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Mashutka [201]
4 years ago
11

Please help I’ll name you brainliest and give five stars please

Physics
1 answer:
OLga [1]4 years ago
7 0

Answer:

i think its 10 or 11 not sure though if its wrong im sorry im trying to help

Explanation:

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Select the options that best complete the statement.Positively charged particle trajectories(always, never, the same as)follow e
SpyIntel [72]

Answer:

a) always. b) electric field lines are defined by the path positive test charges travel.

Explanation:

By convention, field lines always follow the direction that it would take a positive test charge  (small enough so it can´t disrupt the field created by a charge distribution), under the influence of an electric field, at the same point where the test charge is located.

So any positive charge, subject to an electric field influence, moves along the field line that passes through its current position, in the same way that a positive test charge would.

We could say also that the electric force on a positively charged particle is in the same direction as the electric field that produces that force (due to some charge distribution) , which is true, but it doesn´t explain why.

3 0
4 years ago
A car traveling 75 km/h slows down at a constant 0.50 m/s2 just by "letting up on the gas." calculate (a) the distance the car c
Serjik [45]

Solution:

At 1st convert km/h to m/s. 1 km = 1000 m, 1 h = 3600 s, 1 km/m = 1000/3600 = 5/18 m/s  

Initial velocity = 75 * 5/18 = 20.8 m/s  

The car’s velocity decreases from 20.8 m/s to 0 m/s at the rate of 0.5 m/s each second. We have the final velocity, initial velocity, and the acceleration.  

Now according to the equation determine the distance.  

vf^2 = vi^2 + 2 * a * d  

a = -0.5 m/s^2  

0 = 20.8^2 + 2 * -0.5 * d  

so d = 431.64 m  

since we have the final velocity, initial velocity, and the acceleration. Use the following equation to determine time.  

vf = vi + a * t  

0 = 20.8 – 0.5 * t  

Solve for t = 41 seconds  

(c) the distance travels by it during the first and fifth second are.  

d = vi * t + ½ * a * t^2  

d1 = 20.8 * 1 – ½ * 0.5 * 1^2 = 20.55 m  

The easiest way to the distance for the 5th second is:

d = vi * t + ½ * a * t^2, a = -0.5  

d5 = 20.8 * 5 – ½ * 0.5 * 5^2 = 91.5 m  

d6 = 20.8 * 6 – ½ * 0.5 * 6^2 =  106.8m

d6 – d5 = 15.3 m  

this is the required solution.


3 0
3 years ago
an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0
3 years ago
The batteries in the figure above are connected in​
Elena-2011 [213]
The answer is a, series circuit.
5 0
4 years ago
Read 2 more answers
In the livingroom of your house you would like to run a 300 W stereo, a 30 W computer, a 960 W laser light projector, and two 12
choli [55]

Answer:

12.75 A

Explanation:

The total power is 300 + 30 + 960 + 2 × 120 = 1530 W

Electrical power is given by

P = IV

Then

I = \dfrac{P}{V}

With a voltage of 120 V,

I = \dfrac{1530}{120} = 12.75 \text{ A}

8 0
4 years ago
Read 2 more answers
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