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andrey2020 [161]
3 years ago
8

If I Throw a bowling ball and a tennis ball from a building, describe if one would hit the ground first and why.

Physics
1 answer:
olga nikolaevna [1]3 years ago
3 0

Answer:

The bowling ball would

Explanation:

Because it contains more weight! And that will make it fall down quicker.

Hope it helped

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5kg of copper ball and I kg of water are at the same temperature of 31 C. Which
blagie [28]

Answer:

r

Explanation:

8 0
3 years ago
Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.
shusha [124]

Answer:

F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

                                  r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2  }

- Then compute the angle that Force makes with the y axis:

                                 cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

                                 F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

                                F_net = F_1,2 + kq / y^2

- Hence,

                                F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2}  )

- Now for the limit y >>a:

                              F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2}  )

- Insert limit i.e a/y = 0

                              F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2})  \\\\F_n = 3*k*q/y^2

Hence the Electric Field is off a point charge of magnitude 3q.

8 0
4 years ago
Sound source A has a deciBel rating of 50 dB. Sound source B is 1000 times more intense. What is the deciBel rating of B
masha68 [24]

The decibel system of sound intensity operates by a logarithmic scale, meaning that sound intensity increases exponentially in relation to the decibel rating.

For decibels, the equation between intensity and the dB equivalent is:

dB = 10log(i),

where “i” is the intensity of the sound. The ten in front of the log means that an increase in ten dB results in a tenfold increase in sound intensity; for example, a 30 dB sound is ten times softer than a 40 dB sound.

In this case, a sound with a dB of 80 would be 1000 times more intense than a 50 dB sound, so the decibel rating of B is 80.

Hope this helps!

3 0
3 years ago
Mass and Weight: What is the mass of an object that experiences a gravitational force of 685 N near Earth's surface where g
11Alexandr11 [23.1K]

Answer:

m = 69.9 kg

Explanation:

The mass and the weight of an object are two different quantities. Mass is basically the amount of matter that is present in a body. It remains same everywhere in the universe and measured in kilograms.

Weight is basically a force. It is the force by which earth attracts everything towards itself. The weight of an object changes from planet to planet, with the change in value of the gravitational acceleration (g).

Therefore, the relation between mass and weight of an object is given by the following formula:

W = mg

m = W/g

where,

m = mass = ?

W = Weight = 685 N

g = 9.8 m/s²

Therefore,

m = (685 N)/(9.8 m/s²)

<u>m = 69.9 kg</u>

4 0
3 years ago
Identify the frequency used by these broadcast media.<br><br> FM<br> AM<br> Television
Oliga [24]

Answer:

AM radio = 540 - 1650 KHz

FM radio = 88 to 108 MHz

Television = 54 to 216 MHz

7 0
4 years ago
Read 2 more answers
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