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3 years ago

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A 300 gg ball on a 70-cmcm-long string is swung in a vertical circle about a point 200 cmcm above the floor. The string suddenly

breaks when it is parallel to the ground and the ball is moving upward. The ball reaches a height 600 cmcm above the floor. Part APart complete What was the tension in the string an instant before it broke? Express your answer to two significant figures and include the appropriate units. TT = 34 NN

1 answer:

Ratling [72]3 years ago

6 0

Answer:

the tension in the string an instant before it broke = 34 N

Explanation:

Given that :

mass of the ball m = 300 g = 0.300 kg

length of the string r = 70 cm = 0.7 m

At highest point, law of conservation of energy can be expressed as :

$\frac{1}{2} mv^2 = mgh\\\\v = \sqrt{2gh}\\\\v = \sqrt{2*(9.8 \ m/s^2)*(6.00 \ m - 2.00 \ m)}\\\\$

$v = 8.854 \ m/s$

The tension in the string is:

$T = \frac{mv^2}{r}\\\\T = \frac{(0.300 \ kg)*(8.854 \ m/s^2)}{0.70 \ m}\\\\T = 33.59 N\\\\T = 34 \ N$

Thus, the tension in the string an instant before it broke = 34 N

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A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the

meriva

Answer:

a. $2.86 m/s^2$

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

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By Newton's second law

$8F-f=(8M+m)a$

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Answer:

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3 years ago

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She uses a voltmeter, that measures in volts, and an ammeter that measures in amps. Both were correctly placed in her circuit. I

Nesterboy [21]

Answer:

A) the ammeter is x

B)

voltage across R₁ (left resistor) = 0.75 V

voltage across the right one = 0.3 V

C) 1.05 V

Explanation:

From the diagram attached below;

A) Assuming the homes were wired in series, and one of the homes face short circuit then all the houses would face power cut but it doesn't happen. So they must be connected in parallel.

Therefore; The ammeter is connected in series, Hence, the ammeter is x and the voltmeter must be z.

B)

Given that:

x = 0.15 A

z = 0.3 V

Resistor (R) on the left = 5 ohms

Then, voltage across R₁ (left resistor) = 5×(x)

= 5×0.15

= 0.75 V

voltage across the right one = z = 0.3 V

C)

The total voltage of battery = 0.75+0.3 = 1.05 V

6 0

3 years ago

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch

Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: $-3.62\cdot 10^7 N/C$

c) Electric field 8.00 cm outside the paint layer: $-7.27\cdot 10^7 N/C$

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

$\epsilon_0 = 8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius $r$ as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

$q=0$

So we have

$\int E dS=0$

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

$r=R=0.065 m$

The area of the surface is

$A=4\pi R^2$

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

$E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}$

The charge included within the sphere in this case is the charge on the paint layer, therefore

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

So, the electric field is:

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C$

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

$r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m$

Gauss theorem this time becomes

$E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}$

And the charge included within the sphere is again the charge on the paint layer,

$q=-17.0\mu C=-17.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C$

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3 years ago

State Ohms law in easy words

Alexus [3.1K]

Answer:

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points.

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