Solar flares are disturbances on the Sun’s surface that send large amounts of energy away from the Sun’s surface. These disturbances can often interfere with technology here on Earth. What statement BEST describes the cause of solar flares?

Answer:

popu

Explanation:

2u2uwju2i2je82jei

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 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

Answer:

73N

Explanation:Just multiply 1.2^2 by 50

Answer:

  L = 5076.5 kg m² / s

Explanation:

The angular momentum of a particle is given by

         L = r xp

         L = r m v sin θ

the bold are vectors, where the angle is between the position vector and the velocity, in this case it is 90º therefore the sine is 1

as we have two bodies

       L = 2 r m v

let's find the distance from the center of mass, let's place a reference frame on one of the masses

        [tex]x_{cm}[/tex] = [tex]\frac{1}{M} \sum x_{i} m_{i}[/tex]i

        x_{cm} = [tex]\frac{1}{m+m} ( 0 + l m)[/tex]

        x_{cm} = [tex]\frac{1}{2m} lm[/tex]

        x_{cm} = [tex]\frac{1}{2}[/tex]

        x_{cm} = 13.1 / 2 = 6.05 m

let's calculate

          L = 2  6.05  74.3  5.65

          L = 5076.5 kg m² / s

Answer:

5076.5

Explanation:

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]

Answer: 30 to 40 s

Explanation:

Answer:

θ₂/ θ₁= 2.58

Explanation:

In this exercise you are asked to compare the angular diameters of Mars and Jupiter. The angular diameter or angle in radians is

           θ = D / R

where D is the diameter of the body, the distance from Earth to the body of interest and θ is angle in radians

The different distances are tabulated with respect to the Sun

Sun -Earth     1,496 10¹¹ m

Sun- Mars     2.28 10¹¹ m

Sun - Jupiter 7.78 10 m

The Radii of the planets are

Mars     3.37 10⁶ m

Jupiter 6.99 10⁷ m

let's calculate the angles for each body

a) Mars

       θ₁ = 2r / R'

the distance from the ground is

      R ’= D_planet - D_earth

      R ’= 2.28 10¹¹ - 1.496 10¹¹

       R ’= 0.784 10¹¹ m

let's calculate

       θ₁ = [tex]\frac{2 \ 3.37 \ 10^6 }{0.784 \ 10^{11}}[/tex]

        θ₁ = 8.6 10⁻⁵ radians

b) Jupiter

       R ’= 7.78 10¹¹ - 1.496 10¹¹

      R ’= 6.284 10¹¹ m

let's calculate

       θ₂ = [tex]\frac{2 \ 6.99 \ 10^7}{6.284 \ 10^{11}}[/tex]

        θ₂ = 2.22 10⁻⁴ radians

the ratio of the angular diameters is

       θ₂/ θ₁ = [tex]\frac{2.22 \ 10^{-4}}{8.6 \ 10^{-5}}[/tex]

        θ₂/ θ₁= 2.58

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:

Explanation:

Kinetic energy of the mass of 10 kg

= 1/2 m v² , m is mass and v is velocity .

= .5 x 10 x 20²

= 2000 J

The opposing force stops it . so work done by opposing force will be equal to this energy and it will be negative .

So energy transfer will be  - 2000 J .

= 2 kJ .

If distance travelled by mass is d , force 25 N will have a displacement of d . so work done by force of 25 N

= 25 x d

25 d = 2000

d = 80 m .

Hence system travels a distance of 80 m .

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 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:

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:

13.4436

Explanation:

Explanation:

There are two answers

m/(s/s)=m

or

(m/s)/s=m/s²

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:

I cannot fully see the picture, but I'm going to have to tell you to go with towards the sun because it says summer.

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

maybe a spectrograph ?

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:

   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:

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:

Kinetic energy Temperature

Explanation:

Answer:

Position X.

Explanation:

To know the correct answer to the question, it is important that we know the definition of potential energy.

Potential energy can be defined as the energy possessed by an object in relation to its position. Mathematically it can be expressed as:

PE = mgh

Where

PE => is the potential energy.

m => is the mass of the object.

g => is the acceleration due to gravity.

h => is the height to which the object is located.

Considering the formula for potential energy (i.e PE = mgh), we can see clearly that the potential energy (PE) is directly proportional to the height (h). This implies that the greater the height, the greater the potential energy and the smaller the height, the smaller the potential energy.

Considering the diagram given above, we can see that the greatest height attained by the ball is at position X. Thus, the ball will have the greatest potential energy at position X.

Answer:

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:

the required charged is 7.06 × 10⁻¹³ C

Explanation:

Given that;

Radius = 30 microns = 30 × 10⁻⁶

Speed v = 20 m/s

length x = 1.6 cm = 0.016 m

spacing d = 1.0 mm = 0.001 m

Voltage V = 1500 V

from the question, the electric field between the plates is uniform and equal to Voltage divided by the distance between the plates.

Electric field E = V/d

E = 1500 V /  0.001 m

E = 1.5 × 10⁶ V/m

Mass of ink drop m = pv

m = 10³ kg/m³ × [tex]\frac{4}{3}[/tex]πr³

m = 1000 kg/m³ × [tex]\frac{4}{3}[/tex]π × (30 × 10⁻⁶)³

m = 1.131 × 10⁻¹⁰ Kg

Time taken to travel t =  x / sped

t = 0.016 m / 20 m/s

t = 0.0008 s

From the kinematic equation

to form the top of a letter 3 mm ( 0.003 m )high

y = [tex]\frac{1}{2}[/tex]at²

2y = at²

a = 2y/t²

we substitute

a = (2 × 0.003 m) / (0.0008 s)²

a =  9375 m/s²

Now Force F = Eq = ma

so

q = ma / E

we substitute

q = ( 1.131 × 10⁻¹⁰ Kg × 9375 m/s² ) / ( 1.5 × 10⁶ V/m )

q = 7.06 × 10⁻¹³ C

Therefore, the required charged is 7.06 × 10⁻¹³ C