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adoni [48]
3 years ago
10

The half-life of Co-55 is 175 hours. How much of a 4000 g of Cobalt-55 sample would be left after 525 hours?

Physics
1 answer:
Harrizon [31]3 years ago
5 0

We know that whatever amount we start with, half of it decays and forms atoms of other elements in 175 hours.  So in order to figure out how much is left after 525 hours, we'll need to know how many half-lifes pass in that amount of time.

Well, (525 divided by 175) is exactly 3 half-lifes.  So this will be easy.

-- After 1 half-life . . .

. . . . . 50% decays, 50% is still there.

-- After the 2nd half-life . . .

. . . . . (half of the leftover 50%) = another 25% decays, 25% is left.

-- After the 3rd half-life . . .

. . . . . (half of the leftover 25%) = another 12.5% decays, 12.5% is left.

12.5% of 4,000g = (0.125 x 4,000g) = <em>500 g</em> .

============================================

<u>Another way</u>:

After 1 half-life, 1/2 is left.

After 2 half-lifes, 1/4 is left.

After 3 half-lifes, 1/8 is left.

1/8 of 4,000g = (4,000g/8) = <em>500 g </em>.

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goldfiish [28.3K]

Answer:

Mass of the body will be = 6 kg

Explanation:

Given:

Mass of an object = 6 kg

Acceleration due to gravity of the planet = 10 m/s^2

To find the mass of the body on that planet.

Solution:

Mass of the body is defined as the total amount of matter contained in the body.

Thus, mass of a body will always remain constant irrespective of the acceleration due to gravity. This is because it is an independent quantity and does not vary with acceleration due to gravity.

<em>It is the weight of the body that changes with the change in the acceleration due to gravity as it is given by:</em>

W=mg

where m represents mass of the body and g represents the accelration due to gravity.

Hence, the mass of the given body will remain = 6 kg.

6 0
3 years ago
A water discharge of 9 m3/s is to flow through this horizontal pipe, which is 0.98 m in diameter. If the head loss is given as 1
Mashutka [201]

Answer: The power required by the pump to produce a discharge of 9m³/s is 5990joules/secs.

Explanation: The given parameters from the questions are:

Flow rate Q = 9m³/s, Diameter D = 0.98m, acceleration a = 1.0, head loss(Pressure P) is given by the function 10v²/2g.

STEP 1. Find the velocity of water in the pipe from the equation:

Diameter D = (√4.Q/π.v), where v is the velocity, and Q is flow rate

Making v subject of the formula gives:

v = 4Q/π.√D =[ 4 × 9m³/s / 3.142 × (√0.98m)] = 11.69m/s.

STEP 2. Find the pressure from the relationship, P = 10v²/2g, NB. g = a

P = 10 × (11.69m/s)² / 2× 1.0m/s²

P = 683.25N/m² or Pascal.

STEP 3. Find force exerted by the pump;

Recall that Pressure P = Force/Area

But Area A = π.r², where r = D/2

Therefore, A = π.(D/2)²

A = 3.142 × [0.98m/2]² = 0.75m²

Therefore, Force = Pressure × Area

Force F = 683.25N/m² × 0.75m²

F = 512.44N.

STEP 4. Find work done

Work done W by the pump is = Force × distance d moved by the water

W = F . d

Also recall that flow rate Q = Velocity/time.

Q = v/t, we can write t = v/Q.

Time t = 11.69m/s / 9m³/s = 1.298s

Also recall that velocity v = distance d/time t, v = d/t, making d subject of formula gives v × t

Distance d = v × t = 11.69m/s × 1.298s = 15.17m.

Hence,

Work Done W = Force × distance

W = 512.44N × 15.17m = 7775.56Nm or joules.

Lastly, Power P = Work done/ time

P = 7775.56joules/1.298s

P = 5990.4joules/s.

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A golf ball is whacked in a direction 25 degrees south of the east axis. The ball travels 125m. What are the east and north comp
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<u>We are given:</u>

Direction of motion: 25 degrees south of the east axis

Distance covered  = 125 m

<u>East component of the Ball:</u>

<em>this component is denoted by green color in the image</em>

Once we drop a perpendicular from the end of the direction vector on the x-axis, we get a right angled triangle

The magnitude of the side of the triangle on the x-axis denotes the east component of the ball

Using trigonometry, we find that the east component of the ball is:

125 * Cos(25 degrees)

125 * 0.9 = 112.5 i   (here, i denotes rightward direction on the x-axis)

<u />

<u>North Component of the Ball:</u>

<em>this component is denoted by blue color in the image</em>

Using trigonometry, we find that the North component of the ball is:

125* Sin(25 degrees)  (-j)      <em>[j denotes upward movement on the y-axis, since the vector is acting downwards, we have used '-j']</em>

125 * 0.42 (-j)

52.5 (-j) =   -52.5 j

Therefore the direction vector of the ball is 112.5 i - 52.5 j

<em>where 112.5 i is the East Component and -52.5 is the North Component</em>

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