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2 years ago

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A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph

ere? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)

1 answer:

Nonamiya [84]2 years ago

5 0

Answer:

The electric field is $E = 7.5 *10^{6} \ N/C$

Explanation:

From the question we are told that

The radius of the metal sphere is $R = 2.0 \ cm = 0.02 \ m$

The excess charge which the metal sphere carries is $q = 3.0 \mu C = 3.0*10^{-6} \ C$

The distance of the position being to the center is $D = 6.0 \ cm = 0.06 \ m$

The coulomb constant is $k =9*10^{9} \ N \cdot m^2 /C^2$

Generally the electric field is mathematically represented as

$E = \frac{k * q}{D^2}$

substituting values

$E = \frac{9*10^{9} * 30.*10^{-6}}{(0.06)^2}$

$E = 7.5 *10^{6} \ N/C$

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Consider a very small hole in the bottom of a tank 17.0 cm in diameter filled with water to a heightof 90.0 cm. Find the speed a

umka21 [38]

Answer:

Speed of water, v = 4.2 m/s

Explanation:

Given that,

Diameter of the tank, d = 17 cm

It is placed at a height of 90 cm, h = 0.9 m

We need to find the speed at which the water exits the tank through the hole. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 0.9}$

v = 4.2 m/s

So, the speed of water at which the water exits the tank through the hole is 4.2 m/s. Hence, this is the required solution.

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2 years ago

Two wires are used to suspend a sign that weighs 500 N. The two wires make an angle of 100° between each other. If each wire is

vaieri [72.5K]

1) draw a diagram.
2) label diagram. (split the 100 degrees into 50, (which is right down the middle) to make a right angle triangle.)
3) since its a free body diagram, the forces known must be labelled. (force of gravity). this shows that the straight vertical line of the right angle triangle is Fg (force gravity). label it.
4) use trigonometry. rearrange the equation to solve for what needs to be known.

angles known: 50 (split 100 in half to make a right angle triangle), 90 (since its right angle), and 40 (180-90-50 = 40)

sides known: vertical lined up with the 90 degree angle. Fg. --> fg=mg=500N x 9.81m/s^2 = 4905N

use formula: sin or cos

i used sin. sin(40) = 4905 / ?
- times '?' on both sides. : sin(40) x '?' = 4905
-divide both sides by sin(40): '?' = 4905/ sin(40)
--> Solve.

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2 years ago

A 2200 kg truck has put its front bumper against the rear bumper of a 2400 kg SUV to give it a push. With the engine at full pow

leonid [27]

Answer:

a) The maximum possible acceleration the truck can give the SUV is 7.5 meters per second squared

b) The force of the SUV's bumper on the truck's bumper is 18000 newtons

Explanation:

a) By Newton's second law we can find the relation between force and acceleration of the SUV:

$F=ma$

With F the maximum force the truck applies to the SUV, m the mass of the SUV and a the acceleration of the SUV; solving for a:

$a=\frac{F}{m}=\frac{18000}{2400}\approx7.5\,\frac{m}{s^{2}}$

b) Because at this acceleration the truck's bumper makes a force of 18000 N on the SUV’s bumper by Third Newton’s law the force of the SUV’s bumper on the truck’s bumper is 18000 N too because they are action-reaction force pairs.

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2 years ago

A 0.200-kg mass attached to the end of a spring causes it to stretch 5.0 cm. If another 0.200-kg mass is added to the spring, th

ziro4ka [17]

Answer:

A: 4 times as much

B: 200 N/m

C: 5000 N

D: 84,8 J

Explanation:

A.

In the first question, we have to caculate the constant of the spring with this equation:

$m*g=k*x$

Getting the k:

$k=\frac{m*g}{x} =\frac{0,2[kg]*9,81[\frac{m}{s^{2} } ]}{0,05[m]} =39,24[\frac{N}{m}]$

Then we can calculate how much the spring stretch whith the another mass of 0,2kg:

$x=\frac{m*g}{k} =\frac{0,4[kg]*9,81[\frac{m}{s^{2} } ]}{39,24[\frac{N}{m}]} =0,1[m]\\$

The energy of a spring:

$E=\frac{1}{2}*k*x^{2}$

For the first case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,05[m])^{2} =0,049 [J]$

For the second case:

$E=\frac{1}{2} *39,24[\frac{N}{m}]*(0,1[m])^{2} =0,0196 [J]$

If you take the relation E2/E1 = 4.

B.

We have the next facts:

x=0,005 m

E = 0,0025 J

Using the energy equation for a spring:

$E=\frac{1}{2}*k*x^{2}$ ⇒ $k=\frac{E*2}{x^{2} } =\frac{0,0025[J]*2}{(0,005[m])^{2} } =200[\frac{N}{m} ]$

C.

The potential energy of the diver will be equal to the kinetic energy in the moment befover hitting the watter.

$E=W*h=500[N]*10[m]=5000[J]$

Watch out the units in this case, the 500 N reffer to the weighs of the diver almost relative to the earth, thats equal to m*g.

D.

The work is equal to the force acting in the direction of the motion. so we have to do the diference beetwen angles to obtain the effective angle where the force is acting: 47-15=32 degree.

The force acting in the direction of the ramp will be the projection of the force in the ramp, equal to F*cos(32). The work will be:

W=F*d=F*cos(32)*d=10N*cos(32)*10m=84,8J

7 0

2 years ago

Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights

Arlecino [84]

Answer:

275 kPa

Explanation:

mass of the gas=m=1.5 kg

initial volume if the gas=V₁=0.04 m³

initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

V₂=final volume

V₂=2 V₁

As the temperature is constant

T₁=T₂=T

$\frac{P1V1}{T1}$ = $\frac{P2 V2}{T2}$

putting the values in the equation.

$\frac{P1V1}{T1}$ = $\frac{P2 *2V1}{T2}$

P₂= $\frac{P1}{2}$

P₂= $\frac{550}{2}$

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

3 0

2 years ago