If force remains constant and acceleration decreases what must happen to the mass

If the frequency of an electromagnetic wave increases, does the number of waves passing by you increase, decrease, or stay the s

jarptica [38.1K]

Answer:

option(d)

Explanation:

The frequency of a wave is the property of the source of wave.

The velocity of all the electromagnetic waves is same as the speed of light. It only changes as the light passes through one medium to another.

The frequency is defined as the number of waves coming out from the source in 1 second.

As the frequency of wave increases, the number of wave coming per second increases.

So, the number of waves passing by increases but the speed remains same.

Option (d)

3 0

3 years ago

4. Which activity would be BEST for prolonged training within your target heart rate zone?

valina [46]

I am sorry I would answer this questions but I don’t have the options of the heart rates so I can give you the best answer

6 0

3 years ago

a ball rolls off the edge of a 2m high shelf at a speed of 5 m/s and hits the ground the taken to hit the ground is

RUDIKE [14]

Answer:

Explanation:

Use the one-dimensional equation

Δx = $v_0t+\frac{1}{2}at^2$ where delta x is the displacement of the object, v0 is the velocity of the object, a is the pull of gravity, and t is the time in seconds. That's our unknown.

Δx = -2 (negative because where it ends up is lower than the point at which it started),

$v_0=5$ , and

a = -9.8

Filling in:

$-2=5t+\frac{1}{2}(-9.8) t^2$ and simplified a bit:

$-2=5t-4.9 t^2$

this should look hauntingly familiar (a quadratic, which is parabolic motion...very important in physics!!). We begin by getting everything on one side of the equals sign and solving for t by factoring:

$-4.9 t^2+5t+2=0$ (the 0 is also indicative of the object landing on the ground! Isn't this a beautiful thing, how it all just works so perfectly together?)

When you factor this however your math/physics teacher has you factoring you will get that

t = 1.3 sec and t = -.31 sec

Since we all know that time can NEVER be negative, it takes the ball 1.3 sec to hit the ground from a height of 2 m if it is rolling off the shelf at 5 m/s.

3 0

2 years ago

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti

matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

$\phi = \frac{Q}{\epsilon_{o}}$ (Gauss's Law)

Putting the given values into the above formula as follows.

Q = $\phi \times \epsilon_{o}$

= $-953 Nm^{2}/C \times 8.854 \times 10^{-12}$

= $-8.4 \times 10^{-9} C$

= -8.4 nC

Therefore, when the unknown charge is q then,

-14.0 nC + 33.0 nC + q = -8.4 nC

q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0

3 years ago

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago