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3 years ago

A person supporting Eugene Talmadge for governor would have been MOST likely to belong to which occupation?

Physics

2 answers:

lisov135 [29]3 years ago

7 0

Answer:

B) Farmer

Explanation:

tee hee i just got this question :)))

tatyana61 [14]3 years ago

5 0

Answer:

B) Farmer

Explanation:

USA Test Prep tested. Is correct

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Explain three characteristic of a uniform electric field

Trava [24]

Answer:

• They depend solely on the load that generates it

• Two or more electrical charges interact, which can be positive or negative

• The energy source is based on the electrical voltage

3 0

3 years ago

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

vesna_86 [32]

Answer:

$6.63\times 10^8\ N/C$

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

7 0

3 years ago

The force of attraction between two oppositely charged pith is 5mx 10 to the -6th power newtons. If the charge on the two is 6.7

My name is Ann [436]

Answer:

0.28 m

Explanation:

The following data were obtained from the question:

Force (F) = 5×10¯⁶ N

Charge 1 (q₁) = 6.7×10¯⁹ C

Charge 2 (q₂) = 6.7×10¯⁹ C

Electrical constant (K) = 9×10⁹ Nm²C¯²

Distance apart (r) =?

Thus, the distance between the two charges can be obtained as follow:

F = Kq₁q₂/r²

5×10¯⁶ = 9×10⁹ × 6.7×10¯⁹ × 6.7×10¯⁹/r²

5×10¯⁶ = 4.0401×10¯⁷ / r²

Cross multiply

5×10¯⁶ × r² = 4.0401×10¯⁷

Divide both side by 5×10¯⁶

r² = 4.0401×10¯⁷ / 5×10¯⁶

Take the square root of both side

r = √(4.0401×10¯⁷ / 5×10¯⁶)

r = 0.28 m

Therefore, the distance between the two charges is 0.28 m

4 0

3 years ago

What are five atomic models that have been proposed over time ?

Art [367]

Air, water , atomic acid , and pocket rocket

5 0

3 years ago

A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m

Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force $F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k$

Mass of object = 2.00 kg

Initial position $d=(2.75)i-(2.00)j+(5.00)k$

Final position $d=-(5.00)i+(4.50)j+(7.00)k$

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

$W=F\cdot d$

$W=F\cdot(d_{f}-d_{i})$

Put the value into the formula

$W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)$

$W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)$

$W=2.75\times-7.75+7.50\times6.50+6.75\times2.00$

$W=40.93\ J$

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0

3 years ago