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dedylja [7]
3 years ago
8

What is the main function of a protein within the body?

Physics
2 answers:
aksik [14]3 years ago
4 0
The answer looks to be "A" because Protein is used to regulate organs.
Lady_Fox [76]3 years ago
3 0
B. protein promotes energy and digestion
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Hello!
34kurt
Iodine-131 has a half life of 8 days, so half of it is gone every 8 days.
10 grams of iodine-131 is left for 24 days.
At 8 days: 10/2=5 grams left
At 16 days: 5/2=2.5 grams left
At 24 days: 2.5/2=1.25 grams left.
**
Your mistake is that you stopped at 16 days.
8 0
3 years ago
What part of the visible light spectrum produces the most light?
fomenos

Answer: all colors

.......

7 0
3 years ago
Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
stepan [7]
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
v_f=0 is the final speed of the sled
v_i=5 m/s is the initial speed
d is the distance covered

By rearranging the equation, we find d:
d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m
3 0
3 years ago
Sam and his Dad went grocery shopping. As they went up and down the aisle, adding items to their cart, something changed. The ca
Amanda [17]
B)
The speed of the cart changed because it stopped. 

Hope I could help!
-Marshy
8 0
3 years ago
Read 2 more answers
Capacitor C1 is initially charged to V1 and capacitor C2 is initially charged to V2. The capacitors are then connected to each o
o-na [289]

Answer:

<em>20.08 Volts</em>

Explanation:

<u>Parallel Connection of Capacitors</u>

The voltage across any two elements connected in parallel is the same. If the elements are capacitors, then each voltage is

\displaystyle V_1=\frac{Q_1}{C_1}

\displaystyle V_2=\frac{Q_2}{C_2}

They are both the same after connecting them, thus

\displaystyle \frac{Q_2}{C_2}=\frac{Q_1}{C_1}

Or, equivalently

\displaystyle Q_2=\frac{C_2Q_1}{C_1}

The total charge of both capacitors is

\displaystyle Q_t=Q_1\left(1+\frac{C_2}{C_1}\right)

We can compute the total charge by using the initial conditions where both capacitors were disconnected:

Q_t=V_{10}C_1+V_{20}C_2=25\cdot 24+13\cdot 11=743\ \mu C

Now we compute Q1 from the equation above

\displaystyle Q_1=\frac{Q_t}{\left(1+\frac{C_2}{C_1}\right)}=\frac{743}{\left(1+\frac{13}{24}\right)}=481.95\ \mu C

The final voltage of any of the capacitors is

\displaystyle V_1=V_2=\frac{481.95}{24}=20.08\ V

7 0
3 years ago
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