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VladimirAG [237]
2 years ago
6

A car travels three-quarters of the way around a circle of radius 20.0 m in a time of 3.0 s at a constant speed. the initial vel

ocity is west and the final velocity is south. (a) find its average velocity for this trip. (b) what is the car's average acceleration during these 3.0 s? (c) explain how a car moving at constant speed has a nonzero average acceleration.
Physics
1 answer:
schepotkina [342]2 years ago
3 0
20.3 divided by 3.0 will get u velocity and v times 3.0s 
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What voltage is required to move 6A through 20?<br><br>​
Marina CMI [18]

Answer:

120V

Explanation:

Given parameters:

Current  = 6A

Resistance  = 20Ω

Unknown:

Voltage = ?

Solution:

According to ohms law;

           V = IR

Where V is the voltage

            I is the current

            R is the resistance

Now, insert the parameters and solve;

    V  = 6 x 20  = 120V

7 0
2 years ago
Two objects have masses m and 5m, respectively. They both are placed side by side on a frictionless inclined plane and allowed t
poizon [28]

Answer:

(E) The two objects reach the bottom of the incline at the same time.

Explanation:

Given;

first object with mass, m

second object with mass, 5m

The acceleration of gravity for both object is the same = 9.8 m/s²

Since both objects have the same acceleration of gravity, and no external force due friction (frictionless inclined plane), they will reach bottom of the inclined at the time.

Thus, the acceleration due to gravity is constant for all objects regardless of their masses.

Therefore, the correct option is E;

(E) The two objects reach the bottom of the incline at the same time.

5 0
2 years ago
Fiber optic (FO) cables are based upon the concept of total internal reflection (TIR), which is achieved when the FO core and cl
kozerog [31]

Answer:

False

Explanation:

Though fiber active cable is based on the concept of internal reflection but it is achieved by refractive index which transmit data through fast traveling pulses of light. It has a layer of glass and insulating casing called “cladding,”and this is is wrapped around the central fiber thereby causing light to continuously bounce back from the walls of the Cable.

7 0
2 years ago
A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

5 0
2 years ago
This table shows Wayne’s weight on four different planets. Planet Wayne’s weight (pounds) Mars 53 Neptune 159 Venus 128 Jupiter
oee [108]
The correct order is (in decreasing order of gravity strength)
Jupiter - Neptune - Venus - Mars

In fact, Wayne's weight on each planet is given by
W=mg
where m is Wayne's mass, which is a constant value, and g is the gravity strength at the surface of the planet. Therefore, the Wayne's weight W on each planet is directly proportional to the gravity strength of that planet: so the planet with the strongest gravity is the one where Wayne's weight is the greatest (Jupiter, 333 pounds), followed by Neptune (159),  Venus (128) and Mars (53).
8 0
2 years ago
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