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MaRussiya [10]
3 years ago
12

A car of mass 2000 kg is thrusted forward with an engine force of 2000 N and a total frictional

Physics
1 answer:
PolarNik [594]3 years ago
6 0

Answer:

40.3 m/s

Explanation:

We'll begin by calculating the resultant force.

Resultant force = 2000 – 1000 = 1000N

Next we shall determine the acceleration of the car. This is illustrated below:

Mass of the car = 2000kg

Resultant Force = 1000N

Acceleration =.?

Resultant Force = Mass x Acceleration

1000 = 2000 x Acceleration

Divide both side by 2000

Acceleration = 1000/2000

Acceleration = 0.5m/s²

Finally, we shall determine velocity of the car when it has moved 1000m.

Initial velocity (u) = 25m/s

Acceleration (a) = 0.5m/s²

Distance (s) = 1000m

Final velocity (v) =..?

Applying the formula v² = u² + 2as we can obtain the velocity of the car at 1000m as follow:

v² = u² + 2as

v² = 25² + 2(0.5 x 1000)

v² = 625 + 2(500)

v² = 625 + 1000

v² = 1625

Take the square root of both side

v = √1625

v = 40.3 m/s

Therefore, the speed of the car when it has moved 1000m is 40.3 m/s

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DiKsa [7]
<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

8 0
3 years ago
In transduction, the cochlea is part of this structure:
anyanavicka [17]

Answer:

wats up

Explanation:

8 0
2 years ago
How does changes in distance affect the gravitational pull between two objects? Describe and give one example.
maxonik [38]
The formula is

F_grav = G * m1 * m2 / r^2

G m1 and m2 are going to stay the same once chosen no matter what the distance is. The only thing that will change is the distance.

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As the distance decreases, the gravitational force will Increase.

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3 years ago
When a force of 450N pushes on a 20kg box as
ehidna [41]

Answer:

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Explanation:

7 0
2 years ago
You're driving down the highway late one night at 16.0 m/s when a deer steps onto the road 39.0 m in front of you. your reaction
choli [55]
The breaking distance consists of two parts. The first part is the first 0.5 seconds were no breaking occurs. Given values: t time, v₀ initial velocity:
x₁ = v₀*t
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were the final velocity is zero, v = 0 and the initial velocity v₀= 16m/s:
v = a*t + v₀ = 0 => v₀ = -a*t => t = v₀/-a

x₂ = 0.5*a*t² = 0.5*v°²/a

The total breaking distance is the sum of the two parts:
x = x₁ + x₂ = v₀* t + 0.5 * v₀² / a = 16 * 0.5 + 0.5 * 16² / 12 = 8 + 10,7 = 18,7

You can use this result to calculate the remaining distance. You can use the last equation to calculate the maximum speed you could have to avoid a collision.
Use x = 39m and solve for v₀. 
5 0
3 years ago
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