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Art [367]
3 years ago
12

Only negative charges move True False

Physics
2 answers:
Flura [38]3 years ago
7 0
The answer is False, pls. mark me the brainliest if I’m right. THX
stich3 [128]3 years ago
6 0
False

positive charges move also..which makes this statement false
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A 3.00 × 10^−9-coulomb test charge is placed near
viktelen [127]

Answer:

B

Explanation:

Given:-

- The charge of the test particle q = 3.0 * 10^-9 C

- The force exerted by the metal sphere F = 6.0 * 10^-5 N

Find:-

The magnitude and direction of the electric field strength at this location?

Solution:-

- The relationship between the electrostatic force F exerted by the metal sphere on the test-charge and the Electric Field strength E at the position of test charge is given by:  

                                       F = E*q

- Using the data given we can determine E:

                                       E = F / q

                                       E = (6.0 * 10^-5) / (3.0 * 10^-9)

                                       E = 20,000 N/C

- The direction of electric field is given by the net charge of the source ( metal sphere). The metal sphere is negative charge hence the direction of Electric Field strength E is directed towards the metal sphere.

8 0
3 years ago
Read 2 more answers
Two boxes sit side by side on a smooth horizontal surface. The lighter box 5.2 kg, the heavier box has a mass of 7.4kg (a) find
Paladinen [302]

OK.  So you're pushing on the small box, and on the other side of it, the small
box is pushing on the big box. So you're actually pushing both of them.

-- The total mass that you're pushing is (5.2 + 7.4) = 12.6 kg.

-- You're pushing it with 5.0N of force.

-- Acceleration of the whole thing = (force)/(mass) = 5/12.6 = <em>0.397 m/s²</em> (rounded)

-- Both boxes accelerate at the same rate. So the box farther away from you ...
the big one, with 7.4 kg of mass, accelerates at the same rate.

The force on it to make it accelerate is (mass) x (acceleration) =

                                                              (7.4 kg) x (5/12.6 m/s²) =  <em>2.936 N.</em>

The only force on the big box comes from the small box, pushing it from behind. 
So that same  <em>2.936N</em>  must be the contact force between the boxes.

8 0
3 years ago
A person in a kayak starts paddling, and it accelerates from 0 to 0.680 m/s in a distance of 0.428 m. If the combined mass of th
d1i1m1o1n [39]

Answer:

Net force, F = 44.66 N

Explanation:

It is given by,

Initial velocity of the person, u = 0

Final velocity of the person, v = 0.68 m/s

Distance, s = 0.428 m

Combined mass of the person and the kayak, m = 82.7 kg

We need to find the net force acting on the kayak i.e.

F = ma...........(1)

Firstly, we will calculate the value of "a" from third equation of motion as :

v^2-u^2=2as

a=\dfrac{v^2-u^2}{2s}

a=\dfrac{(0.68\ m/s)^2-0}{2\times 0.428\ m}

a=0.54\ m/s^2

Put the value of a in equation (1) as :

F=82.7\ kg\times 0.54\ m/s^2

F = 44.66 N

So, the net force acting on the kayak is 44.66 N. Hence, this is the required solution.

6 0
3 years ago
For how long a time must a tow truck pull with a force of 550N on a stalled 1200kg car to give it a forward velocity of 2.0m/s?
horsena [70]

Answer:

4.36 seconds

Explanation:

According to the question;

  • Force is 550 N
  • Mass of the car is 1200 kg
  • Velocity of the car is 2.0 m/s

We are needed to find the time the car must the tow track pull the car.

  • From Newton's second law of motion;
  • Impulsive force, F = Mv÷t , where m is the mass, v is the velocity and t is the time.

Rearranging the formula;

t = mv ÷ F

Thus;

Time = (1200 kg × 2.0 m/s²) ÷ 550 N

        = 4.36 seconds

Thus, the time needed to pull the car is 4.36 seconds

5 0
3 years ago
Can someone please help me come up with a hypothesis?. The question that I have is "How many sugars are in your smoothies?"
patriot [66]
'I believe that there is 20 grams of sugar in the smoothies.'

I don't know, just an idea.
8 0
3 years ago
Read 2 more answers
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