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umka21 [38]
2 years ago
9

Which of the following technologies would produce the least energy in light

Physics
1 answer:
AysviL [449]2 years ago
8 0

Answer: C. Radio-controlled toy airplane

Explanation:

Just did the quiz

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An object has a height of 0.066 m and is held 0.210 m in front of a converging lens with a focal length of 0.140 m. (Include the
Anvisha [2.4K]

Explanation:

Given that,

Size of object, h = 0.066 m

Object distance from the lens, u = 0.210 m (negative)

Focal length of the converging lens, f = 0.14 m

If v is the image distance from the lens, we can find it using lens formula as follows :

\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{0.14 }+\dfrac{1}{(-0.21)}\\\\v=0.42\ m

(a) Magnification,

m=\dfrac{v}{u}\\\\m=\dfrac{0.42}{(-0.21)}\\\\m=-2

(b) Magnification, m=\dfrac{h'}{h}

h' is image height

-2=\dfrac{h'}{0.066}\\\\h'=-2\times 0.066\\\\h'=-0.132\ m

Hence, this is the required solution.

4 0
2 years ago
Why should the safety net that is used in circuses under trapezoids be little tightened?
bixtya [17]
So that there isn't too much force restricting the structure of the circus and cause disaster
8 0
2 years ago
(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50
Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

Substituting values we obtain

m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg

For the transition between parts B and C, we can write:

m_Bv_B=m_Cv_C

Which means

(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

Since we want the new final speed of the car (v_C) we do:

v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

Substituting values we obtain

v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s

5 0
3 years ago
Help with these please?
Anuta_ua [19.1K]

E=kq/r^2

q=(E*r^2)/k

q=(.086N/C)(1.7m^2)/(8.99*10^9N*m^2/C^2)

q=2.76*10^-11 C

q=2.8*10^-11 C

5 0
3 years ago
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An ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any
ehidna [41]

Answer:

4. total energy

Explanation:

According to Bernoulli's principle at any two points along a streamline flow The total energy that is sum of pressure energy , Kinetic energy  and potential energy of the liquid all taken in per unit volume remains constant. Therefore,

for ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant Total Energy.

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3 years ago
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