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3 years ago

Ice of mass 5 g at 0 °C melts to water at 0 °C.

Physics

1 answer:

amid [387]3 years ago

5 0

Answer:

Q=1670J

Explanation:

Mass of ice: m=5g=0.005kg

Latent heat: lambda=3.34×10⁵J/kg

Heat received by ice: Q=m×lambda

Q=0.005×3.34×10⁵=5×334=1670J

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A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitud

DaniilM [7]

Answer:

(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.

5 0

3 years ago

A 750 kg National Geographic Drone is rising vertically up into the atmosphere at constant speed. The lift force pushing it upwa

Debora [2.8K]

Answer:

a. 0

b. 1.103625 MJ

c. Conserved

d. 1.103625/n MJ where n = The number of forces

Explanation:

The mass of the drone, m = 750 kg

The upward lift force = 125% of the weight of the drone

The time it takes the drone to reach a height of 250 m = 25 seconds

a. The mechanical energy = The kinetic energy + Potential energy

Therefore, given that the drone stars motion from the surface and was initially at rest, the mechanical energy at the surface = 0

b. The mechanical energy at height, h = 150 m, ME₁₅₀ = The potential energy gained = m·g·h

Where;

g = The acceleration due to gravity = 9.81 m/s²

∴ ME₁₅₀ = 750 kg × 9.81 m/s² × 150 m = 1103625 J = 1.103625 MJ

c. The mechanical energy is equivalent to the potential energy of the drone at the 150 m height, therefore, it is conserved

d. The work done by the force = The energy gained

Therefore, where there are <em>n</em> number of forces, the work done by each force = 1.103625/n MJ

5 0

3 years ago

A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

Which of the following is not a sedimentary structure?A) Magma emplacementB) Mud cracksC) Ripple marksD) Cross beds

lawyer [7]

Answer:

A) Magma emplacement

Explanation:

Sedimentary structures forms during deposition of sediments. It can also form after sediments have been deposited. Sedimentary structures can only be found in sedimentary rocks. Some examples include mud cracks, ripple marks, cross stratification, potholes, etc

Magma emplacement is an igneous process which describes the different mechanisms by which magma can be emplaced. It is only typical of igneous rocks.

6 0

3 years ago

Which situation involves kinetic energy being transformed into sound energy?

tekilochka [14]

Answer:

Explanation:

this is because you move ( kinetic energy) your hands and clap the other person's hand, you will definitely hear sound right.

3 0

3 years ago

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