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valentinak56 [21]

2 years ago

12

Assume that the light from the flashlight is light from a star. Identify the spot where the light from this “star” is most conce

ntrated.

2 answers:

Reptile [31]2 years ago

8 0

Answer:

This is the answer

Explanation:

SCORPION-xisa [38]2 years ago

3 0

<span>The spot where the light is most concentrated would be at the center. The light is most dense at the center of the flashlight. It will show brightly at the center where the light is disbursed.</span>

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What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

7 0

1 year ago

A 15-cm-tall closed container holds a sample of polluted air containing many spherical particles with a diameter of 2.5 μm and a

TEA [102]

Answer:

t = 224 s

Explanation:

given,

length of container = 15 cm = 0.15 m

diameter of spherical particle = 2.5 μm

mass of particle = 1.9 x 10⁻¹⁴ kg

viscosity of air = μ = 1.18 x 10⁻⁵ kg/m.s

time taken by the particle to stop = ?

radius of particle = 2.5/2 = 1.25 μm

volume of particle = $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3$

Density = $\dfrac{mass}{volume}$

Density = $\dfrac{1.9 \times 10^{-14}}{\dfrac{4}{3}\pi (1.25 \times 10^{-6})^3}$

ρ = 2322 kg/m³

terminal velocity

$v_t = \dfrac{2}{9}\ \dfrac{gR^2(\rho - \rho_{air})}{\mu}$

$v_t = \dfrac{2}{9}\ \dfrac{9.8 \times (1.25 \times 10^{-6})^2(2322 - 1)}{1.18 \times 10^{-5}}$

v_t = 6693 x 10⁻⁷ m/s

$t = \dfrac{d}{v_t}$

$t = \dfrac{0.15}{6693 \times 10^{-7}}$

t = 224 s

7 0

2 years ago

If a body is moving in the horizontal axis with a velocity Vx= 6m/s and in the vertical axis Vy=8m/s What is the angle Theta abo

cluponka [151]

Answer: C

Explanation: It's a lot of math.

7 0

1 year ago

Maria drove to the store, did some shopping, and then came home. During Maria's trip, when was her displacement equal to zero? A

Aleks [24]

I think it’s a) but I’m not sure.

7 0

1 year ago

Read 2 more answers

An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val

Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 $m/s^2$

as per the question the acceleration due to gravity is found to be 9.42 $m/s^2$ in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is - $\frac{0.381}{9.801} *100$

=3.88735 percent

the error is not so high,so it can be accepted.

now we have to know why this occurs-the equation of time period of the simple pendulum is give as- $T=2\pi\sqrt[2]{l/g}$

$g=4\pi^2\frac{l}{T^2}$

As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801 $m/s^2$

5 0

2 years ago