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notsponge [240]

2 years ago

5

What are the different skills and strategies were you aware of in this situation that indicated to you that it would be appropri

ate to apply past knowledge?

1 answer:

const2013 [10]2 years ago

4 0

Answer:

dad

Explanation:

daddy

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A sample of shale contains 0.03% 238U by weight. Calculate the number of spontaneous fissions in one day in a 106 kg pile of the

kumpel [21]

Answer:

a)mass = 300kg

b) no of atoms = 1.81 x 10^26atoms.

c)Fission Activity = 207Bq

d) no of fissions = 1.79 × 107 d^-1

Explanation:

a)mass= (3×10-4)(10^6kg) = 300kg

b) no of atoms = (300kg) X (6.022×10^23/kg) = 1.81 x 10^26atoms.

c)Fission Activity = (0.69)(300kg) = 207Bq

d) no of fissions = (207)(86400) = 1.79 × 107 d^-1

5 0

3 years ago

A football is kicked straight up from a height of 5 feet with an initial speed of 55 feet per second. The formula h equals negat

makvit [3.9K]

Answer:

3.53 second

Explanation:

The formula for the height is

$h=-16t^{2}+55t+5$

When it hits the ground, the height is zero.

So, put h = 0 in the above equation

$0=-16t^{2}+55t+5$

$16t^{2}-55t-5=0$

$t=\frac{+55\pm \sqrt{55^{2}+4\times 5\times 16}}{2\times 16}$

$t=\frac{+55\pm 57.84}{2\times 16}$

Take positive sign

t = 3.53 second.

Thus, the time taken to hit the ground is 3.53 second.

8 0

3 years ago

A mass of 2.50 kg is in a gravitational field of 14.0 N/kg. What force acts on the mass?

Olenka [21]

To determine the force that acts on the mass, just multiply the mass by the gravitational field. Using the given data,
F = (2.50 kg)(14 N/kg) = 35 N
Therefore, the force that acts on the mass is equal to 35 N.

7 0

3 years ago

Read 2 more answers

Point A and B located at 4 meters and 9 meters from a source of the sound. If IA and IB are intensity at point A and point B, th

elena-14-01-66 [18.8K]

The intensity ratio at point A and B will be 81:16.

<u>Explanation:</u>

Sound waves are known to get faded with increase in the distance. This is because, the intensity of the sound is inversely proportional to the square of the distance of the source from the observer. So, if an observer is standing greater distance from the source of the sound, he/she will find difficulty in hearing the sound.

So, as the distance between the source and observer increases, the intensity of the sound wave decreases.

$I = \frac{1}{r^{2} }$

As here two points A and B are located at 4 m and 9 m distance from the source, then the intensity of sound at A and B will be inversely proportional to their respective square of the distance as shown below.

$I_{A} =\frac{1}{r_{A}^{2} } = \frac{1}{4 \times 4}=\frac{1}{16}$

Similarly,

$I_{B} =\frac{1}{r_{B}^{2} } = \frac{1}{9 \times 9}=\frac{1}{81}$

So, the ratio of intensity at point A and B will be

$\frac{I_{A} }{I_{B} } = \frac{81}{16} =81:16$

Thus, the intensity ratio at point A and B will be 81:16.

7 0

3 years ago

What si unit replaces pound

Artist 52 [7]

The SI unit of mass is the kilogram, and
the SI unit of force is the Newton.

8 0

4 years ago