What is true about all uranium atoms?

Devon places sliced potatoes in a package of aluminum foil. He puts a metal grate over a campfire and sets the package on the gr

sp2606 [1]

when he set the potatoes in the aluminum foil over the campfire because the heat was directly touching them and heating them

7 0

3 years ago

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Define force ? ☹❣️️ Degine weight?

mylen [45]

Explanation:

force is anything that allows a body to be in motion..

6 0

3 years ago

How much heat transfer (in kcal) is required to raise the temperature of a 0.850 kg aluminum pot containing 2.00 kg of water fro

hram777 [196]

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

$\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T +m_w L_v$

Here,

$m_{Al}$ = Mass of Aluminum

$C_p$ = Specific Heat of Aluminum

$C_w$ = Specific Heat of Water

$m_w =$ Mass of water

$L_v =$ Latent of Vaporization

Replacing,

$\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)$

Converting,

$\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})$

$\Delta Q = 440.409kCal$

Therefore is required 440.409kCal

8 0

3 years ago

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$