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Natasha2012 [34]
3 years ago
6

Why can you not put diesel in a regular car?

Physics
2 answers:
goblinko [34]3 years ago
6 0
It's not compatible. It's like putting metal in a microwave; it's not going to go well (sorry I don't have a scientific answer, I'm just thinking out loud). 
Ilia_Sergeevich [38]3 years ago
4 0
It will mess up the car's engine, and get stuck in the carburetor.
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THIS IS MY EXAM HURRY PLS
tangare [24]

Answer:

elements in the same column have the same number of neutrons. elements with similar mass are placed in the same column.

5 0
3 years ago
Read 2 more answers
To which group/family does each of these belong? A. Sulfur _________ B. Sodium _________ C. Argon _________ D. Silicon _________
ArbitrLikvidat [17]

Answer:

A. Sulfur _________ group 16 chalcogens

B. Sodium _________ group 1 alkali metals

C. Argon _________ group 18 noble gases

D. Silicon _________ group 14 carbon family

E. Chlorine _________ group 17 halogens

F. Phosphorus_________ group 15 pnicogens

7 0
2 years ago
Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the
Zepler [3.9K]

Answer:

51.94°

Explanation:

I_0 = Unpolarized light

I_2 = Light after passing though second filter = 0.19I_0

Polarized light passing through first filter

I_1=\frac{I_0}{2}

Polarized light passing through second filter

I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}

The angle between the two filters is 51.94°

5 0
3 years ago
The constant forces F1 = 8 + 29 + 32 N and F2 = 48 - 59 - 22 N act together on a particle during a displacement from the point A
steposvetlana [31]

Answer:

- 600 J

Explanation:

A (20, 15, 0 ) m

B (0, 0, 7) m

\overrightarrow{F_{1}}=8\widehat{i}+29\widehat{j}+32\widehat{k}

\overrightarrow{F_{2}}=48\widehat{i}-59\widehat{j}-22\widehat{k}

Net force

\overrightarrow{F}=\overrightarrow{F_{1}}+\overrightarrow{F_{2}}

\overrightarrow{F}}=\left ( 8+48 \right )\widehat{i}+\left ( 29-59 \right )\widehat{j}+\left ( 32-22 \right )\widehat{k}

\overrightarrow{F}}=56\widehat{i}-30\widehat{j}+10\widehat{k}

\overrightarrow{S}=\overrightarrow{OB}-\overrightarrow{OA}

\overrightarrow{S}=\left ( 0-20 \right )\widehat{i}+\left ( 0-15 \right )\widehat{j}+\left ( 7-0 \right )\widehat{k}

\overrightarrow{S}=-20\widehat{i}-15\widehat{j}+7\widehat{k}

Work done is defined as

W = \overrightarrow{F}.\overrightarrow{S}

W = \left ( 56\widehat{i}-30\widehat{j}+10\widehat{k} \right ).\left (-20\widehat{i}-15\widehat{j}+7\widehat{k}  \right )

W = -1120 + 450 + 70

W = - 600 J

3 0
3 years ago
A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc
hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

E V = log_2(\dfrac{N^2}{t})

N is the diameter of the aperture and t is the time of exposure

now,

log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})

\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}

inserting all the values

\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}

      N₂² = 4

      N₂ = 2 mm

hence , the correct answer is option E

4 0
3 years ago
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