Sugar dissolves in water by ______ or breaking into smaller pieces that spread throughout the water.

A sample contains 16 g of a radioactive isotope. how much radioactive

mafiozo [28]

After one half-life, 8 g of radioactive isotope will remain in the sample.

<h3>What is radioactivity?</h3>

The act of producing radiation spontaneously is known as radioactivity. This is accomplished by an unstable atomic nucleus that want to give up some energy in order to move to a more stable form.

The following formula is used to compute the number of half lives elapsed:

$\rm N=\frac{N_0}{2^n} \\\\ N=\frac{16}{2} \\\\ N= 8 \ gram$

Hence,8 gram of radioactive isotope remains in the sample after 1 half-life.

To learn more about the radioactivity, refer to the link;

brainly.com/question/1770619

#SPJ1

7 0

2 years ago

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Which of the angles shown in this picture is the angle of incidence? A B

Dmitry_Shevchenko [17]

The pictures are not attached, therefore, I cannot give a specific choice.
However, I will try to help you out.

The angle of incidence is defined as the angle formed between the ray of light and the normal to the surface that the ray is falling on.

The angle of incidence can be shown in the attached image.

Therefore, for your question, choose the image on which the above description applies.

Hope this helps :)

6 0

3 years ago

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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

What is the mass in kg of a leopard in a tree if the tree branch is 36 m up and the leopard's gravitational potential energy is

ElenaW [278]

Answer:

83.3kg

Explanation:

GPE = m × g × h

GPE = mass of leopard × 10 × 36m

29988J = 360 × mass

mass = 83.3kg

5 0

3 years ago

A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori

Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²
 That's about 2.634 m/s² 

(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = 2.45 N.

That's the force that's accelerating the little block, so that must be the tension
in the string.

7 0

3 years ago