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devlian [24]
3 years ago
9

Wagonium-292 has a half-life of 1 hour. If you started with an 80 gram sample, how much Wagonium-292 will remain after 4 hours?

A.) 10g B.) 40g C.) 160g D.) 5G
Physics
1 answer:
viva [34]3 years ago
6 0

After 1 hour, 80 g decays to 40 g.

After another hour (total 2 hours), 40 g decays to 20 g.

After another hour (total 3), 20 g decays to 10 g.

After one more hour (total 4), 10 g decays to (D) 5 g.

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A car accelerates at 3 m/s*2. Assuming the car starts from rest, how much time does it need to
uranmaximum [27]
<h3><u>Given</u><u>:</u><u>-</u></h3>

Acceleration,a = 3 m/s²

Initial velocity,u = 0 m/s

Final velocity,v = 12 m/s

<h3><u>To</u><u> </u><u>be</u><u> </u><u>calculated:-</u><u> </u></h3>

Calculate the time take by a car.

<h3><u>Solution:-</u><u> </u></h3>

According to the first equation of motion:

v = u + at

★ Substituting the values in the above formula,we get:

⇒ 12 = 0 + 3 × t

⇒ 12 = 3t

⇒ 3t = 12

⇒ t = 12/3

⇒ t = 4 sec

5 0
4 years ago
Read 2 more answers
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
LUCKY_DIMON [66]

Answer:

e.)At twice the distance, the strength of the field is E/4.

Explanation:

The strength of the electric field at a certain distance from a point charge is given by:

E=k\frac{Q}{r^2}

where

k is the Coulomb's constant

Q is the charge

r is the distance from the point charge

In this problem, the distance from the point charge is doubled:

r' = 2r

So the new electric field strength is

E'=k\frac{Q}{(2r)^2}=k \frac{Q}{4 r^2}=\frac{1}{4} (k\frac{Q}{r^2})=\frac{E}{4}

so, at twice the distance the strength of the field is E/4.

4 0
3 years ago
14. Three identical light bulbs are connected in series, then are disconnected and arranged in parallel. For each of the scenari
Inessa [10]
In order to make things easier to describe and explain, let's call
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b).  In series, the total current is  V / (3R) .
In parallel, the total current is  3V / R .
Changing from series to parallel, the total current in the circuit
increases to 9 times its original value.

c).  In series, the power dissipated by the circuit is 

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In parallel, the power dissipated by the circuit is

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Changing from series to parallel, the power dissipated by
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8 0
3 years ago
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Answer:

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