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2 years ago

After two hours, 1/16 of the initial amount of a certain radioactive isotope remains undecayed. The half-life of the isotope is:

1 answer:

7 0

The half life is 30 minutes.
30 mins- 1/2 left
60 mins- 1/4 left
90 mins- 1/8 left
120 mins- 1/16 left
120 mins= 2 hours

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Two airplanes leave an airport at the same time. The velocity of the first airplane is 650 m/h at a heading of 60.5 ◦ . The velo

Assoli18 [71]

First, illustrate the problem as shown in the attached picture. Next, let's find the distance traveled by planes A and B after 2.9 h.

Distance of A: 650 m/h * 2.9 h = 1,885 m
Distance of B: 560 m/h * 2.9 h = 1,624 m

Then, we use the cosine law to determine the distance x. The angle should be: 85 - 60.5 = 24.5°

x² = 1,885² + 1,624² - 2(1,885)(1,624)(cos 24.5°)
x = √619381.3183
x = 787 m

7 0

3 years ago

To determine an athlete's body fat, she is weighed first in air and then again while she's completely underwater. it is found th

il63 [147K]

The difference in weight is due to the displacement of water (the buoyancy of water is acting on the athlete thus giving her smaller weight).

The amount of weight displaced or the amount of buoyant force is:

Fb = 690 N - 48 N

Fb = 642 N

From newtons law, F = m*g. Using this formula, we can calculate for the mass of water displaced:

m of water displaced = 642N / 9.8m/s^2

m of water displaced = 65.5 kg

Assuming a water density of 1 kg/L, and using the formula volume = mass/density:

V of water displaced = 65.5kg / 1kg/L = 65.5 L

The volume of water displaced is equal to the volume of athlete. Therefore:

V of athlete = 65.5 L

The mass of athlete can also be calculated using, F = m*g

m of athlete = 690 N/ 9.8m/s^2

m of athlete = 70.41 kg

Knowing the volume and mass of athlete, her average density is therefore:

average density = 70.41 kg / 65.5 L

average density = 1.07 kg/L = 1.07 g/mL

5 0

3 years ago

Read 2 more answers

HELPPP IM DESPERATE

damaskus [11]

the first one cuz I know

5 0

2 years ago

interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista

steposvetlana [31]

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

8 0

3 years ago

Two resistors, of R1 = 3.93 Ω and R2 = 5.59 Ω, are connected in series to a battery with an EMF of 24.0 V and negligible interna

son4ous [18]

Answer:

2.521 (A); 14.0924 (V)

Explanation:

more info in the attachment, the answers are marked with red colour.

4 0

2 years ago