1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Likurg_2 [28]
2 years ago
5

PLSEAE HELP ME I WILL GIVE BRIANLYEST!!!!

Physics
1 answer:
Shkiper50 [21]2 years ago
8 0

Answer:

The star map shows directions relative to the position of the North Star.Explanation:

You might be interested in
A 10-cm-long thin glass rod uniformly charged to 8.00 nCnC and a 10-cm-long thin plastic rod uniformly charged to - 8.00 nCnC ar
ch4aika [34]

Complete Question

A 10-cm-long thin glass rod uniformly charged to 8.00 nC and a 10-cm

long thin plastic rod uniformly charged to -8.00 nC are placed side by

side, 4.20 cm apart. What are the electric field strengths E_1 to E_3 at

distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line

connecting the midpoints of the two rods

a.) Specify the electric field strength E1

b.) Specify the electric field strength E2

c.) Specify the electric field strength E3

Answer:

              E_1=7.13*10^5 N/C

             E_2= 2.95*10^{5} N/C

              E_3= 3.84*10^5 N/C

Explanation:

  From the question we are told that

          The length of the thin glass is  L = 10 cm

          The  charge on the glass rod is  q_g = 8.00nC = 8* 10^{-9} C

           The length of the plastic rod is  L_p = 10cm

             The charge on the  plastic rod is q_p =- 8.00nC = -8.0*10^{-9}C

           The distance between the materials  is d = 4.20cm = \frac{4.2}{100} =0.042m

          The various distances to obtain electric field of are r_1 = 1.0cm

                                                                                                r_2 = 2.0cm

                                                                                                 r_3 = 3.0cm

The objective of the solution is to obtain the electric field E_1 , E_2 \ and E_3 at distance d_1 , d_2 \ and \ d_3  from the glass rod  along the line connecting its mid point  

   Generally electric field of a charge rod at a distance of r the line dividing the rod  into half  is mathematically represented as

                              E = k \frac{2Q}{r\sqrt{L^2 + 4r^2} }

For the  r_2 = 1.0cm = \frac{1}{100} = 0.01m

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

                               E_l =  k \frac{2Q }{r \sqrt{L^2 + 4r^2_1} }

The electric filed by the positively charge glass rod on the right  side of the dividing line is mathematically represented as  

                            E_r =  k \frac{2Q }{(0.044 - r_1) \sqrt{L^2 + 4r^2_1} }

The net electric field is,

            E_{net} =E_1= E_l + E_r

                    = k \frac{2Q}{r_1\sqrt{L^2 + 4 r^2_1 } } + k \frac{2Q}{(0.04-r_1) \sqrt{L^2 + 4 (0.044 -r_1)^2} }

Where k is  know as the coulomb's constant  with a constant value of

                  k = 9*10^9 \ kgm^3 s^{-4} A^{-2}

           =(9*10^9) \frac{(2) (8*10^{-9})}{(0.01)\sqrt{(0.01^2 + 4(0.01)^2)} }  + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.01)\sqrt{(0.01)^2 + (4) (0.042 - 0.01)^2} }

                           = 6.44*10^5 + 6.9*10^4

                           E_1=7.13*10^5 N/C

For the  r_2 = 2.0cm = \frac{2}{100} = 0.02m

           The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

                               E_l =  k \frac{2Q }{r_2 \sqrt{L^2 + 4r^2_2} }

The electric filed by the positively charge glass rod on the right  side of the dividing line is mathematically represented as  

                            E_r =  k \frac{2Q }{(0.044 - r_2) \sqrt{L^2 + 4r^2_2} }

The net electric field is,

            E_{net} =E_2= E_l + E_r

                    = k \frac{2Q}{r_2\sqrt{L^2 + 4 r^2_2 } } + k \frac{2Q}{(0.04-r_2) \sqrt{L^2 + 4 (0.044 -r_2)^2} }

Where k is  know as the coulomb's constant  with a constant value of

                  k = 9*10^9 \ kgm^3 s^{-4} A^{-2}

           =(9*10^9) \frac{(2) (8*10^{-9})}{(0.02)\sqrt{(0.02^2 + 4(0.02)^2)} }  + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.02)\sqrt{(0.02)^2 + (4) (0.042 - 0.02)^2} }

            = 1.6*10^{5}+ 1.3*10^{5}

             E_2= 2.95*10^{5} N/C

For the  r_3 = 3.0cm = \frac{3}{100} = 0.03m

           The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

                               E_l =  k \frac{2Q }{r_3 \sqrt{L^2 + 4r^2_3} }

The electric filed by the positively charge glass rod on the right  side of the dividing line is mathematically represented as  

                            E_r =  k \frac{2Q }{(0.044 - r_3) \sqrt{L^2 + 4r^2_3} }

The net electric field is,

            E_{net} =E_3= E_l + E_r

                    = k \frac{2Q}{r_3\sqrt{L^2 + 4 r^2_3 } } + k \frac{2Q}{(0.04-r_3) \sqrt{L^2 + 4 (0.044 -r_3)^2} }

Where k is  know as the coulomb's constant  with a constant value of

                  k = 9*10^9 \ kgm^3 s^{-4} A^{-2}

           =(9*10^9) \frac{(2) (8*10^{-9})}{(0.03)\sqrt{(0.03^2 + 4(0.03)^2)} }  + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.03)\sqrt{(0.03)^2 + (4) (0.042 - 0.03)^2} }

        = 7.2 *10^{4} + 3.1*10^5

      E_3= 3.84*10^5 N/C                

8 0
3 years ago
. Water is flowing at 12m/s in a horizontal pipe under a pressure of 600kpa radius 2cm. a. What is the speed of the water on the
lana66690 [7]

Answer:

Outlet Velocity = 192 m/s

Outlet Pressure = 510 kPa

Explanation:

Givens:

Inlet Velocity, V₁ = 12 m/s

Inlet Pressure, P₁ = 600 kPa = 600,000 Pa

Inlet Radius r₁ = 0.5 cm

Outlet Velocity, V₂ = not given (we are asked to find this)

Outlet Pressure,  P₂ = not given (we are asked to find this)

Outlet Radius, r₂ = 0.5 cm

From these, we can find the following:

Inlet Area, A₁ = π (r₁)² = π(2)² = 4π cm²

Outlet Area, A₂ = π (r₂)² = π(0.5)² = 0.25π cm²

<u>Part A :</u>

Assuming that water is incompressible, we can reason that within the same given time, the amount of volume of water entering the inlet must equal the volume of water exiting the outlet. Hence by the continuity equation (i.e. conservation of mass)

Inlet Volume flow rate = Outlet Volume flow rate

(recall that Volume flow rate in a pipe is given by Velocity x Cross Section Area), Hence the equation becomes

V₁ x A₁ = V₂ x A₂  (substituting the values that we know from above)

12 x 4π = V₂ x 0.25π  (we don't have to change all to SI units because the conversion factors on the left will cancel out the conversion factors on the right).

V₂ = (12 x 4π) / (0.25π)

V₂ = 192 m/s  (Answer)

<u>Part B:</u>

For Part B, if we assume a closed ideal system (control volume method), we can simply apply the energy equation (i.e Bernoulli's equation)

P₁ + (1/2)ρV₁ + ρgh₁ = P₂ + (1/2)ρV₂ + ρgh₂

Because the pipe is horizontal, there is no difference between h₁ and h₂, hence we can neglet this term:

P₁ + (1/2)ρV₁ = P₂ + (1/2)ρV₂  (rearranging)

P₂ = P₁ + (1/2)ρV₁ - (1/2)ρV₂

= P₁  + (1/2)ρ (V₁-V₂)

Assuming that the density of water is approx, ρ = 1000 kg/m³

P₂ = 600,000  + (1/2)(1000) (12-192)

= 600,000  + ( -90,000)

= 510,000 Pa

= 510 kPa (Answer)

8 0
3 years ago
A rocket, blasting into space, accelerates at 2 m/s^2 for 20 seconds. What is its speed at the end of 20 seconds?​
OLEGan [10]

Answer:

  Acceleration  =  (change in speed)  /  (time for the change)

Change in speed = (ending speed) - (beginning speed)

                           =     (3,600 mi/hr) - ( 0 )

                           =       3,600 mi/hr

                           =     same as  1 mile/second.

Acceleration  =  (1 mi/sec) / (10 sec)

                     =     0.1 mi/sec² .

But 1 mile = 5,280 ft,

so

                      0.1 mi/s²  =  528 ft/s²          

Explanation:

have a great day

4 0
2 years ago
What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?
Tanya [424]

Answer:

The minimum coefficient of friction is 0.22

Explanation:

Suppose If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve.

We need to calculate the ideal speed to take a 85 m radius curve banked at 15°.

Given that,

Radius = 85 m

Angle = 15°

Speed = 20 km/h

We need to calculate the ideal speed

Using formula of speed

\tan\theta=\dfrac{v^2}{rg}

v=\sqrt{rg\tan\theta}

Put the value into the formula

v=\sqrt{85\times9.8\tan15}

v=14.9\ m/s

We need to calculate the minimum coefficient of friction

Using formula for coefficient of friction

v^2=\dfrac{rg(\sin\theta-\mu\cos\theta)}{\mu\sin\theta+\cos\theta}

Put the value into the formula

(5.55)^2=\dfrac{85\times9.8(\sin15-\mu\cos15)}{\mu\sin15+\cos15}

\dfrac{30.8025}{85\times9.8}=\dfrac{0.2588-\mu0.966}{\mu0.2588+0.966}

0.9754462\ mu-0.223541=0

\mu=\dfrac{0.223541}{0.9754462}

\mu=0.22

Hence, The minimum coefficient of friction is 0.22

3 0
2 years ago
Determine which factors will increase the reaction rate and which factors will decrease te reaction rate.
Elodia [21]

Answer: increase reaction rate side : increase surface area

Increase temperature

Add a catalyst

Decrease reaction rate side :

Decrease surface area

Decrease temperature

Decrease concentration

Explanation:

Got it right

8 0
3 years ago
Other questions:
  • What keeps the particles of a liquid from spreading out to fill an entire container in the same way that gas particles do?
    7·2 answers
  • Which statement best describes a physical change?
    10·1 answer
  • Which two types of waves can transmit energy through a vacuum?
    10·2 answers
  • Buffalo, New York, and Raleigh, NC, lie approximately on the same meridian. Buffalo has a latitude of 42.9° N, and Raleigh has a
    11·1 answer
  • Compared with the amount of current in the filament of a lamp, the amount of current in the connecting wire is A. definitely les
    9·2 answers
  • A leaky faucet drips once each second. at this rate, it takes 1 hour to fill a 1-liter graduated cylinder to the 200-cm3 mark. w
    7·1 answer
  • What kind of stars make up the galactic nucleus?
    6·2 answers
  • Acetone is a flammable solvent used in the making of plastics. The density is 790kg/m3. to convert 790kg/m3 to g/cm3​
    14·1 answer
  • The propeller of an aircraft accelerates from rest with an angular acceleration α = 7t + 8, where α is in rad/s2 and t is in s
    15·1 answer
  • If you wanted to
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!