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jolli1 [7]
3 years ago
10

Two point charges are placed on the x-axis as follows: charge q1 = 3.95 nC is located at x= 0.198 m , and charge q2 = 4.96 nC is

at x= -0.297 m. What are the magnitude and direction of the total force exerted by these two charges on a negative point charge q3=6.00nCq that is placed at the origin?
Physics
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

F = 2.40 × 10^{-6}  N

Explanation:

given data

charge q1 = 3.95 nC

x= 0.198 m

charge q2 = 4.96 nC

x= -0.297 m

solution

force on a point charge kept in electric field F = E × q       ................1

here E is the magnitude of electric field and q is the magnitude of charge

and

first we will get here electric field at origin

So net field at origin is

E = (Kq2÷r2²) - (kq1÷r1²)           ...............2

put here value

E = 9[(4.96÷0.297²)-(3.95÷0.198²)]

E = 400.72 N/C        ( negative x direction )

so that force will be

F = 6 × 10^{-9} × 400.72

F = 2.40 × 10^{-6}  N

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Answer:

98,000 pa

Explanation:

The formula for water pressure is as follows:

pressure = pgh

Where <em>p </em>is the density of water (in kg/m3), <em>g </em>is the gravitational field strength, and <em>h </em>is the height of the water.

The density of water is 1000kg/m3, the gravitational field strength is 9.8, and the height is 10. Substituting in these values:

pressure = 1000 \times 9.8 \times 10

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7 0
2 years ago
A school bus has a mass (including the driver and passengers) of 1.64 times 10^4 kg and is driving north at a speed of 15.2 km/h
Butoxors [25]

Answer:

Explanation:

Given

mass of bus along with travelers travelling in North direction is m_1=1.6\times 10^4 kg

speed of bus towards North v_1=15.2 km/h\approx 4.22\ m/s

mass of bus travelling in South direction is m_2=1.578\times 10^4 kg

speed of bus v_2=12.2 km/h\approx 3.38\ m/s

mass of each Passenger in south moving bus m_0=64.8 kg

Momentum of North moving bus

P_1=m_1\times v_1

P_1=1.6\times 10^4\times 4.22

P_1=6.768\times 10^4 kg-m/s

Momentum with south moving bus

P_2=m_2\times v_2+n\cdot m_0\times v_2

P_2=(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38

For total momentum to be towards south

P_2-P_1 should be greater than 0

thus for least value of n

P_2=P_1

(1.578\times 10^4+n\cdot 64.8 )\cdot 3.38=6.768\times 10^4

1.578\times 10^4+n\cdot 64.8=2.0023\times 10^4

n=\frac{4243.6686}{64.8}=65.48\approx 66    

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3 years ago
The patellar tendon attaches to the tibia at a 20 deg angle 3 cm from the axis of rotation at the knee. If the force generated i
gregori [183]

Answer:

the resulting angular acceleration is 15.65 rad/s²

Explanation:

Given the data in the question;

force generated in the patellar tendon F = 400 N

patellar tendon attaches to the tibia at a 20° angle 3 cm( 0.03 m ) from the axis of rotation at the knee.

so Torque produced by the knee will be;

T = F × d⊥

T = 400 N × 0.03 m × sin( 20° )

T = 400 N × 0.03 m × 0.342

T = 4.104 N.m

Now, we determine the moment of inertia of the knee

I = mk²

given that; the lower leg and foot have a combined mass of 4.2kg and a given radius of gyration of 25 cm ( 0.25 m )

we substitute

I = 4.2 kg × ( 0.25 m )²

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So from the relation of Moment of inertia, Torque and angular acceleration;

T = I∝

we make angular acceleration ∝, subject of the formula

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we substitute

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Goshia [24]

Answer:

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Explanation:

8 0
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olchik [2.2K]

Answer:

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Explanation:

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For a capacitor, the reactance is given by:

X=\frac{1}{2 \pi f C}

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The reactance has a similar meaning to that of the resistance for a DC current. In fact, we notice that:

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4 0
4 years ago
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