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2 years ago

Why is the picture above not an accurate representation of charges and their field lines?

Physics

2 answers:

pochemuha2 years ago

7 0

Because field lines don’t all go in to the pole. The bottom half go outwards not inwards.

Liula [17]2 years ago

3 0

Answer:

C, There should be more field lines around +2Q than the -Q

Explanation:

just did the test its correct x

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The initial temperature of a bomb calorimeter is 28.50°

solniwko [45]

The answer is 5,170 J

8 0

3 years ago

Read 2 more answers

A football player who weighs 550 N stands indoors wearing her football boots. The boot’s

EleoNora [17]

Answer:

183333 Pa

Explanation:

The weight of the football player is : 550 N ,thus the force the player exerts on the floor is 550 N

The area of blades in contact with the floor is = 30cm² = 0.003 m²

Pressure = Force / Area

Pressure = 550 / 0.003

Pressure = 183333 Pa

3 0

2 years ago

If a car go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 mi/

Taya2010 [7]

Answer:

v = 87.57 m/s

Explanation:

Given,

The initial velocity of the car, u = 0

The final velocity of the car, v = 60 mi/hr

The time period of car, t = 8 s

= 0.00222 hr

The acceleration of the car is given by the formula,

a = (v -u) / t

= 60 / 0.00222

= 27027 mi/hr²

If the car has initial velocity, u = 50 mi/hr

The time period of the car, t = 5.0 s

= 0.00139 hr

Using first equations of motion

v = u + at

= 50 + (0.00139 x 27027)

= 87.57 mi/hr

Hence, the final velocity of the car, v = 87.57 mi/hr

3 0

3 years ago

Two trains A and B of length 400 m each are moving on two parallel tracks with

Vinvika [58]

Answer:

The initial distance between the trains is 1450 m. 

Explanation:

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration is $1m/s^2$ and it accelerated for 50 seconds.

 $a=1 m/s^2$ 

t=50 s 

initial speed u=72km/h 

we have to convert this speed into m/s 

 $u=72 \times 5/18=20 m/s$ 

Distance covered in accelerating phase $S=ut+1/2 at^2$ 

 $=20 \times 50+1/2 \times 1 \times 50^2$ 

 $=1000+1250=2250 m$ 

If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

Here length of train A+length of train $B=400+400=800 m$ 