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storchak [24]
3 years ago
8

Using a dish-shaped mirror, a solar cooker concentrates the sun's energy onto a pot for cooking. A cooker with a 1.2-m-diameter

dish focuses the sun's energy onto a pot with a diameter of 25 cm What is the intensity at the base of the pot
Physics
1 answer:
Alekssandra [29.7K]3 years ago
3 0

Answer:

26500 W/m²

Explanation:

Given that:

The diameter of the dish d = 1.2 m

Assuming the Solar power P capture by the dish = 1300 W

The surface area of the pot is calculated by using the formula:

= \pi r^2

\pi \dfrac{d^2}{4}

where; the diameter = 25 cm

Area = \pi \dfrac{d^2}{4}

Area = \pi \dfrac{0.25^2}{4}

Area = 0.049 m²

The intensity is calculated by using the formula

I = power/area

I = 1300/0.049

I = 26530.6 W/m²

I ≅ 26500 W/m²

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How is the temperature of a gas related to the kinetic energy of its particles
love history [14]

The average kinetic energy of a gas particle is directly proportional to the temperature. An increase in temperature increases the speed in which the gas molecules move. All gases at a given temperature have the same average kinetic energy. Lighter gas molecules move faster than heavier molecules.

7 0
3 years ago
Use the definition of scalar product, a overscript right-arrow endscripts times b overscript right-arrow endscripts = ab cos θ,
makkiz [27]

Answer: \theta=cos^{-1}0.991=7.69^o

The following vectors have been given: \vec{a}=3.0\widehat{i}+3.0\widehat{j}+3.0\widehat{k}\\ \vec{b}=5.0\widehat{i}+7.0\widehat{j}+6.0\widehat{k}

The angle between these two vectors can be found by:

cos\theta=\frac{\vec{a}.\vec{b}}{||\vec{a}|| ||\vec{b}||}\\
||\vec{a}=\sqrt{a_x^2+a_y^2+a_z^2}

\vec{a}.\vec{b}=a_xb_x+a_yb_y+a_zb_z\\ \vec{a}.\vec{b}=3\times5+3\times7+3\times6=15+21+18=54

||\vec{a}||=\sqrt{3^2+3^2+3^2}=\sqrt{27}\\ ||\vec{b}||=\sqrt{5^2+7^2+6^2}=\sqrt{110}

cos\theta=\frac{54}{\sqrt{27}\times\sqrt{110}}\\=0.991\\ \Rightarrow \theta=cos^{-1}0.991=7.69^o

7 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th
vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0
3 years ago
Suppose you are working on green house, which light you would<br> use for their growth?
nikitadnepr [17]

Answer:

Fluorescent lighting is usually used.

Explanation:

6 0
2 years ago
Read 2 more answers
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