722.69 m/s

[tex]V_{max1}[/tex] =  5.9899 m/s

therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s

b)

A' = 2AsinKx

= 2.20sin( 2π/0.6 ( 0.075) rad )

= 2.20 sin(  0.7853 rad ) mm

= 2.20 × 0.706825 mm

A' = 1.555 mm = 1.555×10⁻³

so

[tex]V_{max2}[/tex] = A' × ω

[tex]V_{max2}[/tex] = 1.555×10⁻³ × 2722.69

[tex]V_{max2}[/tex] = 4.2338 m/s

Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s

Answer:

13.4436

Explanation:

Answer:

a = 7.749 ft/s²

Explanation:

First to all, we need to convert all units, so we can work better in the calculations.

The horizontal acceleration is asked in ft/s² so the units of speed will be the same. The Work is in BTU and we need to convert it in ft.lbf in order to get the acceleration and final speed in ft/s:

W = 10 BTU * 778.15 Lbf.ft / BTU = 7781.5 lbf.ft

Now, to get the acceleration we need to get the final speed of the object first. This can be done, by using the following expression:

W = ΔKe  (1)

And Ke = 1/2mV²

So Work would be:

W = 1/2 mV₂² - 1/2mV₁²

W = 1/2m(V₂² - V₁²)    (2)

Finally, we need to convert the mass in lbf too, because Work is in lbf, so:

m = 55 lb * 1 lbf.s²/ft / 32.174 lb = 1.7095 lbf.s²/ft

Now, we can calculate the final speed by solving V₂ from (2):

7781.5 = (1/2) * (1.7095) * (V₂² - 20²)

7781.5 = 0.85475 * (V₂² - 441)

7781.5/0.85475 = (V₂² - 400)

9103.83 + 400 = V₂²

V₂ = √9503.83

V₂ = 97.49 ft/s

Now that we have the speed we can calculate the acceleration:

a = V₂ - V₁ / t

Replacing we have:

a = 97.49 - 20 / 10

Hope this helps

Answer:

A. estrogen

Explanation:

This is released in the female reproductive organ.

Answer:

   F = 196 N

Explanation:

For this exercise we will use Newton's second law,  we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

Y axis  

       N- W = 0

       N = mg

X axis

       F -fr = ma

In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

       F- fr = 0

       F = fr

the friction force has the equation

       fr = μ N

       fr = μ mg

we substitute

        F = μ mg

let's calculate

         F = 0.80 9.8 25

         F = 196 N

Answer:

432.78 Kg

Explanation:

From the question given above, the following data were obtained:

Distance apart (r) = 6.8 m

Force of attraction (F) = 5.4×10¯⁸ N

Mass of Daffy Duck (M₁) = 86.5 kg

Mass of Minnie Duck (M₂) =?

NOTE: Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

The mass of Minnie Duck can be obtained as follow:

F = GM₁M₂ / r²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 6.8²

5.4×10¯⁸ = 6.67×10¯¹¹ × 86.5 × M₂ / 46.24

Cross multiply

6.67×10¯¹¹ × 86.5 × M₂ =5.4×10¯⁸ × 46.24

Divide both side by 6.67×10¯¹¹ × 86.5

M₂ = 5.4×10¯⁸ × 46.24 / 6.67×10¯¹¹ × 86.5

M₂ = 432.78 Kg

Therefore, the mass of Minnie Duck is 432.78 Kg

Answer: B

Explanation:

The vertical component of a vector such as velocity is the magnitude of the vector multiplied by the sine of the angle.

[tex]V_y=48*sin(53)=38.3m/s[/tex]

maybe a spectrograph ?

Answer:

h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Explanation:

Using two rolls of tissue paper : One roll dropped normally while the other drops as some holds onto a sheet of the toilet paper ( I.e. the tissue paper drops rotating about its axis )

Determine the ratio of heights  h1/h2

mass of tissues = same

radius of tissues = same

h1 = height of tissue 1

h2 = height of tissue 2

For the first tissue ( Tissue that dropped manually )

potential energy = kinetic energy

mgh = 1/2 mv^2  

therefore the final velocity ( v^2 ) = 2gH  ----- ( 1 )

second tissue ( Tissue that dropped while rotating )

gh = [tex]\frac{v^2}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] ) ------ ( 2 )

To determine the ratio of heights we will equate equations 1 and 2

hence :

gh = [tex]\frac{2gH}{u}[/tex] ( 3 + [tex]\frac{u^2}{R^2}[/tex] )

∴ h1/h2 = [tex]\frac{2R^2}{3R^2 + h^2}[/tex]

Answer:

Kinetic energy Temperature

Explanation:

Answer:

popu

Explanation:

2u2uwju2i2je82jei

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

Answer:

using W=1/2kW2

k=22N/m w=2.9

w=1/2×22×2.9×2.9

w=92.51Joules

Approximately 93J answer is C