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3 years ago

How do electric field lines indicate the strength of the field?

Physics

2 answers:

Paha777 [63]3 years ago

6 0

I am almost sure it it (c)

KiRa [710]3 years ago

3 0

<h2>Answer:</h2>

The correct answer is the option B which is by how close the lines are.

<h3>Explanation:</h3>

Electric field of lines are a type of pattern used to describe the direction and the strength of the electric field.
These are lines which always projects from Positive ions to negative ions.
Then strength of electric field is shown by the closeness of lines.
Where these lines are close to each other it means strength of electric field is high and vice versa.
These lines show that electric field is a vector property.

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A 25 newton force applied on an object moves it 50 meters. The angle between the force and displacement is 40.0 degrees. What is

zvonat [6]

Work done is when a force is exerted to cause a displacement in a certain object.
the equation for work done ;
work done = force applied * displacement of the object
when the force applied is not in the same direction as that of the displacement of the object then the effect of the force is not its whole value. The force is then applied at an angle to that of the displacement of the object, then the resultant force is the force exerted* cos of the angle between force and displacement, in this instance the angle is 40 °.
the new equation is then;
work done = force cos 40° * displacement
after substitution,
work = 25 N * 0.76 * 50 m
= 957.55 J
round it off
= 9.6 *10² J
the correct answer is B

5 0

3 years ago

Read 2 more answers

WHAT SHOULD I NAME MY DEAD RAT?

olga_2 [115]

Answer:

pablito

Explanation:

6 0

2 years ago

Define the law of universal gravitation in your own words.

Firdavs [7]

Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

7 0

3 years ago

Capacitors, C1 = 1.0 F and C2 = 1.0 F, are connected in parallel to a 6.0 volt battery (ΔV = 6.0V). If the battery is disconnect

Iteru [2.4K]

Answer:

66.2 sec

Explanation:

C₁ = 1.0 F

C₂ = 1.0 F

ΔV = Potential difference across the capacitor = 6.0 V

C = parallel combination of capacitors

Parallel combination of capacitors is given as

C = C₁ + C₂

C = 1.0 + 1.0

C = 2.0 F

R = resistance = 33 Ω

Time constant is given as

T = RC

T = 33 x 2

T = 66 sec

V₀ = initial potential difference across the combination = 6.0 Volts

V = final potential difference = 2.2 volts

Using the equation

$V = V_{o} e^{\frac{-t}{T}}$

$2.2 = 6 e^{\frac{-t}{66}}$

t = 66.2 sec

5 0

3 years ago

A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The

almond37 [142]

Answer:

3688 km/h

Explanation:

Given:-

- The speed of vehicle relative to earth, vs_e = 3760 km/h

- The relative speed of command and motor, v_c/m = 90 km/h

- The mass of command = m

- The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

- Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

- We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e = v_c/m + vc_e

- Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

- Solve for vc_e:

5m*vs_e - 4m*(v_c/m) = 5m*vc_e

vs_e - 0.8*(v_c/m) = vc_e

- Plug in values and evaluate vc_e:

vc_e = 3760 - 0.8*(90)

vc_e = 3,688 km/h

5 0

3 years ago

Read 2 more answers