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Misha Larkins [42]
3 years ago
10

When the k. E of

Physics
2 answers:
Ierofanga [76]3 years ago
8 0

\sqrt{2}Answer:

KE2 = 2 KE1

1/2 M V2^2 = 2 * (1/2 M V1^2)

V2^2 = 2 V1^2

V2 = \sqrt{2} V1

Since momentum = M V  the momentum increases by \sqrt{2}

kolezko [41]3 years ago
6 0

Answer:

i need my points back -_-

Explanation:

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Assume four 1 kohm resistors are connected so that they form a square. what is the equivalent resistance if the resistance is me
borishaifa [10]
If the resistors are arranged in a shape of a square, then they are in a series type of circuit. This circuit arrangement is a non-branching, one-way flow of electrons. The total resistance in a series circuit is the summation of the individual resistances, If you place the ohmmeter (measures resistance) on two non-adjacent sides, then, you are measuring the resistance of two of the resistors.

Resistance = 2(1 kΩ) = 2 kΩ
8 0
3 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
kherson [118]

Explanation:

The Coulomb's law states that the magnitude of each of the electric forces between two point-at-rest charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

F=\frac{kq_1q_2}{d^2}

In this case we have an electron (-e) and a proton (e), so:

F=-\frac{ke^2}{d^2}\\F=-\frac{8.99*10^9\frac{N\cdot m^2}{s^2}(1.6*10^{-19}C)^2}{(933*10^{-9}m)^2}\\F=-2.64*10^{-16}N

In this case, the electric force is negative, therefore, the force is repulsive and its magnitude is:

F=2.64*10^{-16}N

3 0
3 years ago
4
tensa zangetsu [6.8K]

Answer:

C. amount of charge on the source charge.

Explanation:

Electric field lines can be defined as a graphical representation of the vector field or electric field.

Basically, it was first introduced by Michael Faraday and it is typically a curve drawn to the tangent of a point is in the direction of the net field acting on each point.

The number, or density, of field lines on a source charge indicate the amount of charge on the source charge. Therefore, the density of field lines on a source charge is directly proportional to quantity of charge on the source.

8 0
2 years ago
3m/s for 12 seconds how dar would he walk
Sunny_sXe [5.5K]
I think 36m/12s because 3×12 =36
4 0
3 years ago
7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle
jasenka [17]

Weight of the carriage =(m+M)g =142.1\ N

Normal force =Fsin(\theta) + W = 197.1\ N

Frictional force =\mu N=27.59\ N

Acceleration =4.66\ m\ s^{-2}

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

  • So Weight(W) =(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with F_x, force of 110\ N acting vertically downward.Both are downward and Normal is upward so Normal force =Summation\ of\ both\ forces

  • Normal force (N) = Fsin(\theta)+W=110sin(30) + 142.1 =197.1\ N
  • Frictional force (f) =\mu N=0.14\times 197.1 =27.59\ N

To calculate acceleration we will use Newtons second law.

That is Force is product of mass and acceleration.

We can see in the diagram that F_y=Horizontal and F_x=Vertical component of forces.

So Fnet = Fy(Horizontal) - f(friction) = m\times a

  • Acceleration (a) =\frac{Fcos(\theta)-\mu N}{mass(m)} =\frac{(95.26-27.59)}{14.5}= 4.66\ m\ s{^2 }

So we have the weight of the carriage, normal force,frictional force and acceleration.

3 0
3 years ago
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