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strojnjashka [21]
3 years ago
8

which of the following is not necessary in a great summary?your investigayltion, your predicatio, excuses, possible errors?

Physics
1 answer:
zloy xaker [14]3 years ago
5 0

Excuse are not acceptable


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A surface will be an equipotential surface if (there may be more than one correct choice).A. the electric field is zero at all p
aev [14]

Answer:

c. A, C

Explanation:

On equi-potential surface, fields are equal in magnitude at all points . If field is zero at all points , they will be equal so first option is correct.

Field is perpendicular to  equipotential surface at all points.

6 0
2 years ago
Calculate the power of a light bulb that uses 40J of electricity in 4 seconds.
alexdok [17]

Answer:

power=work done /time

power=40J/4s

power=10wat

7 0
3 years ago
I need help with this question I’m not sure where to even begin
Strike441 [17]

We will have the following:

a. We determine the tension force of T2 as follows:

We know that the system must be at equilibrium on the horizontal axis:

\sum F_x=T_1cos(42.5)+T_2cos(36.5)=0

So:

\begin{gathered} T_2cos(36.5)=-(1235N)cos(42.5)\Rightarrow T_2=-\frac{(1235N)cos(42.5)}{cos(36.5)} \\  \\ \Rightarrow T_2=1132.711003...N\Rightarrow T_2\approx1132.7N \end{gathered}

So, the value of T2 is approximately 1132.7 N.

b. We will determine the torques created by T1 and T2 as follows:

T1:

\tau_{T1}=(10m)(1235N)sin(42.5)\Rightarrow\tau_{T1}\approx8343.5N\ast m

T2:

\tau_{T2}=(10m)(1132.7N)sin(36.5)\Rightarrow\tau_{T2}\approx6737.6N\ast m

So the torques of T1 and T2 on the base are approximately 8343.5 N*m and 6737.6 N*m respectively.

c. The torques around that axis generated by the normal force and the weight are both 0 N*, since they are parallel to the axis.

d. We will determine the angular acceleration as follows:

\begin{gathered} \alpha=\frac{\tau_{T1}}{I}\Rightarrow\alpha=\frac{\tau_{T1}}{(1/3mL^2)} \\  \\ \Rightarrow\alpha=\frac{(8343.5N\ast m)}{(1/3(200kg)(10m)^2)}\Rightarrow\alpha=1.251525rad/s^2 \\  \\ \Rightarrow\alpha\approx1.25rad/s^2 \end{gathered}

So, the angular acceleration is approximately 1.25 radians/ s^2.

6 0
1 year ago
What are the chemical symbols for the elements: gold, silver, and tin respectively? A) Ag, Au, and Sn
Grace [21]

Answer:

The answer to your question is B) Au, Ag and Sn

Explanation:

This is a brief explanation of the origin of the symbols

Gold: the symbol (Au) comes from the Latin Aurum

Silver: the symbol (Ag) comes from the Latin Argentum

Tin: the symbol (Sn) comes from the Latin Stannum

Titanium has the chemical symbol Ti-

5 0
2 years ago
Good morning! someone please answer this, ill give you brainliest and your earning 50 points.
wariber [46]

One step of copper mining requires blasting ore to be able to blasted and extracted from mine.

  • <em>This</em><em> </em><em>process</em><em> </em><em>causes</em><em> </em><em>harm</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>Geosphere</em><em> </em><em>(</em><em>Lithosphere</em><em>)</em>
  • <u>In</u><u> </u><u>this</u><u> </u><u>step</u><u>,</u><u> </u><u>they</u><u> </u><u>blast</u><u> </u><u>the</u><u> </u><u>mines</u><u> </u><u>inside</u><u> </u><u>the</u><u> </u><u>earth</u><u>.</u><u> </u><u>Which</u><u> </u><u>causes</u><u> </u><u>destruction</u><u> </u><u>and</u><u> </u><u>harm</u><u> </u><u>to</u><u> </u><u>the</u><u> </u><u>geosphere</u><u>.</u><u> </u>
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Mercury is used to recover minute pieces of gold that is mixed in soil and sediment. Sometimes Mercury can be lost in this process resulting in widespread contamination of local rivers and lakes.

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