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bearhunter [10]
3 years ago
10

An object possessing an excess of 6.0x10^6 electrons has a net charge of

Physics
1 answer:
guajiro [1.7K]3 years ago
7 0
A single electron has a charge of
e=-1.6 \cdot 10^{19}C
Therefore, if we have an excess of N=6.0\cdot10^6 electrons, the total net charge will be the product between the charge of a single electron and the total number of electrons in excess:
Q=Ne=(6.0\cdot10^6)(-1.6 \cdot 10^{-19}C)=9.6 \cdot 10^{-13}C
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When does an object falling vertically through the air reach terminal velocity?
zlopas [31]

Answer:

a. when the acceleration of the objects become negative

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what do you mean by momentum? avehicle is runing with a velocity of 5m/s. it the momentum of the vehicle is 5000kg m/s. what is
Fantom [35]
Momentum is simply a quantity that measures the impact of a moving body over something is due to the mass it posseses or the velocity with which it is moving. Mathematically,ut is the product of mass and velocity of a body. It is represented by capital p...(P) P=mv Solution to the problem: we know P=mv For mass,eliminate m from the equation, m=P/v Put values, m=5000/5=1000kg
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Consider two wheels with fixed hubs. The hub cannot move, but the wheel can rotate about it. The hubs are fixed to a stationary
kykrilka [37]

Answer:

Magnitude of force on wheel B is 4 N

Explanation:

Given that

I=mr^2

For wheel A

m= 1 kg

d= 1 m,r= 0.5 m

F=1 N

We know that

T= F x r

T=1 x 0.5 N.m

T= 0.5 N.m

T= I α

Where I is the moment of inertia and α is the angular acceleration

I_A=1 \times 0.5^2\ kg.m^2

I_A=0.25\ kg.m^2

T= I α

0.5= 0.25 α

\alpha = 2\ rad/s^2

For Wheel B

m= 1 kg

d= 2 m,r=1 m

I_B=1 \times 1^2\ kg.m^2

I_B=1 \ kg.m^2

Given that angular acceleration is same for both the wheel

\alpha = 2\ rad/s^2

T= I α

T= 1 x 2

T= 2 N.m

Lets force on wheel is F then

T = F x r

2 = F x 1

So F= 2 N

Magnitude of force on wheel B is 2 N

3 0
3 years ago
A skier (down a straight course) is able to enter the starting gate with a speed of 1 m/sec and the average acceleration down th
Zigmanuir [339]
He was moving at 31 m/s
3 0
3 years ago
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

5 0
2 years ago
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