Ask question

Ask question

All categories

marishachu [46]

4 years ago

9

Una sustancia radioactiva, cuya temperatura es inconstante, aumenta su temperatura en 30ºF, luego disminuye e 40ºC, luego aument

a en 12R y por último disminuye en 10K; entonces la temperatura final de dicha sustancia: a) No varia b) Aumenta en 24R c) Disminuye en 48R d) Aumenta en 48R e) Disminuye en 24R

1 answer:

irakobra [83]4 years ago

6 0

Answer:

What the heck

Explanation:

You might be interested in

Complete the following table with the appropriate units of measurement. Word bank: gram, meter, kelvin, liter, second, grams/cm3

lara31 [8.8K]

mass gram, time sec, temp kelvin, vol liter, dens grams/cm3

5 0

4 years ago

Read 2 more answers

Calculate the force that the 4kg block exerts on the 10kg block

Kryger [21]

Acceleration=force/mass=28/(10+4)=2m/s^2

force10kg=ma=10*2
force4kg=ma=(10*2)=20
the4 kg is pushing against the 10kg block

vf=vi+at
-10=20*28/14 * t
t=30/2=15sec

i hope this can help you.

8 0

3 years ago

in which kind of radioactive decay would the number of protons in the resulting nucleus be more than in the initial nucleus?

9966 [12]

Answer:

A related type of beta decay

Explanation:

4 0

2 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

3 years ago

Which situation would create a field like the one shown here? A. an interaction between two magnetic south poles B. an interacti

alexgriva [62]

Answer:

you have no picture

Explanation:

if it shows the arrows going towards eachother it is north and south. if it is not it will be either north and north or south and south

4 0

3 years ago

Read 2 more answers