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madreJ [45]
3 years ago
14

IF YOU ANSWER ALL THE QUESTIONS THEN I WILL GIVE YOU BRAINEST!!! 20 POINTS!!!!

Physics
1 answer:
Firdavs [7]3 years ago
5 0

Answer:

1.) A,     2.) D,     3.) C

Explanation:

Hope this Helps! :D

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A satellite m-500 kg orbits the earth at a distance d 215 km, above the surface of the planet. The radius of the earth is re 6.3
AnnZ [28]

Answer:

7.78 * 10³ m/s

Explanation:

Orbital velocity is given as:

v = √(GM/R)

G = 6.67 * 10^(-11) Nm/kg²

M = 5.98 * 10^(24) kg

R = radius of earth + distance of the satellite from the surface of the earth

R = 2.15 * 10^(5) + 6.38 * 10^(6)

R = 6.595 * 10^(6) m

v = √([6.67 * 10^(-11) * 5.98 * 10^(24)] / 6.595 * 10^(6))

v = √(6.048 * 10^7)

v = 7.78 * 10³ m/s

4 0
3 years ago
A teacup on a spinning teacup ride takes 1.5 s to complete a revolution around the center of the platform. What is the centripet
arlik [135]
The correct answer is 53 meters per second squared 
5 0
3 years ago
Read 2 more answers
Which of the following is NOT a possible cause of a genetic disorder in humans?
DanielleElmas [232]
Im pretty sure the answer is mutations in the DNA of genes because i believe the other 3 are correct 
4 0
3 years ago
Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2
Bess [88]

Answer:

The acceleration of M_2 is  a =  0.7156 m/s^2

Explanation:

From the question we are told that

    The mass of first block is  M_1 =  2.25 \ kg

    The angle of inclination of first block is  \theta _1 =  43.5^o

    The coefficient of kinetic friction of the first block is  \mu_1  = 0.205

      The mass of the second block is  M_2 = 5.45 \ kg

     The angle of inclination of the second block is  \theta _2 =  32.5^o

      The coefficient of kinetic friction of the second block is \mu _2 = 0.105

The acceleration of M_1 \ and\  M_2 are same

The force acting on the mass M_1 is mathematically represented as

     F_1 = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

=> M_1 a = T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1

Where T is the tension on the rope

The force acting on the mass M_2 is mathematically represented as    

  F_2 =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

   M_2 a =  M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

At equilibrium

  F_1 =  F_2

So

 T -  M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2

making a the subject of the formula

    a =  \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}

substituting values a =  \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}

    => a =  0.7156 m/s^2

     

3 0
4 years ago
Visible light transmitting through a certain medium.
Verizon [17]
I think its A: Translucent hope this helps if I'm wrong I'm sorry
8 0
3 years ago
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