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yKpoI14uk [10]
2 years ago
13

4. Two scientists are experimenting with pure samples of isotope X, a highly radioactive substance. The first scientist has a sa

mple with a mass of 20 grams. He
measures a half-life of 232 seconds. The second scientist has a sample of the same substance with a mass of 80 grams. What is the half-life that she is most likely
to measure?
A. 58 seconds
B.232 seconds
C.464 seconds
D. 928 seconds
Physics
1 answer:
AlekseyPX2 years ago
4 0

Answer:

D

Explanation:

The substances are the exact same!

The half life should be the same too :)

A half life is how long it takes for a substance to break down and leave your system.

So for 20 grams, it goes away in 232 seconds, but for 80 grams, which is 4 times the amount, you would also multiply the half life by 4

I hope this is correct!

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An object travels at a speed of 7500 cm/sec . how far will it travel in kilometers in one day
DiKsa [7]

Answer:

6480 km

Explanation:

The speed of the object is

v = 7500 cm/sec

We need to convert centimetres into kilometers and seconds into days. We have:

1 cm = 1\cdot 10^5 km

1 s = \frac{1}{60\cdot 60 \cdot 24}d

Using these conversion factors, we find:

v=7500 \frac{cm}{s} \cdot 1\cdot 10^{-5} \frac{km}{cm}\cdot (24)(60)(60) \frac{s}{d}=6480 km

3 0
3 years ago
calculate the amount of work done by a person while taking a bag of mass 100kg to the top of the building hight 10m. The mass of
vredina [299]

Explanation:

Total mass=100+10=110

Total weight=mass×gravitational field strength

=110×10

=1100N

Work done=force×distance

=1100×10

=11000J

<em>Please mark me as brainliest if this helped you!</em>

6 0
2 years ago
Entropy measures _____.<br> a. disorder<br> b. energy<br> c. heat transferred<br> d. force
11111nata11111 [884]
Entropy measures disorder and nothing else
8 0
3 years ago
Read 2 more answers
A block rests on a frictionless table on Earth. After a 40-N horizontal force is applied to the block, it accelerates at 9.2 m/s
Paul [167]

Answer:

4.6 \frac{m}{s^2}

Explanation:

Since the table is frictionless, there is no force of dynamic  friction between table an block when the horizontal force is applied to it on Earth. Exactly the same is true when the table is taken to the Moon. Therefore, the Net Force acting on the object in both cases when the object accelerates, is the external horizontal force.

Notice that on Earth and on the Moon, the weight of the object (vertical and pointing up) is compensated by the normal force of the table on the object (pointing up and of the same magnitude as the weight) that precludes movement in the vertical direction. So in both cases, its acceleration will only be due to the horizontal force.

We use the equation for Net Force to find the mass of the object:

F=m*a\\40 N =m * 9.2 \frac{m}{s^2}\\\frac{40}{9.2} kg=m\\m=\frac{40}{9.2} kg

We use this mass (since the mass of the object is a constant independent of where the object is) to find the acceleration the object will experience when the 20 N horizontal force is applied on it on the Moon:

f=m*a\\20N=\frac{40}{9.2} kg*a\\20*9.2=40*a\\\frac{20*9.2}{40} =a\\a=\frac{9.2}{2} \frac{m}{s^2} \\ a=4.6 \frac{m}{s^2}

3 0
3 years ago
A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu
Len [333]

Answer:

0.4 m/s²

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 1 kg

Radius (r) = 0.2 m

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Tangential acceleration (aₜ) =?

Next, we shall determine the angular acceleration (a) of the sphere. This can be obtained as follow:

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20/10

a = 2 rad/s²

Finally, we shall determine the tangential acceleration (aₜ) of the sphere. This can be obtained as follow:

The tangential acceleration (aₜ) and the angular acceleration (a) are related according to the equation:

Tangential acceleration (aₜ) = Angular acceleration (a) × Radius (r)

aₜ = ar

With the above formula, we can obtain the tangential acceleration (aₜ) as follow:

Radius (r) = 0.2 m

Angular acceleration (a) = 2 rad/s²

Tangential acceleration (aₜ) =?

aₜ = ar

aₜ = 2 × 0.2

aₜ = 0.4 m/s²

Therefore, the tangential acceleration is 0.4 m/s²

6 0
3 years ago
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