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3 years ago

Krishne wants to measure the mass and volume of a thimble. Which tools should she use?

1 answer:

6 0

To measure the mass, you would use a balance. To measure the volume, you can use a variety of ounces, cups, pints, quarts, and gallons. Good luck!

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You rub a clear plastic pen with wool, and observe that a strip of invisible tape is attacted to the pen. Assuming that the pen

gayaneshka [121]

Answer:

option B

Explanation:

When you rub a clear plastic pen with the wool the plastic pen gets charges this phenomenon is known as frictional charging.

Due to rubbing, the pen gets negatively charged.

We know, opposite charge attract each other and the same charge repel each other.

So, when the pen is negatively charged the tape might be positively charged or the tape might be uncharged.

Hence, the correct answer is option B

6 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

melisa1 [442]

Answer:

Weight = 8.162 Newton.

Explanation:

Given the following data;

Mass = 2.2 kg

Acceleration due to gravity = 3.71 N/kg

To find the weight of the textbook;

Weight = mass * acceleration due to gravity

Weight = 2.2 * 3.71

Weight = 8.162 N

Therefore, the weight of the science textbook in mars is 8.162 Newton.

3 0

3 years ago

EXPLAIN HOW ENERGY IS TRANSMITTED THROUGH A MEDIUM

jek_recluse [69]

I say it helped then because TrueType had room

5 0

3 years ago

Read 2 more answers

When did agriculture first began?

timofeeve [1]

Answer:

Humans invented agriculture between 7,000 and 10,000 years ago, during the Neolithic era, or the New Stone Age. There were eight Neolithic crops: emmer wheat, einkorn wheat, peas, lentils, bitter vetch, hulled barley, chickpeas, and flax.

5 0

3 years ago

Read 2 more answers

To practice Problem-Solving Strategy 23.2 for continuous charge distribution problems. A straight wire of length L has a positiv

Lesechka [4]

Answer:

E = k Q / [d(d+L)]

Explanation:

As the charge distribution is continuous we must use integrals to solve the problem, using the equation of the elective field

E = k ∫ dq/ r² r^

"k" is the Coulomb constant 8.9875 10 9 N / m2 C2, "r" is the distance from the load to the calculation point, "dq" is the charge element and "r^" is a unit ventor from the load element to the point.

Suppose the rod is along the x-axis, let's look for the charge density per unit length, which is constant

λ = Q / L

If we derive from the length we have

λ = dq/dx ⇒ dq = L dx

We have the variation of the cgarge per unit length, now let's calculate the magnitude of the electric field produced by this small segment of charge

dE = k dq / x²2

dE = k λ dx / x²

Let us write the integral limits, the lower is the distance from the point to the nearest end of the rod "d" and the upper is this value plus the length of the rod "del" since with these limits we have all the chosen charge consider

E = k $\int\limits^{d+L}_d {\lambda/x^{2}} \, dx$

We take out the constant magnitudes and perform the integral

E = k λ (-1/x) ${(-1/x)}^{d+L} _{d}$

Evaluating

E = k λ [ 1/d - 1/ (d+L)]

Using λ = Q/L

E = k Q/L [ 1/d - 1/ (d+L)]

let's use a bit of arithmetic to simplify the expression

[ 1/d - 1/ (d+L)] = L /[d(d+L)]

The final result is

E = k Q / [d(d+L)]

3 0

3 years ago