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svet-max [94.6K]
2 years ago
14

What is the average speed of a car that travels 30 m in the 3 secs and 2 m 4 secs and 20 m 5 secs

Physics
1 answer:
Doss [256]2 years ago
6 0

Answer:

4.3333333

Explanation:

total distance / total time taken

30 + 2 + 20 / 3 + 4 + 5

52 / 12

4.3 recurring

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Help? Will give thanks if right (thank you)
qwelly [4]

Hello there.

1.986*10^6

10^6=1,000,000

1.986*1,000,000

<u><em>Answer=1,986,000</em></u>

Explanation: First you had to ten to the six power of the ten with six times. It gave us 1000000 should be have six zeros at any time. Then you can multiply by 1.986*1000000=1,986,000 is the right answer. Hope this helps! Thank you for posting your question at here on Brainly. -Charlie

8 0
3 years ago
ANSWER QUCIK<br> PLS THANK YOU
Reil [10]
Electrical energy in the charger and cable

Chemical energy in the battery of the mobile phone
8 0
3 years ago
A 5,400 W motor is used to do work. If the motor is used for 640 s, about how much work could it do?
Gemiola [76]

Answer:

3,500,000 J​

Explanation:

WORK = POWER * TIME

WORK= 5400 * 640

=6456000 J = 3,500,000 J​

4 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim
Bingel [31]

Answer:

given,

speed of the car = 20 m/s

final speed of car = 0 m/s

distance between car and the deer = 38 m

reaction time, t = 0.5 s

deceleration of the car = 10 m/s².

a) distance between deer and car

  distance travel in the reaction time

   d₁ = v x t

   d₁ = 20 x 0.5 = 10 m

   distance travel after you apply brake

   using equation of motion

   v² = u² + 2 a s

   0 = 20² - 2 x 10 x s

    s =  20 m

total distance traveled by the car

D = d₁ + d₂

D = 20 + 10 = 30 m

  distance between car and the deer = 38 m - 30 m

                                                              = 8 m

b) now, maximum speed car.

   distance travel in reaction time

    d₁ = s x t

    d₁ = 0.5 V

distance left between them

   d₂ = 38 - d₁

   d₂ = 38 - 0.5 V

   distance travel after you apply brake

   using equation of motion

    v² = u² + 2 a d₂

    0 = (V)² - 2 x 10 x (38 - 0.5 V)

     V² + 10 V - 760 = 0

now, solving the quadratic equation

  x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

  V = \dfrac{-10\pm \sqrt{10^2-4(1)(-760)}}{2(1)}

         V = 23.01 , -33.01

rejecting the negative term.

hence, maximum speed of the car could be V = 23.01 m/s

 

7 0
3 years ago
A ball is launched from ground level and hits the ground again after an elapsed time of 4 seconds and after traveling a horizont
jeka94

Answer:

1)a. It is constant the whole time the ball is in free-fall.

2)b. = 14 m/s

3) e. = 19.6 m/s

Explanation:

1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.

2) speed = distance/time

horizontal distance = 56m

Time = 4 seconds

Speed = 56m/4s = 14m/s

3) acceleration due to gravity g = 9.8m/s^2

Initial vertical velocity = u

Final vertical velocity = v = -u

Using the law of motion;

v = u + at

a = acceleration = -g = -9.8m/s^2

t = time of flight = 4

Substituting the values;

-u = u - 4(9.8)

-2u = -4(9.8)

u = -4(9.8)/-2

u = 2(9.8) = 19.6 m/s

Initial vertical velocity = u = 19.6 m/s

3 0
3 years ago
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