Ultraviolet rays from the sun are able to reach Earth's surface because-

A lunar lander is making its descent to moon base i. the lander descends slowly under the retro-thrust of its descent engine. th

olga_2 [115]

V = final velocity
u = initial velocity
a = acceleration
s = displacement
Take downward direction as positive.
v^2 = u^2 + 2as
v^2 = 0.8^2 + 2(1.6)(5.5)
v^2 = 18.24
v = 4.27 m/s

5 0

3 years ago

Read 2 more answers

When light passes through a convex lens, it does this.....

Tpy6a [65]

Explanation:

6. Converge or come together

7. convex

3 0

2 years ago

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the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc

sergejj [24]

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

3 0

2 years ago

flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electr

Ksenya-84 [330]

Answer:

$6.29591\times 10^{-6}\ N/C^2$

Explanation:

Flux is given by

$\phi=EAcos\theta$

A = Area

$A=0.4\times 10^{-3}\times 0.6\times 10^{-3}$

E = Electric field = 76.7 N/C

Angle is given by

$\theta=90-20\\\Rightarrow \theta=70^{\circ}$

$\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2$

The flux through the sheet is $6.29591\times 10^{-6}\ N/C^2$

6 0

2 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago