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julsineya [31]
2 years ago
11

A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a consta

nt force (in contrast to the constant power in part a). if such a sports car went from zero to 29.0 mph in time 1.30 s , how long would it take to go from zero to 58.0 mph ?
Physics
1 answer:
-Dominant- [34]2 years ago
7 0

The car would go from  zero to 58.0 mph in 2.6 sec.

Since the force on the car is constant, therefore the acceleration of the car would also be constant.

Now for constant acceleration we can use the equation of motion

Using first equation of motion to calculate the acceleration of the car

v=u+at

29=0+a×1.30       ...... Eq. (1)

Again using the first equation of motion

58=0+a*t             ....... Eq. (2)

Dividing eq. (2) with equation 1

t=2×1.3

t=2.6 sec

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Force applied = F = 628 N 
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<span>Newton's 2nd law of motion : F = Ma </span>
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<span>New mass of the crate = M1 = 3.8M kg </span>
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A 70kg man spreads his legs as shown calculate the tripping force​
shutvik [7]

This is simply defined as the energy of an object as an attribute of a physical movement

Force

This is simply defined as the energy of an object as an attribute of a physical movement

it is important to note that a tripping for will be a force opposing motion therefore

For a 70kg man trip on a force the for it must me acted upon by a Force

For more information on this visit

brainly.com/question/21811998

6 0
1 year ago
At a certain instant the current flowing through a 5.0-H inductor is 3.0 A. If the energy in the inductor at this instant is inc
lys-0071 [83]

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

5 0
2 years ago
An automobile approaches a barrier at a speed of 20 m/s along a level road. The driver locks the brakes at a distance of 50 m fr
AleksAgata [21]

Answer:

μ = 0.408

Explanation:

given,

speed of the automobile (u)= 20 m/s

distance = 50 m

final velocity  (v) = 0 m/s

kinetic friction = ?

we know that,

v² = u² + 2 a s

0 = 20² + 2 × a × 50

a = \dfrac{400}{2\times 50}

a = 4 m/s²

We know

F = ma = μN

ma = μ mg

a = μ g

\mu = \dfrac{a}{g}

\mu = \dfrac{4}{9.81}

μ = 0.408

hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408

5 0
2 years ago
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