Calculate the average speed

A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

Savatey [412]

The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

By substituting the given values, we get

$q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }$

6 0

3 years ago

What are compounds considered to be

bija089 [108]

Answer:A compound is a substance formed when two or more elements are chemically joined.

Explation:Water, salt, and sugar are examples of compounds.

I hope this helps<3

7 0

2 years ago

A temporary license permits the holder to drive for up to _ days while the application is reviewed

Ugo [173]

60 days, tell me if I'm correct please.

7 0

3 years ago

An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete

dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a) $=2 m/s^2$

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

$v=0+2\times 6=12 m/s$

After this object will start moving under gravity

height reached in first 6 s

$s=ut+\frac{at^2}{2}$

$s=0+\frac{2\times 6^2}{2}$

s=36 m

After fuel run out distance traveled in upward direction is

$v^2-u^2=2as_0$

here v=0

u=12 m/s

$a=9.8 m/s^2$

$0-12^2=2(-9.8)(s)$

$s_0=\frac{144}{2\times 9.8}=7.34 m$

$s+s_0=36+7.34=43.34 m$

7 0

3 years ago

Suppose that a dominant allele (P) codes for a polka-dot tail and a recessive allele (p) codes for a solid colored tail. In addi

ra1l [238]

When you put that into a Punnett square you first use the FOIL method because this is a dihybrid case where two traits are considered. This will be the resulting Punnett square:

PL Pl pL pl

PL PPLL PPLl PpLL PpLl

Pl PPLl PPll PpLl Ppll

pL PpLL PpLl ppLL ppLl

pl PpLl Ppll ppLl ppll

Now based on the Punnett square above, you can get the ratio. Remember that the dominant trait will always be expressed when you count phenotype combinations so as long as there is a capital letter in the combination, it will always be the one expressed.

So there are 9 polka dot, long eyelashes(PL), 3 polka dot, short eyelashes (Pl), 3 solid color, long eyelashes (pL), 1 solid short eyelashes(pl).

So your answer is D. 9:3:3:1 ratio

8 0

2 years ago