A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?

Answers

Answer 1

Answer:

Answer:

$E = -9.5* 10^4~N/C$

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

$\int {\vec{E}d\vec{a}} = (Q_(enc))/(\epsilon_0)$

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

$E4\pi r^2 = (-84* 10^(-9))/(8.8* 10^(-12))\nE4\pi (8.9 * 10^(-2))^2 = (-84* 10^(-9))/(8.8* 10^(-12))\n\nE = -9.5* 10^4~N/C$

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A bug flying horizontally at 1.7 m/s collides and sticks to the end of a uniform stick hanging vertically from its other end. After the impact, the stick swings out to a maximum angle of 7.0Â° from the vertical before rotating back. If the mass of the stick is 16 times that of the bug, calculate the length of the stick (in m).

Answers

Answer:12.11 m

Explanation:

Given

Bug speed =1.7 m/s

Let mass of bug is m

mass of rod 16m

maximum angle turned by rod is 7^{\circ}[/tex]

From Energy conservation

kinetic energy of bug =Gain in potential energy of rod

$(1)/(2)mv^2=16mgL(1-cos\theta )$

$L=(1.7^2)/(2* 16(1-cos7))$

L=12.11 m

Newton's law of gravity was inconsistent with Einstein's special relativity becausethe distance, and thus the force, depends on the reference frame chosen
time dilation slows the apparent acceleration, reducing the force
mass, and thus the force, depends on the reference frame chosen

Newton didn't know about Einstein

Answers

Answer:

Mass and thus force depends on the reference frame chosen

Explanation:

This can be explained as Newton's law of gravity provides action which are instantaneous at a distance and involves the evaluation of all the quantities at present time or at the instant they occur.

If the body undergoes a change in its mass distribution there will be an immediate change in its gravitational force without any lag.

Now, if we talk about special relativity, it would be absurd to say that an information can travel faster than light. The effect is in synchronization with the cause in one reference frame where the effect occurs after the cause for some observer in some other reference frame.

In order to observe Newton's law of gravity all the observer's in different reference frames must observe the same phenomena which could only be possible if time were absolute and in special relativity, time is not absolute.

Therefore, Newton's law of gravity was inconsistent with the Einstein's Special Relativity.

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's initial speed is 1.37 m/s and it stops after 2.8 s. Determine the magnitude of the friction force exerted on the box.

Answers

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_(o)+a\cdot t$ ), we expand the previous expression:

$-f = \left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left((W)/(g) \right)\cdot \left((v-v_(o))/(t)\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_(o)$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,(m)/(s^(2))$ , $v_(o) = 1.37\,(m)/(s)$ , $v = 0\,(m)/(s)$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left((52.4\,N)/(9.807\,(m)/(s^(2)) ) \right)\cdot \left((0\,(m)/(s)-1.37\,(m)/(s) )/(2.8\,s) \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

Final answer:

The magnitude of the friction force acting on the box is determined by calculating the box's acceleration, establishing its mass based on its weight information, and applying these values in Newton's second law. The calculated value is 2.62 N.

Explanation:

To determine the magnitude of the friction force, we first have to compute the acceleration of the box. Acceleration (a) can be found using the formula 'final velocity - initial velocity / time'. Since the final velocity is 0 (the box stops), and the initial velocity is 1.37 m/s, and the time is 2.8 s, we get: a = (0 - 1.37) / 2.8 = -0.49 m/s^2. The negative sign indicates deceleration.

Next, we use Newton's second law, which states that the net force acting on an object is equal to its mass times its acceleration. The net force in this case is the frictional force because there is no other force acting on the box in the horizontal direction. However, we do not know the mass of the box, but we do know its weight, and weight = mass x gravitational acceleration (g). So mass = weight/g = 52.4N / 9.8m/s^2 = 5.35 kg.

Lastly, we substitute the mass and deceleration into Newton's second law to find the frictional force (f): f = mass x deceleration = 5.35kg x -0.49m/s^2 = -2.62 N. Again, the negative sign indicates that the force acts opposite to the direction of motion. Thus, the frictional force magnitude is 2.62 N.

Learn more about Friction force here:

brainly.com/question/33825279

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an ice skater, standing at rest, uses her hands to push off against a wall. she exerts an average force on the wall of 120 N and the push lasts 0.8 seconds. The skater's mass is 55 kg. what is the skater's speed after she stops pushing on the wall

Answers

Answer:

The skater's speed after she stops pushing on the wall is 1.745 m/s.

Explanation:

Given that,

The average force exerted on the wall by an ice skater, F = 120 N

Time, t = 0.8 seconds

Mass of the skater, m = 55 kg

It is mentioned that the initial sped of the skater is 0 as it was at rest. The change in momentum of skater is :

$\Delta p=m(v-u)\n\n\Delta p=mv$

The change in momentum is equal to the impulse delivered. So,

$J=\Delta p=F* t\n\nmv=F* t\n\nv=(Ft)/(m)\n\nv=(120* 0.8)/(55)\n\nv=1.745\ m/s$

So, the skater's speed after she stops pushing on the wall is 1.745 m/s.

The most soaring vocal melody is in Johann Sebastian Bach's Mass in B minor. In one section, the basses, tenors, altos, and sopranos carry the melody from a low D to a high A. In concert pitch, these notes are now assigned frequencies of 146.8 Hz and 880.0 Hz. (Use 343 m/s as the speed of sound, and 1.20 kg/m3 as the density of air.)a. Find the wavelength of the initial note.
b. Find the wavelength of the final note.
c. Assume the choir sings the melody with a uniform sound level of 70.0 dB. Find the pressure amplitude of the initial note.
d. Find the pressure amplitude of the final note.
e. Find the displacement amplitude of the initial note.
f. Find the displacement amplitude of the final note.

Answers

Answer:

Detailed step wise solution is attached below

Explanation:

(a) wavelength of the initial note 2.34 meters

(b) wavelength of the final note 0.389 meters

(d) pressure amplitude of the final note 0.09 Pa

(e) displacement amplitude of the initial note 4.78*10^(-7) meters

(f) displacement amplitude of the final note 3.95*10^(-8) meters

Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

$\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)$

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

$t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s$

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

$S_x = v_x t =(6.0 m/s)(90 s)=540 m$

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

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