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Zigmanuir [339]
4 years ago
9

True or false for both?!

Physics
1 answer:
Zepler [3.9K]4 years ago
7 0
I believe the first one is true because a transverse wave generally looks like a right angle. The second one is also true. May I ask what Physics class you are taking, regular, honors, AP Physics 1, AP Physics 2? Also how close are you to finishing it?
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Two identical particles, each with a mass of 4.5 mg and a charge of 30 nC, are moving directly toward each other with equal spee
e-lub [12.9K]

Answer:

   r₁ = 20.5 cm

Explanation:

In this exercise we can use the conservation of energy

the gravitational power energy is always attractive, the electrical power energy is repulsive if the charges are of the same sign

starting point.

        Em₀ = U_g + U_e + K = -G \frac{m_1m_2}{r} +k \frac{q_1q_2}{r} - 2 ( \frac{1}{2}  m v^2)

the two in the kinetic energy is because they are two particles

final point. When it is detained

        Em_f = U_g + U_e = -G \frac{m_1m_2}{r_1} + k \frac{q_1q_2}{r_1}

the energy is conserved

        Em₀ = em_f

the charges and masses of the two particles are equal

         -G \frac{m^2}{r} + k \frac{q^2}{r} + m v^2 = - G \frac{m^2}{r_1} + k \frac{q^2}{r_1}        

         

sustitute the values

-6.67-11 (4.5 10-3) ² / 0.25 - 9, 109 (30 10-9) ² / 0.25 + 4.5 10-3 4² = - 6.67 10- 11 (4.5 10-3) ² / r1 -9 109 (30 10-9) ² / r1

    -5.4 10⁻¹⁵ + 3.24 10⁻⁵ - 7.2 10⁻⁵ = -1.35 10⁻¹⁵ / r₁  + 8.1 10⁻⁶ / r₁

We can see that the terms that correspond to the gravitational potential energy are much smaller than the terms of the electric power, which is why we depress them.

      3.24 10⁻⁵ - 7.2 10⁻⁵ =  8.1 10⁻⁶ / r₁

      -3.96 10⁻⁵ = 8.1 10⁻⁶ / r₁

      r₁ = 8.1 10⁻⁶ /3.96 10⁻⁵

      r₁ = 2.045 10⁻¹ m

      r₁ = 20.5 cm

4 0
3 years ago
. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to
AveGali [126]
No, the car travels 1 metre in 5s at the start which is 0.2m/s, while the second meter it travels one metre in 8 seconds which is 0.125 m/s, the speed changes therefore it is not constant during the two metres the car travels
3 0
3 years ago
What two element is present in biotite mica that is not present in muscovite mica?
lara31 [8.8K]

Potassium iron magnesium aluminum silicate hydroxide and Potassium magnesium aluminum silicate hydroxide.

3 0
3 years ago
Some types of storage tanks are exempt from the federal regulations on USTs, including
MAXImum [283]

Some types of storage tanks are exempt from the federal regulations on USTs, including : A) tanks that store heating oil burned on the premises.

<h3>What are Underground Storage Tank Systems?</h3>

Underground Storage Tank systems are tanks along with the underground piping that has at least 10% of their total volume underground.

Some of the storage tanks that are exempt from the federal regulations on USTs are;

  • Farm and residential tanks of 1,100 gallons or less capacity holding motor fuel used for noncommercial purposes.
  • Tanks storing heating oil used on the premises where it is stored.
  • Tanks of 110 gallons or less capacity.

Learn more about Underground Storage Tank systems here:

brainly.com/question/26571

6 0
3 years ago
3. A box with a mass of 10.0 kg is placed on a hill that includes upwards at an angle of 30.0°.
FrozenT [24]

Given that,

Mass = 10 kg

Angle = 30°

Static friction = 0.582

Kinetic friction = 0.528

We need to calculate the parallel force

Using balance equation

F =mg\sin\theta

Put the value into the formula

F= 10\times9.8\times\sin30

F = 49\ N

Let’s determine the two friction forces

We need to calculate the static friction force

Using static friction force

F_{s}=\mu mg\cos\theta

Put the value into the formula

F_{s}=0.582\times10\times9.8\cos30

F_{s}=49.4\ N

Since the box is already moving, the force of the push must be greater than difference of the force parallel and the static friction force.

We need to calculate the kinetic friction force

Using kinetic friction force

F_{k}=\mu mg\cos\theta

Put the value into the formula

F_{k}=0.528\times10\times9.8\cos30

F_{k}=44.8\ N

Since this friction force is less than 49 N, the box will accelerate as it moves down hill.

Hence, This is required solution.

5 0
3 years ago
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