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Pie
2 years ago
13

What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k

ilograms?
Physics
2 answers:
Lyrx [107]2 years ago
3 0

We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

\sf \: F=ma

Where,

  • F is force
  • m is mass
  • a is acceleration

In our case,

  • F = ?
  • m = 2500 kg
  • a = 20m/s

\tt \: F_{net}  = 2500 \times 20 \\   \tt= 50000

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>

Inessa [10]2 years ago
3 0

Answer:

Explanation:

f =ma

F 2500 kg (20m/s-0m/s/15s)

F=2500kg(1.33m/s²)

F =3,325 N

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Explanation:

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3 years ago
I need help with question #8 please!!
otez555 [7]

Answer: 3.1158

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3 0
2 years ago
Use a common denominator to find-2/3 + -4/5​
Juliette [100K]

Answer:

-22/15

Explanation:

the least common denominator is  15 so first you multiply -2/3 by 5 in both the numerator and denominator making it -10/15

Then you do the same to -4/5 except you multiply the numerator and denominator by 3 giving you -12/15

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6 0
3 years ago
Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to
Degger [83]

The acceleration of the  box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

  • m is the mass of the box
  • a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

  • θ is the inclination of the plane
  • μ is coefficient of friction

F(net) = 170 -  (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the  box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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8 0
1 year ago
At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 V and 16.2 V/m
son4ous [18]

Answer:

A. 30.7cm

B. 1.7*10^{-10}C

C. The electric field is directed away from the point of charge

Explanation:

A.

 because, E=\frac{v}{d} \\\\d= \frac{4.98}{16.2}\\\\ d = 0.307m\\\\d = 30.7 cm

B.

Considering Gauss's law

EA = \frac{Q}{e}\\\\ where, e = pertittivity. space= 8.85* 10^{-12} Fm^{-1} \\\\A = surface. area. with.radius 0.307m\\Q= eEA = (8.85*10^{-12})(16.2)(4\pi)(0.307)^{2}\\\\= 1.7*10^{-10}C

C. The electric field directed away from the point of charge when the charge is positive.

3 0
3 years ago
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