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Delvig [45]
3 years ago
13

A Ford Mustang weighs about 3500 pounds, and can accelerate from 0-60 MPH in about 5 seconds. What force is responsible for this

acceleration? What is its approximate magnitude?
Physics
1 answer:
Hunter-Best [27]3 years ago
5 0

We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

m = 3500lb = 1587.57kg

The values of the velocities are,

\text{Initial Velocity} = V_i = 0

\text{Final Velocity} = V_f = 60mph = 26.822m/s

We know that the acceleration is equivalent to the change of the speed in a certain time therefore

a = \frac{v_f-v_i}{t}

a = \frac{26.822-0}{5}

a = 5.36m/s^2

Now applying the Newton's second law we have,

F= ma

F = (1587.57)(5.36)

F = 8516.36N

Therefore the approximate magnitude is 8516.36N

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Explanation:

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6 0
2 years ago
Will an electromagnet be obtained if steel is placed inside the solenoid?​
iogann1982 [59]

Answer:A solenoid is a simple electromagnetic device consisting of a coiled electric wire, wrapped in a 3D circular pattern. When electric current is passed through the wire, the solenoid acts like a magnet with N and S poles at the ends of the coil.

When a ferromagnetic material rod is permanently placed inside the solenoid, the metal greatly increases the magnetic effect and becomes a permanent electromagnet. Moreover, it can also be used as an electrical switch by drawing in or pushing out a ferromagnetic material like an iron rod. Depending on the directions of the rod and the electrical current the switching action takes place.

Given figure  represents the solenoid as electromagnet and the switching action.

Explanation:

7 0
3 years ago
To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 41m in front of you. Your reaction time
nordsb [41]

Answer:

Maximum speed you could have and still not hit the deer = 24.07 m/s

Explanation:

Let the maximum speed you could have and still not hit the deer be y.

Your reaction time before stepping on the brakes is 0.50s.

Distance traveled during this time = 0.5y

A deer steps onto the road 41m in front of you

Remaining distance to deer = 41 - 0.5y

The maximum deceleration of your car is 10 m/s²

We have equation of motion, v² = u² + 2as

       Initial velocity, u = y m/s

       Final velocity, v = 0 m/s

       Acceleration, a = -10 m/s²

       Displacement, s = 41 - 0.5y

Substituting,

       v² = u² + 2as

        0² = y² + 2 x -10 x (41 - 0.5y)

          20(41 - 0.5y) = y²

           820 - 10 y = y²

            y² + 10 y -820 = 0

             y = 24.07 m/s or y = -34.06 m/s(not possible)

Maximum speed you could have and still not hit the deer = 24.07 m/s

           

7 0
3 years ago
In a game of basketball, a forward makes a bounce pass to the center. The ball is thrown with an initial speed of 4.8 m/s at an
tensa zangetsu [6.8K]
The equation to be used here is the trajectory of a projectile as written below:

y = xtanθ +/- gx²/2v²(cosθ)²
where
y is the vertical distance
x is the horizontal distance
θ is the angle of trajectory or launch angle
g is 9.81 m/s²
v is the initial velcity

Since the angle is below horizontal, let's use the minus equation. Substituting the values:

- 0.8 m = xtan15° - (9.81 m/s²)x²/2(4.8 m/s)²(cos15°)²
Solving for x,
x = 2.549 m

However, we only take half of this distance because it was specified that the distance asked before bouncing. Hence, the horizontal distance is equal to 1.27 m.
5 0
3 years ago
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