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Dennis_Churaev [7]
3 years ago
14

A 12.65 g sample of a radioactive substance is allowed to decay for 17.22 min. At that time, the sample weighed 3.115 g. What is

the half-life for this substance?
Physics
1 answer:
Mademuasel [1]3 years ago
8 0

Answer:

8.61 min

Explanation:

original mass= 12.65

first half life = 12.65/2 = 6.325

second half life = 6.325/2 = 3.1625

Note : 3.1625 is the closest to the value (3.115) given so we work with it

total time for decay =17.22

therefore two decays = 17.22/2= 8.61

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"<span>H-C=N:" is the one answer among the choices given in the question that shows the correct dot diagram. The correct option among all the options that are given in the question is the fourth option or option "D". The other choices can be neglected. I hope that this is the answer that has come to your help.</span>
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3 years ago
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A car in motion has kinetic energy. What happens to the kinetic energy when the car brakes to a stop? The kinetic energy is tran
Pie

Answer:

<em>When a moving car brakes to a stop the </em><em>kinetic energy of the car is converted to heat energy. </em>

Explanation:

A moving car has kinetic energy.

It is given by the equation k=\frac{1}{2} mv^2

Where m denotes mass of the car and v denote sits velocity. When the brakes are applied the velocity becomes zero and the car doesn’t possess kinetic energy anymore.

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7 0
3 years ago
imagine you are going on a rid in a spacecraft next to earth. Your trip takes one whole year. Describe earth's tilt in the north
mylen [45]
You will have to fly around the whole earth to get to your landing station
8 0
3 years ago
Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a
suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N

Part (b)

the magnitude of each charge, if they have equal magnitude

F = \frac{KQ^2}{r^2}

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q =  \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C

4 0
3 years ago
A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed
Mashutka [201]

(a)

consider the motion of the tennis ball. lets assume the velocity of the tennis ball going towards the racket as positive and velocity of tennis ball going away from the racket as negative.

m = mass of the tennis ball = 60 g = 0.060 kg

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v = final velocity of the tennis ball after being hit by racket = - 39 m/s

ΔP = change in momentum of the ball

change in momentum of the ball is given as

ΔP = m (v - v₀)

inserting the above values

ΔP = (0.060) (- 39 - 20)

ΔP = - 3.54 kgm/s

hence , magnitude of change in momentum : 3.54 kgm/s

7 0
3 years ago
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