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aleksklad [387]
3 years ago
5

What is the force of gravitational attraction between an object with a mass of 2kg and another object that has a mass of 2kg and

a distance between them of 2.5m
Physics
1 answer:
kupik [55]3 years ago
4 0

Answer:

1.7x10^-10 N

Explanation:

F = G [(m_1)(m_2)]/(r^2)

F = force

G = Gravitational constant 6.67433x10^-11 (N*m^2)/kg^2

m_1 = mass first object

m_2 = mass second object

r = radius between the 2 objects

F = G[(2 kg*2 kg)/(1.25 m)^2]

F = 1.7x10^-10 N

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A steel rod with a length of l = 1.55 m and a cross section of A = 4.89 cm2 is held fixed at the end points of the rod. What is
Vedmedyk [2.9K]

Answer:

The size of the force developing inside the steel rod is 32039.28 N

Explanation:

Given;

length of the steel rod, L =  1.55 m

cross sectional area of the steel, A = 4.89 cm²

temperature change, ∆T = 28.0 K

coefficient of linear expansion for steel, α = 1.17 × 10⁻⁵ 1/K

Young modulus of steel,  E = 200.0 GPa.

Extension of the steel is given as;

α ∆T L = FL / AE

α ∆T = F/AE

F = AEα ∆T

F = ( 4.89 x 10⁻⁴)(200 x 10⁹)(1.17 × 10⁻⁵)(28.0 K)

F = 32039.28 N

Therefore, the size of the force developing inside the steel rod when its temperature is raised, is 32039.28 N

7 0
3 years ago
Suppose you wanted to be able to see astronauts on the moon. what is the smallest diameter of the objective lens required to res
joja [24]

r = distance of moon from earth = 3.84 x 10⁸ m

R = size of the object on moon = 0.67 m

D = diameter of the lens = ?

\lambda = wavelength of light = 550 x 10⁻⁹ m

Using the equation for Rayleigh criterion

R/r = 1.22 \lambda/D

inserting the values

0.67/(3.84 x 10⁸) = 1.22 (550 x 10⁻⁹) /D

D = 384.6 m

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What is the velocity of galaxy mice
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Explanation:

Both galaxies show bicones of high ionisation gas extending along their minor axes. In NGC 4676A the high gas velocity .

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O A ring made up of equal semicircular section of cast inn and castutool with
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1. Swordfish are capable of stunning output power for short bursts. A 650 kg swordfish has a cross-sectional area of 0.92 m2 and
Oksana_A [137]

Answer:

P_{sp}=178.4W/kg

Explanation:

From the question we are told that:

Mass of fish m_f=650kg

Cross-sectional area A=0.92 m^2

Drag coefficient of \mu= 0.0091

Seawater  density \rho= 1026 kg/m^3.

Speed of  Fish v=30 m/s  

Generally the equation for Drag force F_d is mathematically given by

F_d = \mu * \rho *A v^2 /2

F_d = 0.0091* 0.92* 1026* 30^2/2 \\F_d= 3865. 35 N  

Generally the equation for high speed  Power  P_{sp} is mathematically given by

P_{sp}=3865*35*\frac{v}{m_f}

P_{sp}=F_d*35*\frac{30}{650}

P_{sp}=178.4W/kg

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3 years ago
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