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tiny-mole [99]
3 years ago
6

Calcium-51 has a half-life of 4.5 days. Only 0.75 gram remains of a sample that was initially 12 grams. How old is the sample of

calcium-51?
Physics
2 answers:
Over [174]3 years ago
6 0
How many times did the original sample lose 50% of its radioactivity ?

-- Start with. . . . . . . . . . . . 12 grams.
-- Lose half of it once. . . . . . 6 grams left.
-- Lose half of it again . . . . . 3 grams left.
-- Lose half of it again . . . . . 1.5 grams left.
-- Lose half of it again . . . . . 0.75 gram left.

-- How many times did it lose half ?    4 times.

-- How long does it take to lose half ?  4.5 days. 
                                                       (That's why it's called the 'half-life'.)

-- How long did it take to lose half, 4 times ?

                                           (4 x 4.5 days) = 18 days .
tatiyna3 years ago
3 0

Answer:

18 days

Explanation:

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Explain the relationship between magnetic fields and magnetic force.
andriy [413]

Answer:

Magnetic field is the strength of magnetism created by a magnet, whereas the magnetic force is the force due to two magnetic objects. The concepts of magnetic field and magnetic force are widely used in fields such as classical mechanics, electromagnetic theory, field theory and various other applications.

Explanation:

3 0
3 years ago
Construct a graph of position versus time for the motion of a dog, using the data in the table below. Explain how the graph indi
Lynna [10]

Answer:

The dog is moving at a constant speed

Explanation:

Given that,

Position : 5, 10, 15, 20, 25

Time = 5. 10, 15, 20, 25

We need to draw a position time graph

Using given data

A graph of position and time shows the speed.

According to graph,

The graph indicates that the dog is moving at a constant speed because the graph is straight line.

Hence, The dog is moving at a constant speed

6 0
3 years ago
A small plastic bead has been charged to -15nC.
marishachu [46]

Answer:

Explanation:

q = - 15 n C = - 15 x 10^-9 c

(a) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on proton, F = charge on proton x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (1.67 x 10^-27)

a = 1.6 x 10^14 m/s²

(B) the direction of acceleration is towards the bead, as the force is attractive.

(C) d = 0.9 cm = 0.009 m

electric field on proton, E = Kq/d²

E = 9 x 10^9 x 15 x 10^-9 / (0.009)²

E = 1.67 x 10^6 N/C

Force on electron, F = charge on electron x Electric field

F = 1.6 x 10^-19 x 1.67 x 10^6 = 2.67 x 10^-13 N

acceleration of proton, a = F / m

a = (2.67 x 10^-13) / (9.1 x 10^-31)

a = 3 x 10^17 m/s²

(D) the direction of acceleration is away from the bead, as the force is repulsive.  

8 0
3 years ago
To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit
DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

R = 350ft

a_t = 1.1ft/s^2

PART A )

a_c = \frac{V^2}{R}

a_c = \frac{V^2}{350}

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 5.25ft/s^2

a = \sqrt{a_t^2+a_r^2}

5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

27.5625 = 1.21 + \frac{v^4}{122500}

v=42.3877ft/s

Now calculate the angular velocity of the motorcycle

v = r\omega

42.3877 = 350\omega

\omega = 0.1211rad/s

Calculate the angular acceleration of the motorcycle

a_t = r\alpha

1.1 = 350\alpha

\alpha = 3.1428*10^{-3}rad/s^2

Calculate the time needed by the motorcycle to reach an acceleration of

5.25ft/s^2

\omega = \alpha t

0.1211 = 3.1428*10^{-3}t

t = 38.53s

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a_r = \frac{v^2}{R}

a_r = \frac{21.5^2}{350}

a_r =1.3207ft/s^2

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is 21.5ft/s

a = \sqrt{a_t^2+a_r^2}

a = \sqrt{(1.1)^2+(1.3207)^2}

a = 1.7187ft/s^2

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is 6.75ft/s^2

a = \sqrt{a_t^2+a_r^2}

6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}

45.5625 = 1.21 + \frac{v^4}{122500}

v=48.2796ft/s

3 0
3 years ago
A caterpillar tries to climb straight up a wall two meters high, but for every 2 cm up it climbs, it slides down 1 cm. Eventuall
MaRussiya [10]

Answer:

this question does not make sense could you please make it clearer

3 0
3 years ago
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