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Lapatulllka [165]
3 years ago
7

Which describes a high frequency wave?

Physics
1 answer:
SCORPION-xisa [38]3 years ago
8 0
I think the correct answer from the choices listed above is option A. A high frequency wave is a wave with a low level of energy and a high pitch. Frequency is the number of waves passing per second of time. Hope this answers the question. 
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You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el
nordsb [41]

Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

  • If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
  • Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
  • n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)
4 0
3 years ago
A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op
Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

F cos \theta

where

F is the magnitude of the pull

\theta=30^{\circ} is the angle

- Backward: the force of friction, which is

F_f = 30 N

So, the equation of motion is

F cos \theta - F_f = ma

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

F cos \theta - F_f = 0

Therefore, we can find the pulling force:

F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N

(b) 43.9 N

In this case, the acceleration is

a=0.40 m/s^2

So, the equation of motion in this case is

F cos \theta - F_f = ma

So this time we have to take into account the term (ma).

Using the  same data as before:

m = 20 kg

\theta=30^{\circ}

F_f = 30 N

We find the new magnitude of F:

F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N

6 0
3 years ago
Can someone awnser this
Svetllana [295]

Answer:

The force of gravity exerts a downward force. The floor exerts an upward force. Since these two forces are of equal magnitude and in opposite directions, they balance each other.

6 0
3 years ago
A model rocket is launched directly upward at a speed of 16 meters per second from a height of 2 meters. The function f(t)=−4.9t
Crank

Answer:

Hmax=15.06 meters

Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function

The maxima of the function occurs when the slope is zero. i.e.

\frac{df}{dt} =0\\\frac{df}{dt} =\frac{d}{dt} (-4.9t^2+16t+2)\\\frac{df}{dt} =-4.9*2t+16\\-9.8t+16=0\\t=16/9.8\\t=1.63 secs

Hence the maxima occurs at t=1.63 seconds

The maximum value of f is

f(1.63)=-4.9(1.63^2)+16(1.63)+2\\f(1.63)=15.06\\

hence maximum height is found to be

Hmax=15.06 meters

8 0
3 years ago
What is the volume of an irregularly shaped object that has a mass 3.0 grams and a density of 6.0 g/mL
pogonyaev

Answer: The volume of an irregularly shaped object is 0.50 ml

Explanation:

To calculate the volume, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of object = 6.0g/ml

mass of object = 3.0 g

Volume of object = ?

Putting in the values we get:

6.0g/ml=\frac{3.0g}{\text{Volume of substance}}

{\text{Volume of substance}}=0.50ml

Thus the volume of an irregularly shaped object is 0.50 ml

4 0
3 years ago
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