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Elanso [62]
2 years ago
5

If the car has a mass of 1200 kg, how much force must the engine provide?

Physics
1 answer:
vazorg [7]2 years ago
6 0

Force in Newtons =

(1200)•(the rate of acceleration you want)

It depends on how fast you want to accelerate.

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30 points + brainliest
pishuonlain [190]

Answer:B is the answer

4 0
2 years ago
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Before the start of a trip, air in a tire is at 320 kPa gage pressure and 27°C temperature. At the end of the trip the tire pres
tia_tia [17]

Answer:

The temperature of air in the tire is 55.57 ºC

Explanation:

Please look at the solution in the attached Word file

Download docx
7 0
3 years ago
A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1. 5 second
madreJ [45]

The force involved is = 25200 Newton

Given the values in the questions,

Mass = 1400 kg

Speed or velocity of the car = 27 m/s

Time is given = 1.5 seconds

According to the formula of force,

⇒ Force = Mass x Acceleration

⇒ Force = 1400 x Acceleration ----- equation 1

Now to calculate the value of acceleration we will use,

⇒ Acceleration = (Velocity or Speed) / Time

⇒ Acceleration = 27 / 1.5

⇒ Acceleration = 18 m/s^{2}

Putting the value of acceleration in equation 1,

⇒ Force = 1400 x 18

⇒ Force = 25200 Newton

Therefore, the force involved is = 25200 Newton

To learn more about Force,

brainly.com/question/13191643

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8 0
1 year ago
Two charged spheres on a frictionless horizontal surface are attached to opposite ends of a string & are in static equilibri
deff fn [24]

Answer:

the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

Explanation:

For this exercise we will use coulomb's law

          F = k q₁ q₂ / r²

where they indicate the value of the force F = 2.5 N, the distance between them r = 0.06 m

they also indicate the value of the total load

         q₁ + q₂ = 10 10⁻⁶ C

we substitute the values

         2.5 = 9 10⁹ q₁ q₂ / 0.06²

         2.5 0.06 2/9 10⁹ = q₁ q₂

         1 10⁻¹² = q₁ q₂

         10 10⁻⁶= q₁ + q₂

we have two equations with two i unknowns, so the system can be solved.

         q₁ = 1 10⁻¹² / q₂

we substitute in the other equation

         10 10⁻⁶ = 10⁻¹² / q₂ + q₂

let's solve the equation

         10 10⁻⁶ q₂ = 10⁻¹² + q₂²

         q₂² - 10⁻⁵ q₂ + 10⁻¹² = 0

          q₂ = [10⁻⁵ ± √(10⁻¹⁰ - 4 10⁻¹²)] / 2

          q₂ = [10⁻⁵ + - 0.9798 10⁻⁵] / 2 = 10⁻⁵ (1 + - 0.9798) / 2

          q₂ = 0.9899 10⁻⁵ C

          q₂ = 0.0101 10⁻⁵ C

let's find the charge of the other sphere

         q₂ = 0.9899 10⁻⁵ C

         q₁ = 10⁻⁵ - q₂

          q₁ = 10⁻⁵ - 0.9899 10⁻⁵

          q₁ = 0.0101   10⁻⁵    C

therefore the red sphere has a charge of 9.899 10⁻⁶ C and

the green sphere 0.101 10⁻⁶ C

5 0
3 years ago
a 5.5kg box us pushed across the lunch table. the net force applied to the box is 9.7n. what is the acceleration of the box? ​
lys-0071 [83]

Answer:

1.76 m/s^2

Explanation:

The formula to find the force is F = ma (reference from 2nd law of Newton's Laws of Motion). Since you need to find the acceleration of the box, you need to use the formula a = f/m

a = 9.7 n divided by 5.5 kg = 1.76 m/s^2

4 0
2 years ago
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