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Drupady [299]
3 years ago
10

A 9.5 V battery supplies a 3.5 mA current to a circuit for 7.0 h. Part A How much charge has been transferred from the negative

to the positive terminal
Physics
1 answer:
pychu [463]3 years ago
7 0

Answer:

88.2 C

Explanation:

The current can be defined as the rate of flow of charge in a conductor.

The relation between charge current and time is given as

I = Q/T

I = current, Q= charge and T = time

that is ampere = coulomb / second

The amount of charge passed is from the negative to the positive terminal

shall be given by:

Q = I * t = 3.5mA * 7h * 3600s/h = 88.2 C

Note: take care of the units.  

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PART A:
olga55 [171]

Answer:

4 m/s^{2}

Explanation:

From Equilibrium of forces, The Tension in string is cancelled by the Weight (product of mass and acceleration due to gravity) of the body acting downwards.

The Net force = Mass * Acceleration.

Since Net Force = 20 Newton, Mass = 5kg, therefore;

20 = 5kg * acceleration. Dividing the RHS and LHS of the equation by 5, we have;

Acceleration = \frac{20}{5} which gives 4.

Note: RHS means Right Hand Side.

         LHS means Left Hand Side.

 

7 0
2 years ago
car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of a. The car
Luba_88 [7]

Answer:

0.572

Explanation:

First examine the force of friction at the slipping point where Ff = µsFN = µsmg.

the mass of the car is unknown,

The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.

First the tangential direction

∑Ft =Fft =mat

And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r

Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2

So going backwards and plugging in Ffc =m2atπr/ 2r =πmat

Ff = √(F2ft +F2fc)= matp √(1+π²)

µs = Ff /mg = at /g √(1+π²)=

1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572

7 0
3 years ago
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

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3 years ago
The dimension line is connected to the part being measured by _______________. A. Hidden lines B. Extension lines C. Visible lin
kondaur [170]
B: Extension Lines! You could have just searched this up on google
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Well it transports white blood cells to infected areas.

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