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Neko [114]
3 years ago
5

you discover that Eunice and Bert's go-kart weighs 400 kg and will have a 675 n force actung on it. how fast will it accelerate

(to nearest tenth)​
Physics
1 answer:
alekssr [168]3 years ago
6 0

Answer:

200

Explanation:

subtract it it's 275

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It is known that a shark can travel at a speed of 17 m/s. How far can a shark go in 8 seconds?
o-na [289]

Answer:136 m

Explanation:17*8

6 0
3 years ago
In a Hall-effect experiment, a current of 3A sent lengthwise through a conductor of 1 cm wide, 4 cm long,
Marina CMI [18]

Answer:drift velocity is 1.7*10^-4m/s and number of charge is 2.7*10^-24

Explanation:

4 0
4 years ago
Convert 10 kilowatt to watt​
Nataly [62]

Answer:

0.01 kilowatts cubed

Explanation:

just ask alexa lol

3 0
3 years ago
Read 2 more answers
If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of
EleoNora [17]
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE  = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
 
Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
     FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
     TRUE

C) (KE + PE)beginning = (KE + PE) end.
     TRUE

D) All of the above.
     FALSE

4 0
3 years ago
Read 2 more answers
A glider with a mass of 2 kg is moving rightward at 1 m/s. A second glider, with a mass of 3 kg, is also moving rightward, at 5
Sophie [7]

The velocity of the second glider after the collision is 4.33 m/s rightward.

<h3>Velocity of the second glider after the collision</h3>

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

where;

  • m₁ is mass of first glider
  • m₂ is mass of second glider
  • u₁ is initial velocity of first glider
  • u₂ is initial velocity of second glider
  • v is the final velocity of the gliders

(2)(1) + (3)(5) = (2)(2) + 3v₂

17 = 4 + 3v₂

3v₂ = 17 - 4

3v₂ = 13

v₂ = 13/3

v₂ = 4.33 m/s

Thus, the velocity of the second glider after the collision is 4.33 m/s rightward.

Learn more about linear momentum here: brainly.com/question/7538238

#SPJ1

7 0
2 years ago
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