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GarryVolchara [31]
3 years ago
15

Calculate the speed of the car at each checkpoint by

Physics
1 answer:
lana66690 [7]3 years ago
3 0

Answer: The speed at the first quarter checkpoint is 0.74 m/s. The speed at the second quarter checkpoint is 1.40 m/s. The speed at the third quarter checkpoint is 1.61 m/s. The speed at the finish line is 1.89 m/s.

Explanation: I did the assignment and got it correct :)

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This is a change in the position of a body with respect to time relative to a reference point.
dusya [7]

Answer: MOTION

Explanation:

motion is defined as the displacement of an object with respect to time relative to a stationary object (reference point). A good example of an object that can serve as a reference point includes: a tree or a building. The movement of a body at constant speed towards a particular direction at regular intervals of time can be determined and it's called uniform motion.

There are different types of motion, these includes: simple harmonic motion,

linear motion,

circular motion,

Brownian motion,

Rotatory motion

5 0
2 years ago
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How many degrees are there between the direction of motion and the force of
Aleks04 [339]

Answer:

180°

Explanation:

Friction, if it exists, ALWAYS opposes motion or attempted motion.

5 0
3 years ago
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Which of the following is accurate in describing converging and diverging lenses?
kenny6666 [7]

B. Both of these types of lenses have the ability to produce real images.

3 0
3 years ago
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This is for physical science. Can someone help me?
Radda [10]
A. 320 g
B. 160 g
C. 80 g
D. 40 g
6 0
3 years ago
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Kinematics practice problems Answers: 4. A race car is traveling at +76 m/s when is slows down at -9 m/s2 for 4
arlik [135]

The new velocity after 4 s is 40 m/s

The height of the spaceship above the ground after 5 seconds is 1,127.5 m

The given parameters for the first question;

  • initial velocity of the car, u = 76 m/s
  • acceleration of the car, a = - 9 m/s²
  • time of motion, t = 4 s

The new velocity after 4 s is calculated as;

v = u + at

v = 76 + (-9)(4)

v = 76 - 36

v = 40 m/s

(5)

The given parameters;

  • height above the ground, h = 500 m
  • velocity of spaceship, u = 150 m/s
  • time of motion, t = 5

The height of the spaceship above the ground after 5 seconds is calculated as;

h_y = h_0 + ut - \frac{1}{2}gt^2\\\\h_y = 500 + (150\times 5) -  (0.5\times 9.8 \times 5^2)\\\\h_y = 1,127.5 \ m

Learn more here: brainly.com/question/24527971

5 0
2 years ago
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