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forsale [732]
3 years ago
11

Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the rig

ht as positive and forces pointing to the left as negative. Rank positive forces as larger than negative forces.
1. q1=+1nC
q2=-1nC
q3 =-1nC
2. q1= -1nC
+ q2 = + 1nC
q3= +1nC
3. q1= +1nC
q2= +1nC
q3= +1nC
4. q1= +1nC
q2= + 1nC
q3= -1nC
5. q1= -1nC
q2= - 1nC
q3= -1nC
6. q1=+1nC
q2=-1nC
q3 =+1nC

Physics
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:

Plss see attached file

Explanation:

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An engineer weighs a sample of mercury (ρ = 13.6 × 103 kg/m3 ) and finds that the weight of the sample is 6.0 n. what is the sam
Amanda [17]
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.

Let V =  the volume of the sample.
Then
W = ρVg
or
V =  W/(ρg)
   = (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
   = 4.4972 x 10⁻⁵ m³

Answer: The volume is 44.972 x 10⁻⁶ m³
5 0
3 years ago
a Stone is dropped into a deep well and is heard to hit the water 3.41s after being dropped. determine the depth of the well
tangare [24]
Using SUVAT

U = 0 m/s
A = g = 9.8 ms^-2
T = 3.41 s
S = ?

s = ut + 1/2at^2
s = (0)(3.41) + 1/2(9.8)(3.41^)
s = 1/2(9.8)(3.41^2)
s = 1/2(9.8)(11.6281)
s = 56.97769 m
5 0
3 years ago
A circular conducting loop with a radius of 0.10 m and a small gap filled with a 10.0 ȍresistor is oriented in the xy-plane. If
lidiya [134]

Answer:

The magnitude of the current is 5.45 mA.

Explanation:

Given that,

Resistance = 10.0 ohm

Radius = 0.10 m

Magnetic field = 1.0 T

Angle = 30°

Increase magnetic field = 7.0 T

Time t = 3.0 s

Number of turns = 1

We need to calculate the initial flux

Using formula of flux

\phi=NB_{1}A\cos\theta

Put the value into the formula

\phi=1\times1.0\times\pi\times(0.10)^2\times\cos30^{\circ}

\phi=1\times1.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}

\phi=0.027\ wb

We need to calculate the final flux

\phi=1\times7.0\times\pi\times(0.10)^2\times\dfrac{\sqrt{3}}{2}

\phi=0.1904\ wb

We need to calculate the induced emf

Using formula of emf

\epsilon=\dfrac{\phi_{f}-\phi_{i}}{t}

Put the value into the formula

\epsilon=\dfrac{0.1904-0.027}{3.0}

\epsilon=0.0545\ V

We need to calculate the current

Using formula of current

I=\dfrac{\epsilon}{R}

Put the value into the formula

I=\dfrac{0.0545}{10.0}

I=5.45\ mA

Hence, The magnitude of the current is 5.45 mA.

4 0
3 years ago
If an object has a kinetic energy of 30 j and mass of 34kg how fast is the object moving ?
Murrr4er [49]

Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.

Ek = mv^2 / 2 — multiply both sides by 2

2Ek = mv^2 — divide both sides by m

2Ek / m = V^2 — switch sides

V^2 = 2Ek / m — plug in values

V^2 = 2*30J / 34kg

V^2 = 60J/34kg

V^2 = 1.76 m/s — sqrt of both sides

V = sqrt(1.76)

V = 1.32m/s (roughly)

5 0
3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
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