Why is your State have dimensions

Mkey [24]

I believe the answer is B. let me know If I was right.

3 0

3 years ago

6. When a positive charge is released and moves along an electric field line, it moves to a position of A) lower potential and l

Marysya12 [62]

Answer:

Since you would have to do work on the charge to bring it back to its original position, the charge moves to a position of lower potential and lower potential energy.

4 0

3 years ago

A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from t

VladimirAG [237]

Answer:

271.095 m

Explanation:

✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)

✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)

✓ Let the distance travelled by the sound = s

✓duration of Time that exist between heard of sounds = 0.70s

But we know that

Time = (Distance / Speed)

✓Time it takes the sound to travel through air= s/vi = s/343

✓Time it takes the sound to travel through concrete= s/vc = s/3000

✓ (s/343) - (s/3000) = 0.70

Finding LCM and simplify

[(3000s - 343s)]/1029000 = 0.70

2657s /1029000 = 0.70

Making " s" subject of the formula

s= (1029000 × 0.70)/2657

s=720300/ 2657

s= 271.095 m

Hence, The impact took place at a distance of 271.095 m away from the person.

4 0

3 years ago

"If E = 7.50V and r=0.45Ω, find the minimum value of the voltmeter resistance RV for which the voltmeter reading is within 1.0%

Virty [35]

Answer:

The minimum value is $R_V =44.552\ \Omega$

Explanation:

From the question we are given that

The voltage is $E = 7.5V$

The internal resistance is $r = 0.45$

The objective of this solution is to obtain the minimum value of the voltmeter resistance for which the voltmeter reading is within 1.0% of the emf of the battery

What is means is that we need to obtain voltmeter resistance such that

V = (100% -1%) of E

Where E is the e.m.f of the battery and V is the voltmeter reading

i.e V = 99% of E = 0.99 E = 7.425

Generally

E = V + ir

where ir is the internal potential difference of the voltmeter and

V is the voltmeter reading

Making i the subject of the formula above

$i = \frac{(E-V)}{r}$