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3 years ago

Objects in motion eventually stop moving due to a _____ .can someone help me with this question

Physics

1 answer:

Tom [10]3 years ago

8 0

External force.

Air resistance, friction, grass rubbing on a rolling ball, etc.

If there is no external force on a moving object, it keeps going.

Forever !

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A 2.0-kg mass is projected from the edge of the top of a 20-m tall building with a velocity of 24 m/s at some unknown angle abov

Digiron [165]

Ox:vₓ=v₀

x=v₀t

Oy:y=h-gt²/2

|vy|=gt

tgα=|vy|/vₓ=gt/v₀=>t=v₀tgα/g

y=0=>h=gt²/2=v₀²tg²α/2g=>tgα=√(2gh/v₀²)=√(2*10*20/24²)=√(400/576)=0.83=>α=tg⁻¹0.83=39°

cosα=vₓ/v=v₀/v=>v=v₀/cosα=24/cos39°=24/0,77=31.16 m/s

Ec=mv²/2=2*31.16²/2=971.47 J=>Ec≈0.97 kJ

3 0

2 years ago

The electric potential at the origin of an xy-coordinate system is 40 V. A -8.0-μC charge is brought from x = +∞ to that point.

vredina [299]

Answer:

-320 μJ.

Explanation:

Consider a point with an electrical charge of $q$ . Assume that $V$ is the electrical potential at the position of that charge. The electrical potential of that point charge will be equal to:

$\text{Potential Energy} = q \cdot V$ .

Keep in mind that since both $q$ and $V$ might not be positive, the size of the electrical potential energy might not be positive, either.

For this point charge,

$q = \rm -8.0\; \mu C$ ; (that's -8.0 microjoules, which equals to $\rm -8.0\times 10^{-6}\; J$ )
$V = \rm 40\; V$ .

Hence its electrical potential energy:

$\text{Potential Energy} = q\cdot V = \rm (-8.0\; \mu C) \times 40\; V = -320\; \mu J$ .

Why is this value negative? The electrical potential energy of a charge is equal to the work needed to bring that charge from infinitely far away all the way to its current position. Also, negative charges are attracted towards regions of high electrical potential. Bringing this $\rm -8.0\; \mu C$ negative charge to the origin will not require any external work. Instead, this process will release 320 μJ of energy. As a result, the electrical potential energy is a negative value.

7 0

3 years ago

On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo

ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

$\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i$

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

$\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i$

42 m/s, negative direction (to the west).

5 0

3 years ago

How can a magnetic field be used to generate an electric current?

Irina18 [472]

<span>When a magnet moves near a wire, it's changing field causes the electrons in the wire to flow as electric current.</span>

3 0

3 years ago

the resistivity of a given wire of cross session area 0.7mm² is 49×10⁴.calcuate the resistance of a 2m lenght of wire

sp2606 [1]

R = 1.4GΩ.

The relation between the resistance and the resistivity is given by the equation R = ρL/A, where ρ is the resistivity of a given material, L is the length and A is the cross-sectional area of the material.

To calculate the resistance of a wire of L = 2m, ρ = 49x10⁴Ω.m and A = 0.7mm² = 0.7x10⁻³m² we have to use the equation R = ρL/A.

R = [(49x10⁴Ω.m)(2m)/0.7x10⁻³m²

R = 98x10⁴Ω.m²/0.7x10⁻³m²

R = 1.4x10⁹Ω = 1.4GΩ

6 0

2 years ago