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2 years ago

Is a cookie sheet a conductor or an insulator?

Physics

2 answers:

yanalaym [24]2 years ago

7 0

A cookie sheet is an conductor and helps keep all the heat going into the cookies.

Dahasolnce [82]2 years ago

6 0

Answer:

a conductor

Explanation:

cookie sheets are usually made of tin or another metal, which is a conductor

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Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

3 0

2 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

2 years ago

A 45.2 kg softball player slides across dirt with Uk=0.340. What is her acceleration?

alisha [4.7K]

Answer:

the ball didnt hit my face so

Explanation:

8 0

2 years ago

Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates <0, 4.00 cm> and <0, -4.00 cm

Dmitry [639]

Answer:

Explanation:

Potential energy of the system of charges

= 9 x 10⁹ x [ q₁q₂ / r₁₂ + q₂q₃ / r₂₃ + q₁q₃ / r₁₃ ]

here r₁₂ , r₂₃ , r₁₃ are distance between 1 st and 2 nd charge , 2 nd and 3 rd charge and fist and third charge.

r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

q₁ = 20 x 10⁻⁹ C , q₂ = - 20 x 10⁻⁹ C , q₃ = 10 x 10⁻⁹ C

Potential energy = 9 x 10⁹ x [ - 400 x 10⁻¹⁸ / .08 + -200x10⁻¹⁸ / .04 + 200 x 10¹⁸ / .04 ]

= 9 x 10⁹ x - 400 x 10⁻¹⁸ / .08

= 45 x 10⁻⁶ J .

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

= 9 x 10⁹ x 10 x 10⁻⁹ / .03

= 3000 V .

potential energy of fourth particle = charge x potential

= 3000 x 40 x 10⁻⁹ = 12 x 10⁻⁵ J .

kinetic energy at infinity = 12 x 10⁻⁵ J

1/2 m v² = 12 x 10⁻⁵ J

.5 x 2 x 10⁻¹³ x v² = 12 x 10⁻⁵

v² = 12 x 10⁸

v = 3.46 x 10⁴ m/s

= 9 x 10⁹

5 0

2 years ago

A tank circuit consists of an inductor and a capacitor. Give a simple explanation for why the magnetic field in the induc- tor i

ipn [44]

Answer:

If you pull a permanent magnet rapidly away from a tank circuit, what is likely to happen in that circuit?

Charge will oscillate in the tank's capacitor and inductor.

Explanation:

4 0

2 years ago