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3 years ago

13

“What is the relationship between wavelength and frequency?”

1 answer:

Kisachek [45]3 years ago

7 0

Answer:

They are inversely proportional. If the frequency goes up, the wavelength will go down

Explanation:

You might be interested in

How much work is accomplished when a force of 250 N pushes a box across the floor for a distance of 50 meters?

mote1985 [20]

We use the work formula to solve for the unknown in the problem. The formula for work is expressed as the product of the net force and the distance traveled by the object. We were given both the force and the distance so we can solve work directly.

Work = 250 N x 50 m = 12500 J

Thus, the answer is C.

7 0

3 years ago

400 N of load can be overcome by an effort of 50 N by using a lever. Calculate the mechanical advantage of the lever.

mars1129 [50]

Answer:

load (l)=400N

Effort(E)=50N

mechanical advantage (MA)= load ÷Effort

(ma)=400÷50

(ma)=8

Explanation:

I copy pasted from the answer from the same question. Remember to first check if ur question is there

4 0

3 years ago

How do you do this problem?

mojhsa [17]

Answer:

C

Explanation:

When A and B come in contact with each other, +12 - 12 = 0 so their changes cancel.

Now C has a charge of +12

When A and C come together they each have an equal share of that 12, so each of them has 6

So the answer is

A B C

6 0 6

which is C

4 0

3 years ago

A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through s

KengaRu [80]

Answer:

(A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

Explanation:

Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

For no deflection

$F_{E}=F_{B}$

$Eq=Bqv$

$v = \dfrac{E}{B}$

$v=\dfrac{60000}{0.0500}$

$v=1.2\times10^{6}\ m/s$

(B). We need to calculate the radius

Using magnetic force balance by centripetal force

$Bqv=\dfrac{mv^2}{r}$

$r=\dfrac{mv^2}{Bqv}$

Put the value into the formula

$r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}$

$r=2.0\times10^{6}\ m$

Hence, (A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

6 0

3 years ago

If you are looking for a material that is a good conductor, you will look for a material that will (2 points)

Nataly_w [17]

B)<span>Readily accept electron flow.</span>

7 0

3 years ago

Read 2 more answers