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navik [9.2K]
3 years ago
14

How to solve 2A− B ?

Physics
1 answer:
s344n2d4d5 [400]3 years ago
5 0

Explanation:

you sure that the question is complete?

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4. Derive<br>the relation, P= hd g​
Sergeu [11.5K]

p=F/A

or,P=d×V×G/A (m=d×V)

or,p= d× A×h×g/A (A and A are cut)

or,P=d×H×G

4 0
4 years ago
I need it due in 10 mins ​
ExtremeBDS [4]

Answer:

B. 14.4 N

Rotational speed (Angular Velocity) = 2

The Radius of the circle = 1.2 m

Velocity = Angular velocity × radius = 2×1.2 = 2.4 m/s

Centripetal force= mv²/r = 3 × 2.4×2.4/1.2 = 3 × 2.4 × 2

= 14.4 N

7 0
3 years ago
HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLEASE HELP PLE
Romashka [77]

A_x = 5.0

A_y = -6.3

6 0
3 years ago
What is the value of the equivalent resistance for the three resistors connected in series?
Harman [31]

The value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

To find the answer, we have to know more about the equivalent resistance.

<h3>What is meant by equivalent resistance?</h3>
  • equivalent resistance is the total value of the resistance connected in a circuit.
  • If n resistors are connected in series, then the equivalent resistance will be,

                R_E=R_1+R_2+..........+R_n

  • In our question we have three resistors. Thus, the equivalent resistance will be,

               R_E=R_1+R_2+R_3

Thus, we can conclude that, the value of the equivalent resistance for the three resistors connected in series will be the sum of the three values.

Learn more about the equivalent resistance here:

brainly.com/question/11603204

#SPJ4

7 0
2 years ago
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
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