Saturn has a radius of about 9.0 earth radii, and a mass 95 times the Earth’s mass. Estimate the gravitational field on the surface of Saturn compared to that on the Earth. Show your work.

Answer:

n=5, l=1, m(l) = -1, m(s)= + 1/2

Explanation:

Quantum number are used to describe the position and spin of an electron inside an atom. There are four types of quantum number for describing an electron inside an atom. They are: the principal quantum number, spin quantum number, magnetic quantum number and angular momentum quantum number.

(1).PRINCIPAL QUANTUM NUMBER: denoted by n, and has possible values of n= 1,2,3,4,.... IN HERE, n= 5

(2).ANGULAR MOMENTUM QUANTUM NUMBER: it is denoted by l, and has possible values of l= 0,1,2,3,...,(n-1).

Our l here is one( that is, s-orbital=0, p-orbital=1, d-orbital= 3 and so on)

(3).MAGNETIC QUANTUM NUMBER: The magnetic quantum number, which is denoted by m subscribt l, specifies the exact orbital in which you can find the electron. It has values ranging from -l,...,-1,0,1,...,l.

Here, our value is -1 that is m(l)= -1

(4).SPIN QUANTUM NUMBER: describes the orientation of electrons. Electrons can only have two values here, either a positive one and the half(+1/2) that is the spin up electron or the negative one and half(-1/2) that is the spin down electron.

A set of quantum numbers for an electron in a 5p orbital could be: principal quantum number (n) = 5, angular momentum quantum number (l) = 1, magnetic quantum number (m) = 0, spin quantum number (m_s) = -1/2.

An electron in a 5p orbital can be described using four quantum numbers. Knowing that p orbitals correspond to the angular momentum quantum number l=1, and the principal quantum number n for the fifth energy level is 5, we can deduce the following. The magnetic quantum number, m, ranges from -l to +l, and thus it can have the values -1, 0, or +1 representing different p orbitals in the same shell. The spin quantum number, m_s, can either be -1/2 or +1/2, indicating the orientation of the electron's spin. Thus, an example of a set of four quantum numbers that could be assigned to an electron occupying a 5p orbital could be: (5, 1, 0, -1/2).

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Answer:

1. [tex]v=14.2259\ m.s^{-1}[/tex]

2. [tex]F_T=25.8924\ N[/tex]

3. [tex]\lambda=29.6373\ m[/tex]

Explanation:

Given:

1.

[tex]\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel[/tex]

[tex]v=\frac{6.8}{0.478}[/tex]

[tex]v=14.2259\ m.s^{-1}[/tex]

2.

Now, we find the linear mass density of the slinky.

[tex]\mu=\frac{m}{L}[/tex]

[tex]\mu=\frac{0.87}{6.8}\ kg.m^{-1}[/tex]

We have the relation involving the tension force as:

[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]

[tex]14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }[/tex]

[tex]202.3774=F_T\times \frac{6.8}{0.87}[/tex]

[tex]F_T=25.8924\ N[/tex]

3.

We have the relation for wavelength as:

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

[tex]\lambda=\frac{14.2259}{0.48}[/tex]

[tex]\lambda=29.6373\ m[/tex]

Answer:

h should become four times.

Explanation:

The speed at the bottom depends only on the vertical height difference from the top ... if there is no friction ... then the path you take from top to bottom doesn't matter in any way, only the difference in height between the two

mgh at top = 1/2 mv^2 at bottom, so that v=sqrt[2gh]

You see, the final velocity, v, depends on the square root of h, so double v, quadruple h, or four times higher

Answer:

Glycosidic Bonds

Explanation:

When two adjacent mono-saccharide units link to form di-saccharides and this repeated chain link forms poly-saccharide. These reactions are called dehydration or condensation reactions because upon the formation of the bond the water molecule is released in the form of a byproduct. Glycosidic bonds are covalent bonds that link ring-shaped sugar molecules to other molecules.

Answer:

The rise in temperature is [tex]43.98^{\circ}C[/tex]

Solution:

As per the question:

Mass of hammer, M = 1.30 kg

Speed of hammer, v = 7.3 m/s

Mass of iron, [tex]m_{i} = 14\ g[/tex]

No. of blows, n = 8

Specific heat of iron, [tex]C_{i} = 450\ J/kg.^{\circ}C = 0.45\ J/g^{\circ}C[/tex]

Now,

To calculate the temperature rise:

Transfer of energy in a blow = Change in the Kinetic energy

[tex]\Delta KE = \frac{1}{2}Mv^{2} = \frac{1}{2}\times 1.30\times 7.3^{2} = 34.64\ J[/tex]

For 8 such blows:

[tex]\Delta KE = n\Delta KE = 8\times 34.64\ = 277.12 J[/tex]

Now, we know that:

[tex]Q = m_{i}C_{i}\Delta T[/tex]

[tex]\Delta T= \frac{\Delta KE}{m_{i}C_{i}}[/tex]

[tex]\Delta T= \frac{277.12}{14\times 0.45} = 43.98^{\circ}C[/tex]

The size of the real heart is B) 11.7 cm

Explanation:

In lenses, mirror and other optical systems, the magnification factor is given by:

[tex]M=\frac{y'}{y}[/tex]

where

y' is the size of the image

y is the size of the real object

In this problem, we have:

M = 1.3 is the magnification factor

y' = 15.2 cm is the size of the image (the image of the heart)

Solving for y, we find the actual size of the heart:

[tex]y=\frac{y'}{M}=\frac{15.2 cm}{1.3}=11.7 cm[/tex]

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Answer:

b) a horse pulls a wagon at a constant velocity

Explanation:

As we know that ,work done is the dot product of force vector and displacement vector.

W= F.d

W=work

F=Force

d=Displacement

We also know that

F = m a

m= mass ,a = acceleration

When velocity is constant then rate of change in the velocity will be zero,then we can say that acceleration will be zero.

When a= 0  Then F= 0

W= F.d  ( F=0)

W = 0

Therefore option b i correct because horse is going with constant velocity.

Zero net work is done when a bunch of bananas is placed on a spring scale in the supermarket, as there is no displacement of the bananas despite the force exerted by the scale.

The situation in which zero net work is done is when a bunch of bananas is placed on a spring scale in the supermarket. In this scenario, the force exerted by the scale is upward and counteracts the weight of the bananas, but since there is no displacement or movement of the bananas, the work done is zero. The concept of work in physics is defined as the product of the force applied to an object and the displacement of the object in the direction of the force. Since the bananas don't move, the displacement is zero, resulting in zero work.

In the other scenarios provided:

Answer:

A)

<17, - 1, 0>

B)

0 Ns

C)

<17, - 1, 0>

D)

<4, - 5, 0>

Explanation:

A)

[tex]p_{A,i}[/tex] = initial momentum of object A = 15 i - 8 j + 0 k

[tex]p_{B,i}[/tex] = initial momentum of object B = 2 i + 7 j + 0 k

Total initial momentum of the system is given as the sum of initial momenta of A and B , hence

[tex]p_{i}[/tex] = [tex]p_{A,i}[/tex] + [tex]p_{B,i}[/tex]

[tex]p_{i}[/tex] = (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)

[tex]p_{i}[/tex] = 17 i - j + 0 k

B)

[tex]F_{ext}[/tex] = Net external force on the two objects = 0 N

[tex]t[/tex] = duration of the collision

[tex]I[/tex] = Impulse

Impulse is given as

[tex]I = F_{ext} t \\I = (0) t \\I = 0 Ns[/tex]

C)

[tex]p_{f}[/tex] = final momentum of the system

we know that the Impulse is nothing but change in momentum of the system of objects, hence

[tex]I = p_{f} - p_{i}\\0 = p_{f} - p_{i}\\p_{f} = p_{i}[/tex]

[tex]p_{f}[/tex] = 17 i - j + 0 k

D)

[tex]p_{A,f}[/tex] = final momentum of object A = 13 i + 4 j + 0 k

[tex]p_{B,f}[/tex] = final momentum of object B = ?

Total final momentum of the system is given as

[tex]p_{f} = p_{A,f} + p_{B,f}[/tex]

17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + [tex]p_{B,f}[/tex]

[tex]p_{B,f}[/tex] = 4 i - 5 j + 0 k

So final momentum of the object is <4, - 5, 0>

The total initial momentum of this system just before the collision is 17 i - j + 0 k.

This is defined as a term which quantifies the overall effect of a force acting over time.

Initial momentum of A = 15 i - 8 j + 0 k

initial momentum of B = 2 i + 7 j + 0 k

Total = sum of initial momentum of A and B

= (15 i - 8 j + 0 k) + (2 i + 7 j + 0 k)

= 17 i - j + 0 k

Impulse = Ft where F is force and t is time

Net external force on the two objects = 0 N

Impulse = (0)t

              = 0Ns

Impulse = change in momentum

= 17 i - j + 0 k

Final momentum of object A = 13 i + 4 j + 0 k

Final momentum of object B = ?

Total final momentum = Final momentum of object A + Final momentum of object B

17 i - j + 0 k = ( 13 i + 4 j + 0 k ) + Final momentum of object B

= 4 i - 5 j + 0 k

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Answer:

    [tex]x= 32.544[/tex]

Explanation:

given,

mean weight of bag (μ) = 32

standard deviation (σ) = 0.32

percentage of bag heavier = 4.5%

weight of the bag less than 4.5 % = 100 - 4.5

                                                        = 95.5%

we have to determine the z- value according to 95.5% or 0.955

using z-table

     z-value = 1.70

now, using formula

       [tex]Z = \dfrac{x-\mu}{\sigma}[/tex]

       [tex]1.70 = \dfrac{x-32}{0.32}[/tex]

       [tex]x-32 = 1.70\times {0.32}[/tex]

      [tex]x= 32.544[/tex]

The most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.

To determine the most a bag of baby carrots can weigh and not need to be repackaged, we need to find the weight corresponding to the upper 4.5% of the normal distribution.

The weights are normally distributed with a mean of 32 ounces and a standard deviation (σ) of 0.32 ounces.

We first find the z-score that corresponds to the top 4.5% of the distribution.

Using a z-table or calculator,

Next, we use the z-score formula:

Solving for X (the weight we need):

Multiplying both sides by 0.32:

Adding 32 to both sides:

Thus, the most a bag of baby carrots can weigh and not need to be repackaged is approximately 32.53 ounces.

Final answer:

The acceleration of the crate on the frictionless ramp is 2.39 m/s², while the acceleration on the ramp with friction is 0.49 m/s².

Explanation:

To find the acceleration of the crate, we first need to determine the force component parallel to the incline. This can be done using the formula F_parallel = F * sin(theta), where F is the applied force and theta is the angle of the ramp. Substituting the values, we have F_parallel = 7.0 N * sin(20°) = 2.39 N. The acceleration can then be found using the equation a = F_parallel / m, where m is the mass of the crate. Substituting the values, we have a = 2.39 N / 1.0 kg = 2.39 m/s².

If the ramp has a friction force of 1.9 N, it will oppose the motion of the crate. In this case, we need to subtract the friction force from the parallel force to find the effective force. The effective force, Feff, is given by Feff = F_parallel - friction force. Substituting the values, we have Feff = 2.39 N - 1.9 N = 0.49 N. The acceleration can then be found using the equation a = Feff / m. Substituting the values, we have a = 0.49 N / 1.0 kg = 0.49 m/s².

Answer:

B:the changes in the characteristics within a population that lead to survival.

Explanation:

Answer:

B

Explanation:

study Island

Answer:

[tex]B = 1.37 \times 10^{-10} T[/tex]

Explanation:

As we know that in electromagnetic waves the two fields will induce each other

So here we have

[tex]B = \frac{E}{c}[/tex]

we have

[tex]E = 0.041 V/m[/tex]

[tex]c = 3\times 10^8 m/s[/tex]

so we have

[tex]B = \frac{0.041}{3 \times 10^8}[/tex]

[tex]B = 1.37 \times 10^{-10} T[/tex]

Meaning of geosynchronous satellite:

A satellite above Earth placed in a geosynchronous circular orbit, inclined at an angle with the equatorial plane of Earth and orbiting at an orbital period equal to that of earth’s rotational period is termed as a geosynchronous satellite. Geosynchronous orbit is a High Earth orbit at a distance of 42,164 kilometres from the centre of earth such that the gravitational pull from earth is fair enough for the object in orbital motion to match with the speed of earth’s rotational period (24 hrs)

As an observer from ground, a geosynchronous satellite appears to be at a stationary point, at any part of the day; since it's velocity synchronizes with the angular velocity that of earth. A geosynchronous satellite orbiting in an equatorial plane without any inclinations is termed as a geostationary satellite.  

Answer:

The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]

Explanation:

Given:

Mass of student [tex]m_1=71\ kg[/tex]

Mass of skateboard [tex]m_2=2.8\ kg[/tex]

Distance between student and skateboard [tex]d=2.75\ m[/tex]

Acceleration of student [tex]a=0.65\ ms^{-2}[/tex]

Finding velocity [tex]v_1[/tex] of the student  before jumping on skateboard

Using equation of motion

[tex]v_1^2=v_0^2+2ad[/tex]

here [tex]v_0[/tex] represents the initial velocity of the student which is [tex]=0[/tex] as he starts from rest.

So,

[tex]v_1^2=0^2+2(0.65)(2.75)[/tex]

[tex]v_1^2=3.575[/tex]

Taking square root both sides:

[tex]\sqrt{v_1^2}=\sqrt[1.7875}[/tex]

∴ [tex]v_1=1.89[/tex]

Finding velocity [tex]v[/tex] of student and skateboard.

Using law of conservation of momentum.

[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]

Where [tex]v_2[/tex] is initial velocity of skateboard which is [tex]=0[/tex] as it is at rest.

Plugging in values.

[tex]71(1.89)+(2.8)(0)=(71+2.8)\ v[/tex]

[tex]134.19=73.8\ v[/tex]

Dividing both sides by [tex]73.8[/tex]

[tex]\frac{134.19}{73.8}=\frac{73.8\ v}{73.8}[/tex]

∴ [tex]v=1.82[/tex]

The velocity of the student and skateboard together [tex]=1.82\ ms^{-1}[/tex]

Answer:

No change in the specific heat.

Explanation:

Specific heat is an intrinsic property that it has not depend upon on the amount of substance or energy added, it depends upon the material.

So, when twice the amount of energy is transferred, specific heat of the material does not change rather the energy that is twice in the amount to 1 kg of that material cause it to warm 2.0° C.

Answer:

Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.

Explanation:

By conservation of the linear momentum:

[tex]m*V_{1o}+m*V_{2o}=m*V_{1f}+m*V_{2f}[/tex]  Since both masses are the same, and expressing the equation for each axis x,y:

[tex]V_{1ox}+V_{2ox}=V_{1fx}[/tex]      eqX

[tex]0=V_{1fx}+V_{2fx}[/tex]                 eqY

From eqX:  [tex]V_{1fx}=1m/s[/tex]

From eqY:  [tex]V_{1fy}=-1.37m/s[/tex]

The module is:

[tex]V_{1f}=\sqrt{1^2+(-1.37)^2}=0.35m/s[/tex]

The angle is:

[tex]\theta=atan(-1.37/1)=-53.87\°[/tex]   This is 53.87° due south of east

The acceleration at the bottom of the loop is 34m/[tex]\bold{s^2}[/tex]

Explanation:

Given:  

Mass=52kg

Force=1750N

To calculate:

The acceleration at the bottom of the loop

Formula:

Force=Mass x Acceleration

1750=52 x Acceleration

1750/52=Acceleration

Therefore acceleration at the bottom of the loop is 34m/[tex]s^2[/tex]

Roller coasters are mainly based upon acceleration theory they have two types of acceleration one is at the top of the loop and the other is at the bottom of the loop.

Then the net forces and the values are given. In many problems the roller coaster concept is included and it gives another level of clarity to the problems including the net forces

The two possible frequencies could be ; 642 Hz or 712 Hz

Explanation:

For beats; f₁-f₂=fb where f₁=frequency of the first guitar, f₂=frequency of the second guitar, fb =frequency of beat

In this case, the two possibilities are;

f₁-f₂=fb or f₂-f₁=fb

f₁=677 Hz  and fb=35 Hz

then;

f₁-x=fb

677-x=35

x=677-35=642 Hz

or

x-f₁=fb

x-677=35

x=35+677 =712 Hz

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Final answer:

Both Technician A and Technician B are correct.

Explanation:

Technician A is correct. Boring and honing the cylinders is done to achieve the proper size, shape, and finish required for the installation of the pistons. This ensures a precise fit between the cylinders and pistons, allowing for optimal performance and reduced wear.

Technician B is also correct. Chamfering the bolt holes helps to ensure smooth insertion of the bolts and prevents damage to the threads. Cleaning with a thread chaser removes any debris or contaminants that could compromise the integrity of the assembly.

Therefore, the correct answer is C) Both technicians A and B.

Answer:

b. isothermal

Explanation:

When a thermodynamic process occurs and the temperature remains constant, it is said to be an isothermal process.

The prefix "iso" means "equal", so the word isothermal indicates that the temperature remains equal in the process, and that is what happens in the situation described, the volume is varied (compressed) while maintaining the sample of nitrogen gas remains at a constant temperature.

The object initially had potential energy of 108,410 Joules when placed 6.5 meters higher. Since it maintains the same speed, this energy was used to overcome friction. Thus, the object lost 108,410 Joules to friction.

To answer your question, we first need to calculate the initial potential energy of the object when it was 6.5 m higher. Potential energy (PE) is given by PE = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height. So, the initial potential energy is PE = (1700 kg)(9.8 m/s²)(6.5 m) = 108,410 Joules.

Since the object reaches the top of the loop with the same velocity in both scenarios, we can say that its kinetic energy remained the same. This implies all of the initial potential energy went into overcoming friction. Therefore, the object lost 108,410 Joules of energy to friction.

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The object lost 111,230 joules of energy to friction during its ascent.

The question is asking how much energy, in joules, the object lost to friction if it starts 6.5 m higher than the previous part and still reaches the top of the loop with the same velocity. To calculate this, we need to find the change in gravitational potential energy, which is given by the formula: ΔPE = mgh.

The mass of the object is 1700 kg, and the change in height is 6.5 m. Substituting these values into the equation, we have:

ΔPE = (1700 kg)(9.8 m/s^2)(6.5 m) = 111,230 J.

Therefore, the object lost 111,230 joules of energy to friction during its ascent.

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Answers:

a) 13 s

b) 0.362 m/s

Explanation:

We have the following data:

[tex]m=660000 kg[/tex] is the mass of the mass damper

[tex]L=42 m[/tex] is the length of the pendulum

[tex]A=75 cm \frac{1m}{100 cm}=0.75 m[/tex] is the amplitude

This can be solved by the following equation:

[tex]T=2 \pi \sqrt{\frac{L}{g}}[/tex] (1)

Where:

[tex]T[/tex] is the period

[tex]g=9.8 m/s^{2}[/tex] is the acceleration due gravity

[tex]T=2 \pi \sqrt{\frac{42 m}{9.8 m/s^{2}}}[/tex] (2)

[tex]T=13 s[/tex] (3)

The velocity in a pendulum is maximum [tex]V_{max}[/tex] when the pendulum is in its mean position and the amplitude is maximum. So, the equation in this case is:

[tex]V_{max}=A \frac{2 \pi}{T}[/tex] (4)

[tex]V_{max}=0.75 m \frac{2 \pi}{13 s}[/tex] (5)

[tex]V_{max}=0.362 m/s[/tex]

The total force on the spring is the sum of the weight of the 5-kg object and the additional force due to the elevator's upward acceleration. This force leads to a stretch in the spring, calculated using Hooke's law, of 0.064 m or 6.4 cm.

To solve this problem we'll use Hooke's law (F = -kx), where F is the force, k is the spring constant (in our case, 802 N/m), and x is the displacement or stretch of the spring.

Here, the force isn't just from the weight of the 5.0-kg object (mg), but also from the additional force due to the upward acceleration of the elevator (ma). By adding these two forces together, we can find the total force on the spring (F = mg + ma).

First, let's calculate the weight of the 5-kg object: F_weight = mg = 5.0 kg * 9.81 m/s^2 = 49.05 N.

Next, let's calculate the additional force due to the elevator's acceleration: F_acceleration = ma = 5.0 kg * 0.41 m/s^2 = 2.05 N.

Adding these two forces together: F_total = F_weight + F_acceleration = 49.05 N + 2.05 N = 51.1 N.

Now, we can use Hooke's law to find the stretch of the spring x = F_total / k = 51.1 N / 802 N/m = 0.064 m, or 6.4 cm when converted to centimeters.

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The spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.

The amount by which the spring stretches can be determined using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring. In this case, the force exerted by the spring is equal to the weight of the object attached to it. The equation for this can be written as F = kx, where F is the force, k is the spring constant, and x is the displacement. Rearranging this equation, we can solve for x:

x = F / k

Plugging in the values given in the question, we have:

x = (m * a) / k

Substituting the values m = 5.0 kg, a = 0.41 m/s^2, and k = 802 N/m:

x = (5.0 kg * 0.41 m/s^2) / 802 N/m

x ≈ 0.020 m

Therefore, the spring stretches by approximately 0.020 meters relative to its unstrained length when the elevator is accelerating upward at 0.41 m/s^2.

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The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence [tex]F_{G} = F_{C}[/tex].

Gravitational force between Earth and Satellite: [tex]F_{G} = \frac{G \times M_e \times M_s}{R^2}[/tex]

Centripetal force of Satellite :[tex]F_C = \frac{M_s \times V^2}{R}[/tex]

Where G = Gravitational Constant

[tex]M_e[/tex]= Mass of Earth

[tex]M_s[/tex]= Mass of satellite

R= Radius of satellite’s circular orbit

V = Speed of satellite

Equating  [tex]F_G = F_C[/tex], we get  

Speed of Satellite [tex]V =\frac{\sqrt{G \times M_e}}{R}[/tex]

Thus the speed of satellite depends only on the mass of Earth.

Answer:

maximum amplitude  = 0.13 m

Explanation:

Given that

Time period T= 0.74 s

acceleration of gravity g= 10 m/s²

We know that time period of simple harmonic motion given as

[tex]T=\dfrac{2\pi}{\omega}[/tex]

[tex]0.74=\dfrac{2\pi}{\omega}[/tex]

ω = 8.48 rad/s

ω=angular frequency

Lets take amplitude = A

The maximum acceleration given as

a= ω² A

The maximum acceleration should be equal to g ,then block does not separate

a= ω² A

10= 8.48² A

A=0.13 m

maximum amplitude  = 0.13 m

The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m

Given data :

period ( T ) = 0.74 secs

acceleration due to gravity ( g ) = 9.8 m/s62

Time period of simple harmonic motion ( T ) = [tex]\frac{2\pi }{w}[/tex]

First step : solve for w

w = 2π / T

   = 2π / 0.74

   = 8.49 rad/sec

Next step : determine the maximum amplitude ( A )

a = w² A

where ;  a = maximum acceleration = 9.8 , w = 8.49

therefore

A = a / w²

   = 9.8 / ( 8.49 )²

   = 0.136  ≈ 0.14 m

Hence we can conclude that The maximum amplitude of the motion for which the block does not separate from the plate is : 0.14 m.

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Answer:

6.62 m

About 44.14 grapefruits lined in a single line would be the length of the diameter of star Aldebaran

Explanation:

First we know,

Diameter of the Sun - 1.39 million km (Ds)

Diameter of Aldbaran - 61.40 million km (Da)

Thus,

[tex]\frac{Da}{Ds}[/tex] =  [tex]\frac{61.40}{1.39}[/tex] ≈ 44.14

We know that in this scale the size of the sun is that of a grapefruit.

The average diameter of a grapefruit is about 15cm

Thus  Ds in this scale is 15cm

Thus Da in this scale,

Da = Ds *44.14

Da = 0.15 * 44.14 ≈ 6.62 m

Thus the diameter of Aldebaran in this scale is 6.62 m

A molecule of glucose is an example of potential energy rather than kinetic energy.

Potential energy is a form of stored energy that is dependent on the relationship between different system components. Example are when a spring is compressed or stretched, its potential energy increases. If a steel ball is raised above the ground as opposed to falling to the ground, it has more potential energy.

The energy an object has as a result of motion is known as kinetic energy. A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.

From the given option, crawling beetle foraging for food, light flashes emitted by firefly water, rushing over Niagara Falls and a molecule of glucose. Molecule of glucose have potential rather than kinetic energy

because there is no motion in the molecule of glucose.

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A molecule of glucose is an example of potential energy because it has energy stored within its chemical bonds waiting to be used, whereas the other examples are instances of kinetic energy in action.

Potential vs. Kinetic Energy

When discussing energy, it is vital to understand the difference between potential energy and kinetic energy. Potential energy is stored energy, waiting to be released, whereas kinetic energy is the energy of motion. An example of potential energy is a molecule of glucose, which contains energy in its chemical bonds that is not currently in motion but has the possibility to be used in the future. This contrasts with kinetic energy, which would include objects or substances that are actively moving or releasing energy, such as a crawling beetle, light flashes from a firefly, or water rushing over Niagara Falls.

To answer the student's question, the example representing potential energy rather than kinetic energy is:

The examples of a crawling beetle, light from a firefly, and rushing water over Niagara Falls all represent kinetic energy as they are all in motion and actively using or releasing energy.

Impulse : 7.5.10⁻¹⁰N.s

Momentum is the product of the mass of an object and its velocity

Whereas impulse is defined is the product of the force with the time interval when the force is acting on the object.

the pollen grains to be ejected out of the flower in 0.30 ms and acceleration of 2.5 × 10⁴ m/s², then

Δt = 0.3 ms = 3.10⁻⁴ s

a = 2.5 × 10⁴ m/s²

mass of pollen: 1.0 × 10⁻⁷g = 1.0 x 10⁻¹⁰ kg

then:

A force that works according to Newton's second law:

[tex]\rm F=1.0\times 10^{-10}\times 2.5\times 10^4\\\\F=2.5\times 10^{-6}\:N[/tex]

Impulse (I) :

[tex]\rm I=2.5 \times 10^{-6}\times 3\times 10^{-4}\\\\I=7.5\times 10^{-10}N.s[/tex]

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Final answer:

To calculate the impulse delivered to a pollen grain, the mass is converted to kilograms and then multiplied with the given acceleration to find the force. This force is then multiplied by the time over which it acts to obtain the impulse, which is 75 in units of 10⁻¹⁰ kgm/s.

Explanation:

The question asks us to calculate the impulse delivered to a pollen grain by the forceful opening of a bunchberry plant flower. To find the impulse (I), which is the change in momentum, we can use the formula I = Ft, where F is the force and t is the time that the force is applied. Alternatively, since the question provides the acceleration (a) and initial time, we can utilize the link between force, mass, and acceleration (F = ma) and then apply that result to find the impulse using the time.

Firstly, we need to convert the mass of the pollen grain to kilograms (kg) which is the standard unit of mass in physics equations:
1.0 × 10⁻⁷ g = 1.0 × 10⁻¹⁰ kg.
Now, we calculate the force on the pollen grain:
F = ma = (1.0 × 10⁻¹⁰ kg) × (2.5 × 10⁴ m/s²) = 2.5 × 10⁻⁵ N.
Next, with force and the time of application known, we can find the impulse:
I = Ft = (2.5 × 10⁻⁵ N) × (0.30 × 10⁻³ s) = 7.5 × 10⁻⁹ kg·m/s.
In terms of 10⁻¹⁰ kg·m/s, the impulse is 75.

Answer:

1.19cm^3 of glycerine

Explanation:

Let Vo= 150cm^3 for both aluminum and glycerine, using expansion formula:

Volume of spill glycerine = change in volume of glycerine - change in volume of aluminum

Volume of glycerine = coefficient of volume expansion of glycerine * Vo* change in temperature - coefficient of volume expansion of Aluminum*Vo* change temperature

coefficient of volume expansion of aluminum = coefficient of linear expansion of aluminum*3 = 23*10^-6 * 3 = 0.69*10^-4 oC^-1

Change in temperature = 41-23 = 18oC

Volume of glycerine that spill = (5.1*10^-4) - (0.69*10^-4) (150*18) = 4.41*10^-4*2700 = 1.19cm3

Answer:

[tex]\frac{dm}{dt} = 10352 kg/s[/tex]

Explanation:

As we know that thermal energy given to the water to raise its temperature is given as

[tex]Q = ms\Delta T[/tex]

now rate of energy given to the system is

[tex]\frac{dQ}{dt} = \frac{dm}{dt} s \Delta T[/tex]

so we have

[tex]\frac{dQ}{dt} = 1.3 \times 10^8[/tex]

s = 4186 J/kg C

now we have

[tex]1.3 \times 10^8 = \frac{dm}{dt} (4186) 3[/tex]

[tex]\frac{dm}{dt} = 10352 kg/s[/tex]

With the given data, the mass of the warmed water is approximately 1.0×10⁴ kg each second.

To determine the mass of water that is warmed as it flows through the power plant, we can use the following steps:

Calculate the amount of heat transferred to the water:

The power station dissipates 1.3×10⁸ J of heat each second.

Use the specific heat capacity formula:

The formula to calculate heat transfer is

where:

Rearrange the formula to solve for mass (m):

Plugging in the values, we get:

Therefore, the mass of warmed water flowing through the plant each second is approximately 1.0×104 kg.