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maw [93]
3 years ago
12

You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the

gas pedal. You can ignore friction, but you can't ignore air resistance. Identify all the forces acting on the car?
Physics
1 answer:
wariber [46]3 years ago
5 0

Answer:

Weight, air drag, normal.

Explanation:

Since the car is only rolling downhill, the forces that will come into play are: it's own weight (the gravitational force the Earth exerts upon the car), the air resistance (the force the air exerts upon the car, since it says it can't be ignored) and finally the normal force (the contact force the ground exerts on the car). There are no other forces sources.

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What is the largest possible magnitude of the acceleration of the electron due to the magnetic field?
Alina [70]
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<span>An electrons moves at 2.0x10^6 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.4x10^-2 T. 

The </span> largest possible magnitude of the acceleration of the electron due to the magnetic field is <span>= 2.6 × 10 ¹⁶ m / s ²</span>
6 0
3 years ago
Which of the following indicates a heat transfer?
lesya [120]

Answer:

TEMPERATURE CHANGES

Explanation:

WELL ITS BASIC. WHEN TEMPERATURE CHANGES. IT MEANS THAT HEAT IS BEING TRANSFERRED AS I HEARD THAT COLD CANNOT BE TRANSFERRED

3 0
3 years ago
A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
nordsb [41]

Complete Question

A 10 gauge copper wire carries a current of 20 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm2.) mm/s

Answer:

The drift velocity is v  = 0.0002808 \ m/s

Explanation:

From the question we are told that

    The current on the copper is  I  = 20 \ A

     The cross-sectional area is  A =  5.261 \ mm^2 =  5.261 *10^{-6} \ m^2

The number of copper atom in the wire is  mathematically evaluated

      n  =  \frac{\rho *  N_a}Z}

Where \rho is the density of copper with a value \rho =  8.93 \ g/m^3

          N_a is the Avogadro's number with a value N_a  = 6.02 *10^{23}\ atom/mol

         Z  is the molar mass of copper with a value  Z =  63.55 \ g/mol

So

     n  =  \frac{8.93 * 6.02 *10^{23}}{63.55}

     n  = 8.46 * 10^{28}  \  atoms /m^3

Given the 1 atom is equivalent to 1 free electron then the number of free electron is  

         N  = 8.46 * 10^{28}  \  electrons

The current through the wire is mathematically represented as

         I  =  N * e * v * A

substituting values

        20 =  8.46 *10^{28} * (1.60*10^{-19}) * v *  5.261 *10^{-6}

=>     v  = 0.0002808 \ m/s

       

8 0
3 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
3 years ago
Calculate the total energy of 4.0 kg object moving horizontally at 20 m/s 50 meters above the surface.
Serhud [2]

Answer:

Correct answer:  E total = 2,800 J

Explanation:

Given:

m = 4 kg   the mass of the object

V = 20 m/s  the speed (velocity) of the object

H = 50 m the height of the object above the surface

E total = ? J

The total energy of an object is equal to the sum of potential and kinetic energy

E total = Ep + Ek

Ep = m g H   we take g = 10 m/s²

Ep = 4 · 10 · 50 = 2,000 J

Ek = m V² / 2

Ek = 4 · 20² / 2 = 2 · 400 = 800 J

E total = 2,000 + 800 = 2,800 J

E total = 2,800 J

God is with you!!!

4 0
3 years ago
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