What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3

Answer:

The bulk modulus of the liquid is 1.229 x 10¹⁰ Pa

Explanation:

Given;

density of liquid, ρ = 1400 kg/m³

frequency of the wave, f = 390 Hz

wavelength, λ = 7.60 m

The speed of the sound is given by;

v = fλ

v = 390 x 7.6

v = 2964 m/s

The bulk modulus of the liquid is given by;

[tex]v = \sqrt{\frac{B}{\rho}}\\\\v^2 = \frac{B}{\rho}\\\\B = \rho v^2[/tex]

where;

B is bulk modulus

B = (1400)(2964)²

B = 1.229 x 10¹⁰ N/m²

B = 1.229 x 10¹⁰ Pa

Therefore, the bulk modulus of the liquid is 1.229 x 10¹⁰ Pa

Answer:

So the answer is yes, we can the back be shaped like a spinning rod

spinal column that is approximated by a long and narrow rod,

Explanation:

The bone system of the body is very well modeled in physics, the back has a spinal column that is approximated by a long and narrow rod, this rod is fixed in the lower part to the coccyx and has a weight in the upper part (head), this rod has longitudinal vertical movement and twisting movement around the lower part of the bar.

So the answer is yes, we can the back be shaped like a spinning rod

1.085m

Explanation:

Using

a= lambda/sinစ

Sinစ= (587.5*10^-9) x 0.75*10^-3

= 0.000783

Sinစ=0.875*10^-3/d

0.000783= 0.875/d

d= 1.085m

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) ×  (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

Answer:

A. We know that amplitude at x is

Asin (kx)

But k= 2πf/v

k= 2*3.132*235/193= 7.65

So A = 0.35*sin( 7.65x 0.18)= 0.00841m

C

Vmax = Amplitude x angular velocity

= 0.0084 x 2πf

= 0.0084* 2*3.142* 235= 12.4m/s

D. Maximum acceleration = omega² x Amplitude

= (2πf)²* 0.00841= 183.40m/s²

Answer:

They oscillates perpendicularly to one another, the oscillation of one field generates the other field.

Explanation:

In a light wave, an oscillating electric field of a light wave produces a magnetic field, and the magnetic field also oscillates to produce an electric field. The magnetic field and the electric field of a light wave both oscillates perpendicularly to one another. The resultant energy and direction of the wave generated as a result of these oscillating fields is propagated perpendicularly to both fields.

Answer:

a set up where current flows without a voltage difference

Explanation:

because a circuit is a set up of different components, and throughout the circuit the voltage is the same, even with more components

Answer:

a set-up where current flows due to a voltage difference

a closed path of conductors

Explanation:

Answer:

3.948m/s

Explanation:

To solve this we need to apply Doppler effect theory

So

To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by

f1 = f0 x (V - Vo)/(V - Vs)

f0 = 30.0 kHz

V = 344 m/s

Vs will now be the speed of the bat and

Vo will be the speed of the object which is = 0 m/s

So substituting we have

f1 = 30 x 10^3 x (344- 0)/(344- Vs)

Next to find the frequency reflected by wall we use

f2 = f1 x (V + Vs)/(V + Vo)

So substituting the value of f1 calculated above we have

f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]

f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)

But the beat frequency detected by bat is 700 Hz,

So we say

f2 - f0 = 700 Hz

30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700

(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023

344 + Vs = 344 x 1.023 - Vs x 1.0233

Vs = 344 x ( 1.023 - 1)/(1 + 1.023)

So finally

Vs = Speed of source that is the bat is = 3.949m/s

Answer:

(A.) (- 4.33, 6.33 , 0); (B.) (- 3.66, 7.5, 0); (C.) average at (A) (- 4.33, 6.33 , 0) ; (D.) (- 0.2165, 0.3165, 0)

Explanation:

Given the following :

Time - - - - - - - 6.6s - - - - - - - - - 6.9s - - - - - 7.2s

Position - (1.8,5.0,0) - (0.5,6.9,0) - - (−0.4,9.5,0)

(a) Between 6.6 s and 6.9 s, what was the bee's average velocity?

Vavg = Distance / time

[(0.5,6.9,0) - (1.8,5.0,0)] / 6.9 - 6.6

Vavg = [(0.5 - 1.8), (6.9 - 5.0), (0 - 0)] / 0.3

Vavg = - 1.3 / 0.3, 1.9/0.3, 0/3

Vavg = (- 4.33, 6.33 , 0)

b) Between 6.6 s and 7.2 s, what was the bee's average velocity?

Vavg = [(−0.4,9.5,0) - (1.8,5.0,0)] / 7.2 - 6.6

Vavg = - 2. 2/0.6, 4.5/0.6, 0/0.6

Vavg = (- 3.66, 7.5, 0)

c.) Of the two averages (- 4.3, 6.3 , 0) is closer to the instantaneous Velocity at 6.6s

D.) (d) Using the best information available, what was the displacement of the bee during the time interval from 6.6 s to 6.65 s?

Displacement = Velocity * time

Vavg between 6.6 to 6.9 ; time = (6.65 - 6.6) = 0.05 s

= (- 4.33, 6.33 , 0) * 0.05

= (- 0.2165, 0.3165, 0)

Answer:

First to find deceleration of the train we use

v²= u²+ 2as

56²= 73²+ 2(0.5)a

a= -2193mi/hr²

Then we find time in which the train does the intersection

Using

S= ut+ 1/2 at²

1= 73t-1/2(1293)t²

t =68.5s

But since the train is to intersect in 3.9s the time will be the difference which is

65.68s

So finding acceleration

S= ut + 1/2at²

1.3mi= 45/3600mi/s(65.58s)+ 1/2a(65.5)²

So a= 1.179ft/s²

To find velocity we use

V= u + at

= 45/3600mi/s + (-2.33E-4mis²)(65.58s)

V= 0.0271mi/s

= 97.6ft/s

Answer:

Shortest distance = 0.1914 m

Explanation:

Given that,

Amplitude of the wave is 0.0957 m

Frequency of the wave is 3.75 Hz

Wavelength of the wave is 1.97 m

We need to find the shortest transverse distance between a maximum and a minimum of the wave.

The distance between maximum point in positive axis and the baseline is equal to amplitude.

Shortest distance = 2 A

D = 2 × 0.0957

D = 0.1914 m

So, the shortest transverse distance between a maximum and a minimum of the wave is 0.1914 m.

Answer:

resultant force = 14 N  ( East direction)

Explanation:

A =   √( (4*7)² + (4*7)² )

A = 39.6 N

B = 4 * 7

B = 28 N

C = 2 * 7

C = 14 N

∑ y forces = Ay - B = (4*7) - 28 = 0

∑ x forces = Ax - C = (4*7) - 14 = 14 N

so the resultant force = 14 N  ( East direction)

Thus the response above is correct.

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Explanation:

Given:

v₀ = 30 m/s

v = 0 m/s

a = -7.0 m/s²

Find: t and Δx

v = at + v₀

0 m/s = (-7.0 m/s²) t + 30 m/s

t = 4.3 seconds

v² = v₀² + 2aΔx

(0 m/s)² = (30 m/s)² + 2 (-7.0 m/s²) Δx

Δx = 64 meters

Answer:

The Gravitational force.

Answer:

v=10m/s

Explanation:

[tex]a=\frac{vf-vi}{t}\\ 5m/s^{2}=\frac{vf-0}{2}\\ vf=5*2\\vf=10 m/s[/tex]

Answer: -5 m/s^2

Explanation: a = v - u/t

                         = 32 - 47/3

                         = -15/3

                         = -5 m/s^2

Answer:

The peak value of the electric field is 489.64 V/m

Explanation:

Given;

power of the laser, P = 1.0 mW = 1 x 10⁻³ W

Radius of the beam, R = 1.0 mm = 1 x 10⁻³ m

Area of the beam = πr² = π(1 x 10⁻³ )² = 3.142 x 10⁻⁶ m²

The average intensity of the light = P / A

The average intensity of the light = ( 1 x 10⁻³) / (3.142 x 10⁻⁶)

The average intensity of the light = 318.27 W/m²

The peak value of the electric field is given by;

[tex]E_o = \sqrt{\frac{2I_{avg}}{c\epsilon_o}}\\\\E_o = \sqrt{\frac{2(318.27)}{(3*10^8)(8.85*10^{-12})}}\\\\E_o = 489.64 \ V/m[/tex]

Therefore, the peak value of the electric field is 489.64 V/m.

Answer:

Hence, there will be a collision

Explanation:

First we calculate total distance covered by the speedy sue's car before coming to rest:

2as = Vf² - Vi²

where,

a = deceleration = - 1.9 m/s²

s = distance covered = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 35 m/s

Therefore,

2(-1.9 m/s²)s = (0 m/s)² - (35 m/s)²

s = 322.37 m

Now, we calculate time taken by car to stop:

Vf = Vi + at

0 m/s = 35 m/s + (-1.9 m/s²)t

t =  18.42 s

Now, we calculate distance traveled by van in this time:

s₁ = V₁t

where,

s₁ = distance traveled by van = ?

V₁ = speed of van = 5.2 m/s

Therefore,

s₁ = (5.2 m/s)(18.42 s)

s₁ = 95.78 m

Now, for collision to occur, the following relation must be satisfied:

s ≥ 160 m + s₁

using values:

322.37 m > 160 m + 95.78 m

322.37 m > 255.78 m

Hence, there will be a collision

Answer:

The average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex] is 32.5 m/s

Explanation:

v = u +at bhjklj kj h

x = (u + v / 2 )t

Average velocity is given by

[tex]V_{avg} = \frac{Displacement}{Time} \\V_{avg} = \frac{x_{2} - x_{1} }{t_{2} - t_{1} }[/tex]

From the question,

Initial speed, u = 28 m/s

Final speed, v = 37 m/s

Time, t = 10 secs

From the formula

[tex]x = (\frac{u + v}{2})t\\[/tex]

where [tex]x[/tex] is the displacement

Put the given values into the equation to find the displacement [tex]x[/tex]

[tex]x = (\frac{28 + 37}{2}) 10\\ x = (\frac{65}{2})10\\ x = (\frac{32.5}{10})\\ x = 325 m[/tex]

Now, for the average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex]

[tex]V_{avg} = \frac{Displacement}{Time} \\V_{avg} = \frac{x_{2} - x_{1} }{t_{2} - t_{1} }[/tex]

[tex]V_{avg,x} = \frac{325 - 0}{10 - 0}\\V_{avg,x} = \frac{325}{10}\\ V_{avg,x} = 32.5 m/s[/tex]

Hence, the average [tex]x[/tex] component velocity [tex]V_{avg,x}[/tex] is 32.5 m/s

The average x component of velocity during the maneuver of the truck is 32.5 m/s.

Velocity:

The term velocity of any object is defined as the ratio of displacement covered by any object to the time taken by the object to displace.

Given data:

The magnitude of initial speed is, u = 28 m/s.

The magnitude of final speed is, v = 37 m/s.

The time interval is, t = 10 s.

The displacement covered by the truck is calculated as,

[tex]d = \dfrac{v+u}{2} \times t\\\\ d = \dfrac{37+28}{2} \times 10\\\\ d = 325 \;\rm m[/tex]

Now, the expression for the x-component of the average velocity is given as,

[tex]v_{x}=\dfrac{d}{t}[/tex]

Solving as,

[tex]v_{x}=\dfrac{325}{10}\\\\ v_{x}=32.5 \;\rm m/s[/tex]

Thus, we can conclude that the average x component of velocity during the maneuver of truck is 32.5 m/s.

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Answer:

Acceleration, [tex]a=45\ m/s^2[/tex]

Explanation:

The velocity of a particle is given by :

[tex]V=5t^2[/tex]

t is time in seconds

The acceleration in terms of velocity is given by :

[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(5t^2)}{dt}\\\\a=5\times \dfrac{t^3}{3}[/tex]

We need to find the acceleration of the particle at t = 3 s. Put t = 3 s in the expression of a.

So,

[tex]a=\dfrac{5\times 3^3}{3}\\\\a=45\ m/s^2[/tex]

So, the acceleration of the particle at t = 3 s is [tex]45\ m/s^2[/tex].

Explanation:

Density = mass / volume

ρ = 5.7 g / 0.3 mL

ρ = 19 g/mL

Yes, it's pure gold.

Yes, it's pure gold.

According to the question:

Density = mass / volume.

ρ = 5.7 g / 0.3 mL.

ρ = 19 g/mL.

Hence, Yes it's pure gold.

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#SPJ2

Answer:

C. Newton's 3rd law

Explanation:

Swimmers must stroke downward in the water to stay afloat and propel forward. This movement is equal and opposite to the force the water exerts against the swimmer to stop them from moving.

Answer:

c. Newton's Third Law

Explanation:

I got it correct on the test

Answer:

Perfectly elastic collision

Explanation:

In a closed system, an elastic collision is a type of collision between two bodies, where the total momentum and kinetic energy are conserved.

We are told from the problem that the ball bounces back with its original velocity. For the ball to bounce back to the thrower in the first place, this is our first hint that the collision is elastic. If the collision was inelastic, the ball would most likely have stuck to the wall.

In addition to that, the velocity of the ball remains fairly unchanged even after the collision. This confirms that the kinetic energy it had before the collision is the same as the kinetic energy it has after the collision.  

As a result of this, the collision is perfectly elastic

Answer:

What forces act on a marble rolling down a ramp?

Answer: Gravity acts vertically downward. A normal force acts from the ruler toward the marble/ball in a direction that is perpendicular to the plane of the ruler. Friction acts in the direction opposite to which the marble/ball is moving. ... Friction slows down the marble/ball.

Answer:

B

Explanation:

Answer:

false

Explanation:

Answer:

Explanation:

a )

59° south of west  means (180 + 59 )° counterclockwise from the +x-axis.

so angle θ = 239 ° .

r = 3.5 miles .

polar coordinates ( r ,  θ )

= ( 3.50 , 239° )

As the smallest positive angle in degree is 26 then 26 x 9 = 234

so rounding off 239 to 234 , the angle will be 234

polar coordinates ( 3.50 ,  234 )

b )

Cartesian coordinates (in miles) for the same point

x coordinate = r cosθ = 3.5 cos 234 =   -  2.06 mi

y coordinate  = r sinθ = 3.5 sin 234 = - 2.83  mi

Answer:

Answer:

B .There are multiple paths the electrons can take through the circuit, and it is possible for the electron to pass through one circuit component but not another.

Explanation:

Because in a parallel circuit, all components are connected across each other, forming exactly two sets of electrically common points.