half-life questionnnnn:

Answer:

I believe the answer is speed up.

Explanation:

this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll

Explanation:

a) Here is the free-body diagram. Note that I included the components of the weight mg (shown in dotted arrows) for use in the other parts of the problem.

b) The component of the weight parallel to the plane (shown in the diagram as a dotted arrow along the x-axis) is [tex]mg\sin15[/tex] and it is equal to

[tex]mg\sin15 = (25\:\text{kg})(9.8\:\text{m/s}^2)\sin15 = 63.4\:\text{N}[/tex]

c) Applying Newton's 2nd law to the y-axis, we can write

[tex]y:\;\;\;N - mg\cos15 = 0 \Rightarrow N = mg\cos15[/tex]

[tex]N = (25\:\text{kg})(9.8\:\text{m/s}^2)\cos15 = 236.7\:\text{N}[/tex]

d) The component of the weight mg into the plane is the same as the normal force, hence it's also 236.7 N.

e) To solve for the coefficient of friction, we apply Newton's 2nd law to the x-axis:

[tex]x:\;\;\;mg\sin15 - F_f = 0[/tex]

[tex]\Rightarrow F_f = mg\sin15\;\;(2)[/tex]

where [tex]F_f[/tex] is the frictional force defined as [tex]F_f = \mu N[/tex] so we can use Eqn(1) on Eqn (2) to write

[tex]\mu (mg\cos15) = mg\sin15[/tex]

Solving for [tex]\mu,[/tex] we get

[tex]\mu = \dfrac{\sin15}{\cos15} = \tan15 = 0.27[/tex]

Answer:

Weight of object = 11.2 N

Apparent weight = 3.83 N     when immersed

Weight of water displaced = 11.2 - 3.83 = 7.37 N      

d (density) W / V       weight / volume      the weight density

Wo = Vo do    weight of object

Ww = Vo dw    where Ww is weight of equivalent volume of water = 7.37

Wo / Ww = do / dw    dividing previous equations

do = 11.2 / 7.37 dw = 1.52 dw

The density of the object is 1.52 that of water

The density of water is 1000 kg / m^3 * 9.8 m/s^2 = 9800 N/m^3

So the weight density is 14900 N/m^3

An irregularly shaped object weighs 11.20 N in air. When immersed in water, the object has an apparent weight of 3.83 N. It's density can be calculated as 1523 kg/m³.

To find the density, the given values are,

Weight in air = 11.20 N

Weight in water = 3.83 N

density of water = 1000 kg/m³

According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.

Loss in weight of the object = Weight of object in air - weight of object in water

Loss in weight = 11.20 - 3.83 = 7.37 N

Volume of body x density of water x g = 7.37

Let V be the volume of body

V x 1000 x 9.8 =7.37

V = 7.5× 10⁻⁴ m³

Weight in air = Volume of body x density of body x g

11.20 = 7.5× 10⁻⁴ x d x 9.8

d = 1523 kg/m³.

Thus, the density of body is 1523 kg/m³.

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To create this comic strip you can use a narration describing each step and illustrate each one with one image or drawing.

Creating a comic strip involves using images and short texts to explain a specific idea or phenomenon. In the case of the process for a meteor to enter Earth you can use the following ideas.

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Answer:

Nuclear medicine or PET scanning (Positron Emission Tomography)

Explanation:

Not super sure what your question is looking for but I think it can be either. PET is a technology used in the nuclear medicine field, and nuclear medicine is a broad field of using the technology the question described.

Answer:

Explanation:

                            F = ma

mgsinθ - μmgcosθ = ma

      gsinθ - μgcosθ = a

                   μgcosθ = gsinθ - a

                              μ = (gsinθ - a) / gcosθ

                              μ = (9.81sin24 - 0.245) / 9.81cos24

                              μ = 0.4178906...

                              μ = 0.418

The coefficient of kinetic friction will be equal to 0.418.

Friction is the force that prevents solid surfaces, fluid layers, and material elements from sliding against each other. There are various kinds of friction: Dry friction is the force that opposes the relative lateral motion of two in-touch solid surfaces.

Given that a 1.95 kg box sits on an incline of 24 ° with the horizontal. If the box accelerates down the incline at 0.245 m/s 2

The coefficient of kinetic friction will be calculated as,

F = ma

mgsinθ - μmgcosθ = ma

gsinθ - μgcosθ = a

μgcosθ = gsinθ - a

Solve for the value of the coefficient of friction,

μ = (gsinθ - a) / gcosθ

μ = (9.81sin24 - 0.245) / 9.81cos24

μ = 0.4178906...

μ = 0.418

Therefore, the coefficient of kinetic friction will be equal to 0.418.

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Answer:

Weathering and erosion

Explanation:

Answer:

True;   ar = v^2 / R      Radial acceleration because it moves in a circular path

    at = α R = tangential acceleration because its speed changes

a = at + ar    total acceleration equals sum of radial and tangential

Answer:

Energy is conserved.

Explanation:

Without friction, kinetic energy and gravity potential energy are always swapping back and forth.

The block slides down an inclined friction-less track to a infinite hill that the block begins to climb and slow down, in which the energy is conserved.

Friction refers to the force that prevents hard surface, fluid layers, as well as material elements from sliding against each other.

There are various kinds of friction: Dry friction is the force that opposes the relative lateral motion of two in contact solid surfaces.

It is thought to be caused by the interactions of tiny bumps on surfaces rubbing against one another.

The bumps on each surface bend and exert force on one another, making it difficult for the surfaces to slide over one another.

A block's speed increases while its acceleration remains constant as it slides down a frictionless incline. After being placed on a ramp inclined at an angle with the horizontal, a crate remains stationary.

Thus, the block slides beneath an inclined friction-less track to an unlimited hill, where it commences to climb as well as slow down, preserving energy.

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Answer:

18.36 m/s

Explanation:

We can solve this using conservation of energy. The energy in the system will be conserved since there are no outside forces acting upon it so the potential energy and kinetic energy will be equal. Giving us this formula to start:

1/2mv^2=mgh

m=mass

g=gravity

h=height

v=velocity

We can start by figuring out the total height the rock travels which we can do by subtracting the height of the frisbee by the height the rock started at.

19m-1.8m=17.2m

Now we can plug in our variables to solve for velocity.

First we negate mass since its on both sides and cancels out leaving us with.

1/2v^2=gh

Plug in.

1/2v^2=(9.8)(17.2)

1/2v^2=168.56

v^2=337.12

v=18.36m/s

Answer:

1/2 m v^2 + 1/2 I ω^2 = m g h       conservation of energy

I = 2/5 m R^2     inertia of solid sphere

1/2 m v^2 + 1/5 m ω^2 R^2 = m g h

1/2 v^2 + 1/5 v^2 = g h

v^2 = 10 g h / 7 = 1.43 * 9.80 * 19 m^2/s^2 = 266 m^2/s^2

v = 16.3 m/s

v = R ω

ω = 16.3 / .6 = 27.2 / sec

Answer:

The answer would be A. The waves that heat a cup of water in a microwave oven

Answer:

pressure in liquids is given as:

P= hpg

where h is the depth

where p is the density

where g is 10

Explanation:

From the formula above

p = 10 X 1000 X 10

p = 100000N/m

The tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex]

The given parameters:

The net force on the block system is calculated by applying Newton's second law of motion as follows;

Total mass of the block-system = 2 kg + 5 kg = 7 kg

The acceleration of the block-system;

[tex]a = \frac{T_i}{7}[/tex]

The tension in the cable between the blocks, T₂, is calculated as;

[tex]T_2 = m_2 a\\\\T_2 = 2a\\\\T_2 = 2 \times \frac{T_i}{7} \\\\T_2 = \frac{2T_i}{7}[/tex]

Thus, the tension in the cable between the blocks, T₂, is [tex]\frac{2T_i}{7}[/tex].

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Answer:

I use mercury thermometer to measure the temperature above 80°C because the boiling point of alcohol is 78°C . So, it can't measure the temperature above 78°C.

18,000 degrees in one second i believe

Answer:a)   λ = 4.862 10⁻⁷ m,  b)  λ = 4.341 10⁻⁷ m

Explanation:

The spectrum of hydrogen can be described by the expression

in the case of the initial state n = 2 this series is the Balmer series

a) Find the wavelength for n = 4

let's calculate

          = 1,097 10⁷ ()

         \frac{1}{ \lambda} = 1.097 10⁷ 0.1875 = 0.2056 10⁷

          λ = 4.862 10⁻⁷ m

b) n = 5

    \frac{1}{ \lambda} = 1,097 10⁷ ()

    \frac{1}{ \lambda} = 1.097 10⁷ 0.21 = 0.23037 10⁷

     λ = 4.341 10⁻⁷ m

Answer:

analyze, assess or interpret an historical event, era, or phenomenon,.

Explanation:

Secondary sources are works that analyze, assess or interpret an historical event, era, or phenomenon, generally utilizing primary sources to do so. Secondary sources often offer a review or a critique. Secondary sources can include books, journal articles, speeches, reviews, research reports, and more.

Answer:

Explanation:

acceleration phase

average speed was 21.9/2 = 10.95 m/s

distance covered is 10.95 m/s(16.8 s) = 183.96 m

distance at top speed 21.9 m/s(16.4 s) = 359.16 m

distance while decelerating (0² - 21.9²)/(2(-1.8)) = 133.225 m

total = 183.96 + 359.16 + 133.225 = 676.345 = 676 m

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

From the parameters given:

By replacing the given values into the above equation, we can determine the acceleration.

∴

625 N = 105 kg × acceleration.

[tex]\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}[/tex]

acceleration = 5.95 N/kg

acceleration = 5.95 m/s²

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The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:

[tex]v = 2\pi\cdot r\cdot f[/tex] (1)

Where:

If we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:

[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]

[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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Answer:  Your answer Is A)

Explanation:    

Its direction of deflection shows it possitively charged

It brings one element into another by bombardment(transmutation)

Hope you could get an idea from here.

Doubt clarification - use comment section.

Answer:

Explanation:

15 km(1000 m / km) / (20 min(60 s/min)) = 12.5 m/s

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

Answer:

26 g's

Explanation:

Hope this helps~

Have a great day

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