A person would be driving 4 blocks west from the starting point to make the shorter distance while maintaining same displacement.
<u>Explanation:</u>
The measurement of an object’s position change from its point is called displacement. It is usually calculated from starting to the end points and represented by ‘delta s’. In the given scenarios, the person drove in the way that he finals the driving by 4 blocks away from the west.
Means, the persons drive to 8 blocks north and then to 8 blocks south get cancelled. Hence, to make the shorter distance with maintaining same displacement he would be driving 4 blocks west from the starting point.
Answer:
Explanation:
There are a couple of ways you could do this.
The easiest is to use E*R1/(R1 + R2)
- E = 10 volts
- R1 = 590 ohms
- R2 = 840 ohms
So the result would be
E_590 = 10 * 590/(590 + 840)
E_590 = 10 * 590/ (1430)
E_590 = 4.13 volts rounded.
You could do this a slightly longer way.
R = 1430 (total ohms in series.
E = 10 volts
I = ???
I = E/R
I = 10 / 1430
I = 0.00699
Now use this current to figure out the voltage drop.
E = I * R
I = 0.00699 amps
R = 590 ohms
E = 0.00699 * 590
E = 4.13 volts
Pick the way of doing it you like best.
Answer:
a = 0,1[m/s^2]
Explanation:
First we need to indentify the initial data.
And using this kinematic equation we have:
![v = 4[m/s]\\v_{0}= 2 [m/s] \\t = 20[s]\\\\v= v_{0}+a*t\\a=\frac{v-v_{0}}{t} \\a= \frac{4-2}{20} \\a=0.1[m/s^{2}]](https://tex.z-dn.net/?f=v%20%3D%204%5Bm%2Fs%5D%5C%5Cv_%7B0%7D%3D%202%20%5Bm%2Fs%5D%20%5C%5Ct%20%3D%2020%5Bs%5D%5C%5C%5C%5Cv%3D%20v_%7B0%7D%2Ba%2At%5C%5Ca%3D%5Cfrac%7Bv-v_%7B0%7D%7D%7Bt%7D%20%5C%5Ca%3D%20%5Cfrac%7B4-2%7D%7B20%7D%20%5C%5Ca%3D0.1%5Bm%2Fs%5E%7B2%7D%5D)
Here's the <em><u>solution</u></em>,
Let the <em><u>shares</u></em> be 2x, 3x and 5x according to their shared <em><u>investment</u></em> ratio,
now, we know :
=》2x + 3x + 5x = 30000
=》10x = 30000
=》x = 3000
So, value of x = 3000
hence, the <em><u>share</u></em> of each <em><u>shareholder</u></em> is :
- kamal = 2x = 6000
- kamala = 3x = 9000
- Krishna = 5x = 15000
i hope it helped.....
Answer: 1 How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-. 2 34 °C? 4. How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of ... A saturated solution of KClO3 was made with 300 g of H2O at. 34 °C.
Explanation:
