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pochemuha
3 years ago
6

Which of Newton's Three Laws does the following statement satisfy?

Physics
2 answers:
Firlakuza [10]3 years ago
7 0
The answer is C) Newton's Third Law
earnstyle [38]3 years ago
3 0
Newtons third law states that when one body exerts a force on another, the second exerts a force on the first. These two forces are always equal in magnitude and opposite in directions. Option C is the correct answer
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Without using a micrometer screw gauge, how do I find the average diameter of a long piece of thin wire using a metre rule and a
Mice21 [21]

Answer:

Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.

Explanation:

Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.

For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.

5 0
2 years ago
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0
4 years ago
Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a
Alex17521 [72]

Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

7 0
3 years ago
a person does 100 joules of work in pulling back the string of a bow. what will be the initial speed of a 0.5 kg arrow when it i
eduard

If all the energy she put into bending the bow is completely
transmitted to the arrow, then the arrow has the 100 joules
of kinetic energy when it leaves the bow.

                 Kinetic energy = (1/2) (mass) (speed)²

                         100 J        = (1/2) (0.5 kg) (speed²)

Divide each side by 0.25 kg:    100 J / 0.25 kg = speed²

 [ joule ] = [ newton-meter ] = kg-m²/sec²

                                                100 kg-m²/sec² / 0.25 kg = speed²

                                                 400 m²/sec² = speed²

Take the square root of each side:     speed = √400 m/s

                                                                                20 m/s

                                                                    (about 44.7 mph)                                           

3 0
3 years ago
A charge of Q is fixed in space. A second charge of q was first placed at a distance r1 away from Q. Then it was moved along a s
topjm [15]

Answer:

\Delta U = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

Explanation:

The electrostatic potential energy is given by the following formula

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Now, we will apply this formula to both cases:

U_1 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_1^2}\\U_2 = \frac{1}{4\pi\epsilon_0}\frac{Qq}{r_2^2}

So, the change in the potential energy is

\Delta U = U_2 - U_1 = \frac{Qq}{4\pi\epsilon_0}(\frac{1}{r_2^2}-\frac{1}{r_1^2})

7 0
3 years ago
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