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4 years ago

in an elements square on a periodic table, the number with the greatest numerical value represents the

1 answer:

6 0

In an element's square on a periodic table, the numerical value indicates the atomic mass or atomic weight of a certain element. It is also similar to the number of protons in a specific atom. Each element varies from one another. So the greatest numerical value represents the most atomic mass among elements.

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A stubborn 130 kg pig sits down and refuses to move. to drag the pig to the barn, the exasperated farmer ties a rope around the

TEA [102]

<span>Unless the pig moves static friction acts on it once the pig starts moving kinetic friction comes in to play so when the pig is not moving=frictional force acting on it =normal force*co-efficient of static friction.</span>

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4 years ago

What happens during

alukav5142 [94]

Answer:

Explanation:

8 0

3 years ago

Read 2 more answers

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

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2 years ago

Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours

Nostrana [21]

Answer:

Approximately $116\; \text{miles}$ for the train from Boston to NYC Penn Station.

Approximately $105\; \text{miles}$ for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

$\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}$ .

$\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}$ .

Calculate average speed of each train:

$\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$ .

$\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}$

Assume that it takes a time period of $t$ for the trains to pass by each other after departure. Distance each train travelled would be:

$s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t$ .

$s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t$ .

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

$v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}$ .

Rearrange and solve for $t$ :

$(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}$ .

$\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}$ .

Distance each train travelled in $t = (539 / 279)\; \text{hour}$ :

$\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}$ .

$\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}$ .

8 0

2 years ago

If some electrons are transferred from object A to object B,

Igoryamba

Answer: A. Object A will have a positive charge.

Explanation: If the number of protons and electrons are the same, their net charges cancel each other out, and you have a neutral charge. If electrons are transferred to another object, the amount of positive charge will outweigh the amount of negative charge. As a result, you are left with an overall positive charge in object A. Meanwhile, object B is now negative.

6 0

3 years ago