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2 years ago

Which of the following surfaces reflects the most light

Physics

2 answers:

Svetlanka [38]2 years ago

8 0

Answer:C

Explanation:

The best reflecting surface is a mirror or a polished surface.

Aluminum is polished better than any of them above option

Gre4nikov [31]2 years ago

3 0

Answer:

Explanation:

it could be the yellow cloth but we see are shadow on the sidewalk most of the time and shadows are made from reflecting light so your answer should be D

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A tiny 0.0250 -microgram oil drop containing 15 excess electrons is suspended between to horizontally closely-spaced metal plate

qaws [65]

Answer:

(a) 12 × 10⁻³ C = 12 mC (b) The lower plate

Explanation:

Given

mass of oil drop, m = 0.025 μg = 0.025 × 10⁻⁶

radius of plates, r = 6.50 cm = 6.5 × 10⁻² m

k = 1/4πε₀ = 9.0 × 10⁹ Nm²/C²

electric charge, e= 1.6 × 10⁻¹⁹ C

charge on oil drop, q = 15e

charge on plates, Q = ?

First, we find the charge density of the plates, D = Q/A where Q = charge on plates and A = area of plates. Since the plates are circular, the area is given by A=πr² where r = radius of plates. D=Q/πr²

Also, the electric field, E between the plates is given by E = D/ε =Q/Aε = Q/ε₀πr².

The force on the oil drop due to the electric field between the plates is given by F = qE = qQ/ε₀πr².

Since the oil drop is suspended between the plates, it means that the electric force due to the field on the oil drop balances the weight of the oil drop. So, since weight of oil drop W = mg where g = 9.8 m/s². F =W (for oil drop suspension).

So, qQ/ε₀πr²=mg

So, Q=mgε₀πr²/q

From k = 1/4πε₀, ε₀=1/4πk

So, Q = mgπr²/4πkq = mgr²/4kq = (0.025 × 10⁻⁶ × 9.8 × (6.5 × 10⁻²)²)÷(4 × 9 × 10⁹ × 15 × 1.6 × 10⁻¹⁹)= 0.012 C = 12 mC

(b) The lower plate must be positive because, the direction of the electric field must be upwards, so as to balance out the weight of the oil drop so as to suspend it.

6 0

2 years ago

A very delicate sample is placed 0.150 cm from the objective lens of a microscope. The focal length of the objective is 0.140 cm

iVinArrow [24]

Answer:

m = 14*26 = 364

Explanation:

overall magnification is given as m

$m = m_{o}* m_{e}$

mo magnification of objective lens

me magnification of EYE lens

where mo is given as

$m_{o} = \frac{v_{o}}{-u _{0}}$

and me as

$m_{e} = 1+\frac{D}{f_{e}}$

d is distant of distinct vision = 25.0 cm for normal eye

fe = focal length of eye piece

focal length of objective lense is 0.140 cm

we know that

$\frac{1}{v_{0}}-\frac{1}{u_{0}}=\frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{u_{0}} + \frac{1}{f_{0}}$

$\frac{1}{v_{0}} = \frac{1}{0.150} + \frac{1}{0.14}$

$\frac{1}{v_{o}} = 2.1 cm$

$m_{o} = \frac{2.1}{0.150} = 14$

$m_{e} = 1+\frac{25}{1}$

$m_{e} =26$

$m = m_{o}* m_{e}$

m = 14*26 = 364

4 0

3 years ago

. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it

kap26 [50]

Answer:

29.4 N/m

0.1

Explanation:

a) From the restoring Force we know that :

F_r = —k*x

the gravitational force :

F_g=mg

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force

xi , x2 are the displacement made by the two masses.

Givens:

m1 = 1.29 kg

m2 = 0.3 kg

x1 = -0.75 m

x2 = -0.2 m

g = 9.8 m/s^2

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m

Plugging known infromation to get :

l= x1 — (F_1/k)

= 0.1

3 0

2 years ago

What does the statement "all motion is relative" mean?

tekilochka [14]

The statement that motion is relative is an important concept in physics. The meaning behind this statement is that the motion of an object is relative to either the frame of reference of the observer, or to another distinct frame of reference.

4 0

3 years ago

Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that t

ArbitrLikvidat [17]

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T $_H$ in the cycle is twice the minimum absolute temperature T $_L$ in the cycle

T $_H$ = 0.5T $_L$

now, we find the efficiency of the Carnot cycle engine

η $_{th$ = 1 - T $_L$ /T $_H$

η $_{th$ = 1 - T $_L$ /0.5T $_L$

η $_{th$ = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η $_{th$ = 1 - W $_{net$ /Q $_H$

where W $_{net$ is net work done, Q $_H$ is is the heat supplied

we substitute

0.5 = 60 / Q $_H$

Q $_H$ = 60 / 0.5

Q $_H$ = 120 kJ

Now, we apply the first law of thermodynamics to the system

W $_{net$ = Q $_H$ - Q $_L$

60 = 120 - Q $_L$

Q $_L$ = 60 kJ

now, the amount of heat rejection per kg of steam is;

q $_L$ = Q $_L$ /m

we substitute

q $_L$ = 60/0.025

q $_L$ = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q $_L$ = h $_{fg$ = 2400 kJ/kg

now, at h $_{fg$ = 2400 kJ/kg from saturated water tables;

T $_L$ = 40 + ( 45 - 40 ) ( $\frac{2400-2406.0}{2394.0-2406.0}\\}$ )

T $_L$ = 40 + (5) × (0.5)

T $_L$ = 40 + 2.5

T $_L$ = 42.5°C

Therefore, The temperature of the steam during the heat rejection process is 42.5°C

4 0

2 years ago