Burns are a particular safety hazard when handling

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We can begin by identifying key characteristics of both graphs.

Graph A.

Looking at the graph, we can see that the maximum distance (amplitude) is 10 cm (0.1 m). Additionally, its period (T) is 2 seconds (one full cycle).

We also know that:
[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

We can use this equation to compare with the other graph. Notice how the period does NOT depend on how far the spring is stretched. We can eliminate choice A for this reason.

Graph B.

The amplitude is 20 cm (0.2 m), and each period is 4 seconds.

We can now eliminate choice B because the springs are identical, so their spring constants are equal. Distance stretched has no impact on the spring constant.

For the other choices, we must look at forces and work.

Recall that:
Spring potential energy = [tex]\frac{1}{2}kx^2[/tex]

Kinetic energy = [tex]\frac{1}{2}mv^2[/tex]

Using the work-energy theorem:

[tex]\frac{1}{2}kx^2 = \frac{1}{2}mv^2[/tex]

Even if the mass of Block B is greater, its displacement is larger than that of Block A. Since displacement is squared in this equation, it would have a greater effect on the speed. Thus, Choice C is incorrect.

Using Hooke's Law:
[tex]\Sigma F = -kx\\\\ma = -kx\\[/tex]

[tex]a = \frac{-kx}{m}[/tex]

If the mass is greater, the acceleration will be smaller. Choice D is correct.

The torque experienced by the root of the tooth about point A when the steel band exerts the given force is 0.713 Nm.

The torque experienced by the root of the tooth about point A is calculated as follows;

τ = rFsinθ

where;

τ = (0.012) x (80) x (sin 48)

τ = 0.713 Nm

Thus, the torque experienced by the root of the tooth about point A when the steel band exerts the given force is 0.713 Nm.

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Answer:

our acceleration is 0 because there is a constant speed..

a = final speed - initial speed / time

final speed = initial speed ( since speed is constant)

therefore acceleration is 0

Answer:

it states that the atom will rest confortably in the ground state and will not collapse for lack of energy

Answer:

234 90 +energy

Explanation:

because alpha decrease by 4 2helium

i. The electron density of maganese is 2.44 × 10²⁸ electrons/m³

ii. The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

To solve the question, we need to know what electron density is.

This is the number of free electrons per unit volume of material.

To find electron density, we need to find the atom density which is the number of atoms per unit volume, n.

So, the atom density of manganese is n = Nρ/A where

So, n = Nρ/A

n = 6.022 × 10²³ atoms/mol × 7430 kg/m³/0.05494 kg/mol

n = 44743.46 × 10²³ atoms/mol × kg/m³/0.05494 kg/mol

n = 81440.59 × 10²³ atoms/m³

n = 8.144059 × 10²⁷ atoms/m³

n ≅ 8.14 × 10²⁷ atoms/m³

The electron density of maganese is 2.44 × 10²⁸ electrons/m³

So, the electron density of manganese n' = n" × n where

So, n' = n" × n

= 3 electrons per atom × 8.14 × 10²⁷ atoms/m³

= 24.42 × 10²⁷ electrons/m³

= 2.442 × 10²⁸ electrons/m³

≅ 2.44 × 10²⁸ electrons/m³

So, the electron density of maganese is 2.44 × 10²⁸ electrons/m³

The current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

The current flowing through the wire is given by i = n'eV/t where

So, substituting the values of the variables into the equation, we have

i = n'eV/t

i = 2.44 × 10²⁸ electrons/m³ × 1.602 × 10⁻¹⁹ C × 27 × 10⁻⁶ m³/5 s

i = 105.54 × 10³ C/5 s

i = 21.11 × 10³ C/s

i = 21.11 × 10³ A

i = 21.11 kA

So, the current that flows through a cylindrical manganese wire of volume 27 cm3 is 21.11 kA

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The speed of the wave with the given frequency and wavelength is 1.2m/s.

Given the data in the question;

Wavelength is  the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

It is expressed as;

[tex]\lambda = \frac{v}{f}[/tex]

Where v is velocity/speed and f is frequency.

Now, we can easily get the speed of the wave by substituting our given values into the expression above.

[tex]\lambda = \frac{v}{f} \\\\0.4m = \frac{v}{3s^{-1}}\\ \\v = 0.4m * 3s^{-1}\\\\v = 1.2ms^{-1}\\\\v = 1.2m/s[/tex]

Therefore, the speed of the wave with the given frequency and wavelength is 1.2m/s.

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Answer:

it is Newton's third law of motion?

Hope This Helps

The instant acceleration of the object is 10 m/s² and the displacement of the object is 228 m.

The instant acceleration of the object is determined by taking the derivative of the velocity as follows;

v = 8 + 2.5t²

a = dv/dt

a = 5t

a(2) = 5(2)

a(2) = 10 m/s²

The displacement of the object is determined by taking the integral of velocity as follows;

v =  8 + 2.5t²

x = ∫v

x = 8t + 2.5t³/3

x(0) = 0

x(6) = 8(6) + 2.5(6)³/3

x(6) = 228 m

Thus, the displacement of the object is 228 m.

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Answer:

4

Explanation:

We know that :

Solving

I believe the answer is 16.

 Have a blessed day!

The relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

The relative velocity of an object is the velocity of the object observed with respect to rest frame of another object.

The distance traveled by each object is determined from the prouduct of velocity and time of motion.

d = vt

where;

Thus, if the ball moved the same distance when it was on the flat bed truck, then the relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.

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Answer:

ill say math too!

Explanation:

Answer:

A

Explanation:

the atomic number means the number of protons so there are 50 protons

the number of positive charges (protons) should equal the number of negative charges (electrons) so there should be 50 electrons

the mass number = neutrons number + protons number

and we knew already that we have 50 protons so the number of neutrons is 125-50=75 neutrons

Answer:

The temperature, the density, and the element of stars.

Explanation:

The spectral line also can tell us about any magnetic field of the star. The width of the line can tell us how fast the material is moving. We can learn about winds in stars from this.

Answer:

very fast

Explanation:

Answer:

5.057 m/s^2

Explanation:

Force of kinetic friction = .3  = F /normal force

  .3 = F /(50* 9.81)     F of friction = 147.15

Net force =   400 - 147.5  =  252.85  N

F = m * a

252.85 = 50 * a       a = 5.057 m/s^2

Answer:

After 1 decay it has .9375 of its value left  

.9375^n = 1/2

What is the value of n (number of decays)

n log .9375 = -.693

n =.693 / .0645 = 10.7

After 10.7 periods of 50 sec it will have 1/2 of its original nuclei left

1/2 life = 10.7 * 50 sec = 537 sec = 8.95 min

Answer:

Yes, single-use batteries are now made of common metals deemed non-hazardous by the federal government and can be disposed of in your regular trash in all states except California, where it is illegal to throw away all types of batteries.

Answer:

m(P4) =  46.175 (grams)

m (KClO3) = 149 (grams)

Explanation:

1) n(P4) = n(P4O10);

m(P4)/M(P4) = m(P4O10)/M(P4O10);

m(P4) = M(P4)*m(P4O10)/M(P4O10)
          = 123.90*105.8/283.89
          = 46.175 (grams)

2) Analogously, 10n(P4O10) = 3n(KClO3)

m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)
                = 10*122.55*105.8/283.89/3
                = 149 (grams).

149 gm of each reactant is needed to produce 105.8 gm of tetra phosphorous decaoxide .

In a chemical equation, the substance or substances to the left of the arrow are referred to as reactants. A substance that is present when a chemical reaction first begins is known as a reactant.

Products refer to the material or substances to the right of the arrow. A substance that is present after a chemical reaction is complete is called a product.

Given Data:

m(P4) =  46.175 (grams)

m (KClO3) = 149 (grams)

Then it is given by:

1) n(P4) = n(P4O10);

m(P4)/M(P4) = m(P4O10)/M(P4O10);

The equation becomes :

m(P4) = M(P4)*m(P4O10)/M(P4O10)

         = 123.90*105.8/283.89

         = 46.175 (grams)

2) Analogously, 10n(P4O10) = 3n(KClO3)

m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)

               = 10*122.55*105.8/283.89/3

               = 149 (grams).

Thus, 149 gm of each reactant is needed to produce 105.8 gm of tetra phosphorous decaoxide .

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# SPJ2

Answer:

Transverse wave

Explanation:

I hope this helps you! :)

[tex]\qquad\qquad\huge\underline{{\sf Answer}}♨[/tex]

Gravitational Potential of an object with mass m, and height of h metre is :

[tex]\qquad \sf  \dashrightarrow \:mgh[/tex]

Now, if the mass of object is 2100 kg, height of 18 m is placed in there, then it's gravitational potential energy is ~

[tex]\qquad \sf  \dashrightarrow \:2100 \times 10 \times 18[/tex]

[tex]\qquad \sf  \dashrightarrow \:378000 \: \: joules[/tex]

[tex]\qquad \sf  \dashrightarrow \:3.78 \times 10 {}^{5} \: \: joules[/tex]

Answer:

0.1111 W/m²

Explanation:

If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,

I ∝ (1/r²)

I = k/r²

Since k can be the constant of proportionality. k = Ir²

We can write this relation as

I₁ × r₁² = I₂ × r₂²

I₁ = 0.25 W/m²

r₁ = 16 m

I₂ = ?

r₂ = 24 m

0.25 × 16² = I₂ × 24²

I₂ = (0.25 × 16²)/24²

I₂ = 0.1111 W/m²

Answer:

Explanation:

The position between point charge q to x = -8.0 m and y = 1.5 m is:

[tex]r=\sqrt{x^{2}+y^{2}}=\sqrt{(-8)^{2}+(1.5)^{2}}=\sqrt{66.25}\approx 8.13[/tex] meter

Then the magnitude of the electric field E is:

[tex]E=\frac{1}{4\pi\epsilon_{o}} \frac{q}{r^{2}}=(9\times 10^{9}) \frac{6.0\times 10^{-9}}{66.25}\approx 0.81 N/C[/tex]

For the vector of E:

[tex]\tan\theta=\frac{1.5}{-8}=-0.1875 \rightarrow \theta = -10.61^{0}[/tex]

[tex]E_{x}=E\cos (-10.16)=(0.81)(0.984)=0.79704 N/C[/tex]

[tex]E_{y}=E\sin\theta = -0.81(0.184)=-0.14904 N/C[/tex]

[tex]\vec{E}=0.79704\hat{i}-0.14904\hat{j}[/tex] in N/C

Question :-

Answer :-

Explanation :-

As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .

For calculating the Force , we will use the Formula :-

[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]

Therefore , by Substituting the given values in the above Formula :-

[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]

[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]

[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]

Hence :-

[tex] \underline {\rule {185pt}{4pt}} [/tex]