You are driving a 2420.0-kg car at a constant speed of 14.0 m/s along a wet, but straight, level road. As you approach an intersection, the traffic light turns red. You slam on the brakes. The car's wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 26.8 m.
What is the coefficient of kinetic friction between your tires and the wet road?

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

[tex]\displaystyle{t=D/v}[/tex]

and from Hubble's Law:

[tex]v =H_0D[/tex]

Therefore:

[tex]\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}[/tex]

With the given value for the Hubble's constant we have:

[tex]H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s[/tex]

and thus,

[tex]t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years[/tex]

Answer:

T = 2.4 s

Explanation:

given,

angle at which ball is thrown = 29°

speed relative to ground = 24.3 m/s

ball is caught at a distance = 51.0463

acceleration due to gravity = 9.8 m/s²

time for which ball was in the air = ?

now,

velocity of ball in x-direction

V_x =  v cos θ

V_x =  24.3 x cos 29°

V_x = 21.25 m/s

velocity in y direction

V_y =  v sin θ

V_y =  24.3 x sin 29°

V_y = 11.78 m/s

distance on the ground when ball will reach maximum height

x = 51.0463/2 = 25.52 m

at top most point velocity is equal to zero

time for which the ball was in air

v = u + a t

0 = 11.78 - 9.8 t

[tex]t = \dfrac{11.78}{9.8}[/tex]

t = 1.20

this time is taken to travel half distance

total time = 2 x 1.20

T = 2.4 s

time for which ball was in air is T = 2.4 s

Answer:

(a) 7 m

(b) 1  m

Explanation:

Given:

The magnitude of displacement  vector 'a' is 3 m

The magnitude of displacement vector 'b' is 4 m.

The vector 'c' is the vector sum of vectors 'a' and 'b'.

(a)

Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:

[tex]|\overrightarrow c|=|\overrightarrow a|+|\overrightarrow b|\\\\|\overrightarrow c|=3 + 4 = 7\ m[/tex]

Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.

(b)

When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.

So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

[tex]|\overrightarrow c|=||\overrightarrow a|-|\overrightarrow b||\\\\|\overrightarrow c|=|3 - 4| = 1\ m[/tex]

Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.

Answer:

(a) [tex]K = \frac{I}{2\pi a}[/tex]

(b) [tex]J = \frac{I}{2\pi as}[/tex]

Explanation:

(a) The surface current density of a conductor is the current flowing per unit length of the conductor.

                                   [tex]K = \frac{dI}{dL}[/tex]

Considering a wire, the current is uniformly distributed over the circumferenece of the wire.

                                   [tex]dL = 2\pi r[/tex]

The radius of the wire = a

                                    [tex]dL = 2\pi a[/tex]

The surface current density [tex]K = \frac{I}{2\pi a}[/tex]

(b) The current density is inversely proportional

                                     [tex]J \alpha  s^{-1}[/tex]    

                                     [tex]J = \frac{k}{s}[/tex]           ......(1)

k is the constant of proportionality

                                     [tex]I = \int\limits {J} \, dS[/tex]

                                     [tex]I = J \int\limits \, dS[/tex]     ........(2)

substituting (1) into (2)

                                     [tex]I = \frac{k}{s} \int\limits\, dS[/tex]

                                     [tex]I = k \int\limits^a_0 \frac{1}{s}  {s} \, dS[/tex]

                                     [tex]I = 2\pi k\int\limits\, dS[/tex]

                                     [tex]I = 2\pi ka[/tex]

                                     [tex]k = \frac{I}{2\pi a}[/tex]

substitute [tex]J = \frac{k}{s}[/tex]

                                     [tex]J = \frac{I}{2\pi as}[/tex]

In a wire with current I and radius a, the surface current density K with uniform distribution is I/(2πaL). If volume current density varies inversely with the radius, the current is more dense at the center and less dense towards the surface, described by I/(πa²L).

The subject here relates to the physical properties of the current flowing down a wire with a certain radius. (a) If the current I is uniformly distributed over the surface of the wire with a radius a, then the surface current density K can be given by the total current (I) divided by the surface area of the wire. In this context, the surface area of a cylindrical wire can be calculated by the formula 2π * a * L, where L is the length of the wire. Therefore, the surface current density K is I/(2πaL).

(b) If the current is distributed such that the volume current density J is inversely proportional to the radius, it implies that the current is more dense at the center of the wire and less dense as you approach the surface. The volume current density J can be described by the formula I/(πa²L). This complex distribution would likely require calculus to derive an exact relationship between current density and radius.

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Explanation:

Levitt and Dubner’s argument effectively uses logical, concrete evidence to arrive at conclusions about morality and cheating practices.the realities they present about sumo and the bagel business Their utilization of factual evidence in various illustrations shows morality and cheating practices are more common in high incentive situations where telling lies or an act of fooling people was rewarded awesomely. They use statistical evidence and different examples support the fact that Levitt and Dubner have arrived at a generalisation on moral grounds.

Answer:

Levitt and Dubner’s argument uses logic to present evidence to arrive at their final conclusions about morality and human cheating practices. The examples  they used were of Sumo wrestler's in Japan and the bagel business of a self-employed man who provided bagel's and cream cheese to office worker's on the "honor system". Levitt and Dubner used factual evidence as well as some illustrations to present their claims. They concluded that although cheating exists when there is high incentive, most people are  inherently honest when incentives are not a factor.

Explanation:

Answer:27.35 m/s

Explanation:

Given

Time of Flight of Projectile T=4.08 s

Range of Projectile =76.2 m

Time Of Flight of Projectile is given by

[tex]T=\frac{2u\sin \theta }{g}----------1[/tex]

where u=initial Velocity

[tex]\theta =[/tex]Launch angle

g=acceleration due to gravity

Range is given by [tex]R=\frac{u^2\sin 2\theta }{g}------2[/tex]

divide 1 and 2

[tex]\frac{R}{T}=\frac{u^2\sin 2\theta }{g}\times \frac{g}{2u\sin \theta }[/tex]

[tex]\frac{R}{T}=u\cos \theta ------3[/tex]

[tex]u\sin \theta =\frac{Tg}{2}------4[/tex]

squaring and adding 3 & 4 we get

[tex]u^2(\cos ^2\theta +\sin ^2\theta )=(\frac{R}{T})^2+(\frac{Tg}{2})^2[/tex]

[tex]u^2=(19.992)^2+(18.676)^2[/tex]

[tex]u=\sqrt{748.49}[/tex]

[tex]u=27.35 m/s[/tex]                              

The initial speed is the speed of the object at the beginning of the measurement or the starting speed.

The initial speed of the projectile is 27.35 m/s.

The initial speed is the speed of the object at the beginning of the measurement or the starting speed.

Given information-

The time taken by projectile to returns its original height is 4.08 s.

The distance traveled by it is 76.2 m.

Air resistance can be neglected.

The time of flight of a projectile motion can be given as,

[tex]T=\dfrac{2u\sin \theta}{g}[/tex]

Let the above equation is equation 1.

Here, [tex]u[/tex] is the initial velocity, and [tex]\theta[/tex] is the angle of launch.

Rewrite the above equation as,

[tex]u\sin \theta =\dfrac{Tg}{2}[/tex]  

Let the above equation is equation 2.

Now the range of the projectile motion can be given as,

[tex]R=\dfrac{u^2\sin (2\theta) }{g}\\R=\dfrac{u^22\cos \theta\sin \theta }{g}\\[/tex]

Divide this equation by the equation 1 as,

[tex]\dfrac{R}{T}=u\cos \theta[/tex]

Square and add the the above equation and equation 2 as,

[tex]u^2(\cos^2 \theta+\sin^2 \theta)=\dfrac{R}{T}+\dfrac{Tg}{2}\\u^2(1)=\dfrac{76.2}{4.08}+\dfrac{4.08\times9.81}{2}\\u=27.35\rm m/s[/tex]

Hence, the initial speed of the projectile is 27.35 m/s.

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Answer:

The orbital velocity of the moon is, V = 1018 m/s

Explanation:

Given data,

The radius of the moon's orbit, R = 3.85 x 10⁸ m

The mass of the Earth, M = 5.98 x 10²⁴ kg

The formula for orbital velocity is,

                              V = √(GM/R²)

Substituting the values,

                               V = √(6.673 x 10⁻¹¹ x 5.98 x 10²⁴ / 3.85 x 10⁸ )

                                  = 1018 m/s

Hence, the orbital velocity of the moon is, V = 1018 m/s

The orbital speed of the Moon can be calculated using a specific formula that takes into account the mass of the Earth, the mass of the Moon, and the radius of the Moon's orbit.

The orbital speed of an object in orbit around another object can be calculated using the formula:

v = √[G * (M+E) / r]

Where v is the orbital speed, G is the gravitational constant (approximately 6.67 × 10^-11 N.m^2/kg^2), M is the mass of the larger object (in this case, the Earth), E is the mass of the orbiting object (in this case, the Moon), and r is the radius of the orbit.

Substituting the given values into the formula, we can calculate the orbital speed of the Moon.

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Because they perform specific tasks repeatedly throughout your program, as needed

Answer:

Because they perform specific tasks repeatedly throughout your program, as needed

Explanation:

Answer:

C. Yes, the water could be changing the phase.

Explanation:

Answer:

Changes the element of the atom.

Explanation:

The elements are classified by the number of protons they have in their nucleus, so if the number of protons is changed (added or removed), that atom will become one of a different element.

For example, hydrogen has only 1 proton in its nucleus, and helium has 2. So if a proton is added to a hydrogen atom, it becomes a helium atom, and consequently its atomic number, wich is determided by the protons in an element, will also change.

Answer:

Opposite spin neutralizes the magnetic fields.

Explanation:

The reason for the significance of the unpaired electrons with respect to the magnetic properties is because electrons have opposite spin and when the electrons are paired then as a result their opposite spins neutralizes the effect of their magnetic field thus resulting in no field effect.

Thus single electrons which are unpaired contributes to the magnetic properties of the material as compared to the paired electrons.

If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.

A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.

HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : [tex]H^{+} - Cl^{-}[/tex]). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.

Answer:

15.65 °C

Explanation:

cold temperature (Tc) = -15.5 degree C = 273.15 - 15.5 = 257.65 kelvin

minimum coefficient of performance (η) = 8.25

find the maximum hot reservoir temperature of such a generator (Th)

η = \frac{Tc}{Th-Tc}

Th = Tc x (\frac{1}{η} + 1)

Th = 257.65 x (\frac{1}{8.25} + 1)

Th = 288.8 K

Th = 288.8 - 273.15 = 15.65 °C

Answer:

i) C decreases

ii) Q remains constant

iii) E remains constant

iv) ΔV increases

Explanation:

i)

We know, capacitance is given by:

[tex]C=\frac{\epsilon_0.A}{d}[/tex]

[tex]\therefore C\propto \frac{1}{d}[/tex]

In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.

ii)

As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.

iii)

Electric field between the plates is given as:

[tex]E=\frac{\sigma}{\epsilon_0}[/tex]

where:

charge density, [tex]\sigma=\frac{Q}{A}[/tex]

As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.

iv)

i) Capacitance is decreases

ii) Charge Q remains constant

iii) Electric field E remains constant

iv) Change in potential ΔV is  increases

The capacitance is computed as,

                    [tex]C=\frac{\epsilon A}{d}[/tex]

Where A is area of plates and d is distance between plates.

Following information is to be considered.

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Answer: 2/3 of the total volume

Explanation:

See attachment for details

Answer:

32 bottles

Explanation:

If we create a free body diagram on the child we have his weight and the bouyant force

W-B=0

They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced

W=B=Ww

Mg=mg

32 kg=mass of water displaced

1 kilogram per liter (kg/L) is the density of water, this means that 32 Liters of water are displaced and since the bottles can retain 1 liter, the child needs 32 bottles

The child needs a minimum of 32 soda bottles to stay dry on the raft.

To determine the minimum number of soda bottles needed for the child to stay dry on the raft, we need to consider the buoyant force exerted by the bottles. The buoyant force is equal to the weight of the displaced water. Since the child wants to stay dry, the buoyant force should be greater than or equal to the weight of the child.

The weight of the child can be calculated using the formula: weight = mass × gravity. Given the mass of the child is 32 kg, and the acceleration due to gravity is 9.8 m/s², we can find that the weight of the child is 32 kg × 9.8 m/s² = 313.6 N.
Next, we need to find the volume of one soda bottle. Since the empty soda bottles have a total volume of 1.0 L, we can assume that each bottle has a volume of 1.0 L ÷ the number of bottles needed. The mass of the water displaced by one bottle can be calculated using the formula: mass = density × volume. Given that the density of water is 1000 kg/m³, and 1 L = 0.001 m³, we can find that the mass of water displaced by one bottle is 1000 kg/m³ × 0.001 m³ = 1 kg.

To find the minimum number of bottles needed, we can set up the equation: buoyant force = weight of child. The buoyant force is equal to the mass of water displaced by one bottle × gravity × the number of bottles needed. Using the values we found earlier, we have: 1 kg × 9.8 m/s² × the number of bottles needed = 313.6 N. Solving for the number of bottles needed, we find that the minimum number of soda bottles the child needs is 313.6 N / (1 kg × 9.8 m/s²) = 32 bottles (rounded up to the nearest whole number).

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Answer:

a)1.385  rad/s

b) Before: 1080 J. After 498.46 J

Explanation:

The moments of inertia of the turn table, with the shape of uniform disk is:

[tex] I_1 = 0.5mr^2 = 0.5*120*2^2 = 240 kgm^2[/tex]

The angular momentum of the turn table before the impact is

[tex]A_1 = \omega_1I_1 = 3*240 = 720 radkgm^2[/tex]

The moments of inertia of the system after the impact is (treating the parachute man is a point particle)

[tex]I_2 = I_1 + Mr^2 = 240 + 70*2^2 = 240 + 280 = 520 kgm^2[/tex]

According to angular momentum conservation law:

[tex]A_1 = A_2[/tex]

[tex] 720  = \omega_2I_2[/tex]

[tex]\omega_2 = \frac{720}{I_2} = \frac{720}{520} = 1.385 rad/s[/tex]

(b) Before the impact:

[tex]K_1 = 0.5*I_1*\omega_1^2 = 0.5*240*3^2 = 1080 J[/tex]

After the impct

[tex]K_2 = 0.5*I_2*\omega_2^2 = 0.5*520*1.385^2 = 498.46 J[/tex]

The kinetic energies are not equal because the impact is causing the turn table to lose energy.

Answer:

R= - 3388.74

Explanation:

Given that

V₁= 76 k  ( in z-direction)

θ = 48°

V₂ = 60 cos48° i - 60 sin48°  k

The dot product of two vector given as

We know that dot product of two vector  is scalar and cross product of two vector is vector.

R= V₁ . V₂

We have to remember

i.i= j.j = k.k = 1

i.j = j.k = k.i = 0

Now

R= V₁ . V₂

R= (76 k ).(  60 cos48° i - 60 sin48°  k)

R= 0 - 60 x 76  sin48°

R= - 3388.74

The scalar product of the vectors is - 3388.74

Given information:

V₁= 76k  since it is in z-direction

Now vector V₂  makes an angle θ = 48° with x-axis so, it can be resolved as follows:

V₂ = 60 cos48°i - 60 sin48° k

The scalar product of vectors is the product of the projection of one vector with the other vector.

The scalar product or the dot product of two vectors is given as

V= V₁ . V₂

The dot product of the x,y,and z direction components follow the below mentioned rule:

i.i= j.j = k.k = 1

i.j = j.k = k.i = 0

So, the required scalar product

V = V₁ . V₂

V = (76k ).(60cos48° i - 60sin48°  k)

V = 0 -60 x 76sin48°

V = - 3388.74

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Answer:

Ground state

Explanation:

At ground state all electrons are at the lowest energy level. At this level all the electrons, molecules or ions are said to be ground level. When electron get enough energy to jump they move to higher level. Any level higher than ground level is known as excited level. And energy of electron at excited state is higher than ground state. So the state at which all the electrons at their lowest possible energy level is the ground state.

Answer:

The micturition reflex can be voluntarily controlled by the relaxation of the external urethral sphincter.

The micturition reflex, or the process of urination, can be voluntarily controlled by the external urethral sphincter and the sacral micturition center. The external urethral sphincter is a skeletal muscle that we can consciously control, while the sacral micturition center is a group of neurons that normally act reflexively unless allowed for voluntary control by the brain. As the bladder fills, signals are sent through the sacral pelvic nerves which activate parasympathetic neurons to proceed with urination.

The micturition reflex, which is another term for urination or voiding, can be voluntarily controlled by the external urethral sphincter and the sacral micturition center. The external urethral sphincter is a skeletal muscle that can be consciously controlled to maintain urinary continence, while the sacral micturition center, a group of neurons located in the sacral region of the spinal cord, acts reflexively unless its action is modified by higher brain centers for voluntary urination. In response to a filled bladder, these centers trigger the relaxation of both the internal and external urethral sphincters, the contraction of the detrusor muscle, and inhibit the somatic motor neurons for urination to occur.

Children learn to voluntarily control the urination process as they mature, thereby overriding the micturition reflex and delay voiding, a process known as potty training. Voluntary micturition requires (1) an intact spinal cord and (2) a functional pudendal nerve arising from the sacral micturition center. The sacral pelvic nerves play a crucial role in bladder control. Upon receiving signals of bladder stretch, they activate the parasympathetic neurons to release acetylcholine, which triggers detrusor muscle contraction and bladder emptying.

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Answer:

a) v=1.81 m/s; B) v=4.95 m/s

Explanation:

using momentum conservation

m1v1+m2v1=m1v2+m2v2

A)

5.2*672+700*0=5.2*428+700v2

The initial velocity of the block is 0, solving for v2

v2=1.81 m/s

B) Now both the bullet and the block travel together

m1v1+m2v1=(m1+m2)v2

5.2*672+700*0=(5.2+700)v

v=4.95 m/s

a. To find the resulting speed of the block after the bullet strikes it, we can use the law of conservation of momentum. b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block.

a. To find the resulting speed of the block, we can use the law of conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The initial momentum of the bullet is given by m1 * v1, and the final momentum is given by (m1 + m2) * vf, where m1 is the mass of the bullet, v1 is the initial velocity of the bullet, m2 is the mass of the block, and vf is the final velocity of the bullet and block combined.

Using these values, we can solve for vf and then find the resulting speed of the block.

b. The speed of the bullet-block center of mass can be found by calculating the weighted average of the speeds of the bullet and block. Since the mass of the bullet is much smaller than the mass of the block, the center of mass will be closer to the speed of the block.

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Answer:

Lack of proper roof ventilation

Explanation:

The icicles hanging off the eaves of a roof are due to lack of proper roof ventilation. This means after snowfalls the attic of the roof receives warm air which melts the snow above. This is caused because the upper part of roof is above 32⁰C (which melts the snow) and lower part of your roof is at a lower temperature which refreezes the snow dripping down to form icicles.

Answer:

1.0x10^-4

Explanation:

First, in order to do this, we need to calculate the volume of 1 simple atom of Ar. Using the formula of the volume of a sphere we have the following

Converting A to cm:

0.97 * 1x10^-8 = 9.7x10^-9 cm

Now the volume:

V = 4/3π(9.7x10^-9)³

V = 3.82x10^-24 cm³

We know that 1 cm³ is 1 mL, and 1 L is 1000 mL so:

V = 3.82x10^-24 mL / 1000 = 3.82x10^-27 L

Now, using avogadro's number, we should get the total volume of all atoms of Ar so:

3.82x10^-27 * 6.02x10^23 = 2.3x10^-3 L

Finally, at STP the volume of an ideal gas is 22.4 L so:

2.3x10^-3 / 22.4 = 1.03x10^-4

With two significant figure, it would be 1.0x10^-4

Answer:

h=20.66m

Explanation:

First we need the speed when the cord starts stretching:

[tex]V_2^2=V_o^2-2*g*\Delta h[/tex]

[tex]V_2^2=-2*10*(-31)[/tex]

[tex]V_2=24.9m/s[/tex]   This will be our initial speed for a balance of energy.

By conservation of energy:

[tex]m*g*h+1/2*K*(h_o-l_o-h)^2-m*g*(h_o-l_o)-1/2*m*V_2^2=0[/tex]

Where

[tex]h[/tex] is your height at its maximum elongation

[tex]h_o[/tex] is the height of the bridge

[tex]l_o[/tex] is the length of the unstretched bungee cord

[tex]800h+21*(79-h)^2-63200-24800.4=0[/tex]

[tex]21h^2-2518h+43060.6=0[/tex] Solving for h:

[tex]h_1=20.66m[/tex]  and [tex]h_2=99.24m[/tex]  Since 99m is higher than the initial height of 79m, we discard that value.

So, the final height above water is 20.66m

Answer: using the conservation of potential energy stored in spring giving that at maximum amplitude velocity becomes zero.

Mgd= 1/2k(d-l)^2..... equation 1

M= 80kg=mass , g= 10m/s^2 =gravity, d=?=length of fully extended bungee rope, l=31m= length of bungee rope before extension, k=42N/m= spring constant

Simplifying equation above gives

2Mgh/k= d^2 - 2dl + l^2 ....eq 2

Substituting figures into the equ above gives

0 = d^2 - 100.1d +961 ...equ 3

Equ3 can be solved since it is a quadratic equation

d= (-b +or- square root (b^2 - 4ac))/2a ....equa4

Where a=1, b= -100.1, c= 961

Substituting figures into eequa4

d= 89.34m

So therefore the height above the river to me when bungee is fully extended is= 110 - 89.34

= 20.66

Explanation: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k.

Assuming that it performs simple harmonic motion.

Answer:

The work required to move a planet's satellite is W = 2854.61 J

Explanation:

Given data,

The mass of the satellite, m = 1820 kg

The radius of the circular orbit, r =  2R

The radius of the planet, r = 5.37 x 10⁶ m

The mass of the planet, M = 7.5 x 10²⁴ kg

The formula for work done from the 2R to 3R is,

                            W = [tex]\int_{2R}^{3R}\frac{GMm}{r^{2}}dr[/tex]

                            W = GMm/3R - GMm/2R

                             W = (-0.17)GMm/R  J

The negative sign indicates that the energy stored in the satellite as the potential energy.

Substituting the values

                              W = (-0.17) 6.673 x 10⁻¹¹ X 7.51 x 10²⁴ X 1820 / (7.37 x 10⁶)²

                                  = -2854.61 J

Hence, the work required to move a planet's satellite is W = 2854.61 J

Answer:

[tex]Q=35011.6\ J[/tex]

Explanation:

Given:

Now, total heat energy required get to the final state:

[tex]Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)[/tex]

where:

[tex]\Delta T_w=[/tex] change in temperature of water from 0 to 73 degree C

[tex]\Delta T_i=[/tex] change in temperature of ice from -30 to 0 degree C

[tex]\therefore Q=50(2.06\times 30+333+4.184\times 73)[/tex]

[tex]Q=35011.6\ J[/tex]

Answer:

The heat energy required is 350.1J

Explanation:

Given data

Mass of ice =50g---kg =50/1000= 0.05kg

Temperature of ice T1= - 30°c

Temperature of ice T2=0°c

Temperature of liquid T3=0°c

Temperature of water T4= 73°c

The heat of fusion = 333 J/g

heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK

and liquid water = 4.184 J/gK

It will be a good idea to first understand the path this process will follow

Ice at - 30°c - - ice at 0°c

Ice at 0°c - - - - liquid at 0°c fusion

Liquid at 0°c -- - liquid at 73°c

First, you have to calculate the heat absorbed by ice going from -30 C to 0 C. Use the equation:

q = m c (T2-T1)

q= 0.5*2.06(0-(-30))

q= 30.9J

Then, calculate the heat required to melt that ice at 0C. Use the equation:

q = m *(heat of fusion)

q=0.5*333

q=166.5J

Then, calculate the heat required to raise the temperature of water from 0°C to 73°C. Use

q = m c (T4-T3)

q= 0.5*4.184(73-0)

q= 152.7J

Finally, we will sum up the heat required

Total heat energy required = 30.9+166.5+152.7= 350.1J

Answer:

Final velocity will be -5.33 m/sec

Explanation:

We  have given mass of the rugby player m = 106 kg

Initial speed = 7.75 m/sec

Backward force [tex]F=1.85\times 10^4N[/tex]

Time is given as [tex]t=7.5\times 10^{-2}sec[/tex]

Impulse is given by impulse = force × time

So impulse [tex]=-1.85\times 10^4\times 7.5\times 10^{-2}=-1387.5N-s[/tex] ( as force is backward )

We know that impulse is given by change in momentum

So [tex]m(v_f-v_i)=-1387.5[/tex]

[tex]106\times (v_f-7.75)=-1387.5[/tex]

[tex]v_f=-5.33m/sec[/tex]

Final answer:

The photoelectric effect supports the particle theory of light as it shows that light's ability to eject electrons depends on its frequency, indicative of light's particle-like behavior through photons.

Explanation:

The photoelectric effect does not support the wave theory of light, but rather supports the particle theory of light. According to classical wave theory, energy of light is related to its intensity or amplitude. However, the photoelectric effect demonstrates that the kinetic energy of ejected electrons from a metal surface depends on the light's frequency rather than its intensity.

This phenomenon can be explained by considering light to consist of particles called photons. Each photon carries a quantized amount of energy determined by the equation E = hv, where E is energy, h is Planck's constant, and v is frequency. If the frequency of light is above a certain threshold, it can dislodge electrons from the metal because the photons have sufficient energy, showing light's particle-like nature.

Answer:

Part a)

[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]

Part b)

[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]

Part c)

[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]

Explanation:

As per Newton's II law we know that

F = ma

so we will have

[tex]a = \frac{F}{m}[/tex]

so we will have

[tex]a = \frac{F_1 + F_2}{m}[/tex]

Part a)

[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}[/tex]

[tex]a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}[/tex]

[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]

Part b)

[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}[/tex]

[tex]a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}[/tex]

[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]

Part c)

[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}[/tex]

[tex]a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}[/tex]

[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]

Answer:

Angular displacement of the wheel, [tex]\theta=267.41\ rad[/tex]

Explanation:

It is given that,

Angular acceleration of the wheel, [tex]\alpha =6.7\ rad/s^2[/tex]

Final speed of the wheel, [tex]\omega_f=74.5\ rad/s[/tex]

Time taken, t = 4.5 s

Initially, it is required to find the initial angular velocity of the wheel. Using the first equation of rotational kinematics as :

[tex]\omega_f=\omega_o+\alpha t[/tex]

[tex]\omega_o[/tex] is the initial speed of the wheel

[tex]\omega_o=\omega_f-\alpha t[/tex]

[tex]\omega_o=74.5-6.7\times 4.5[/tex]

[tex]\omega_o=44.35\ rad/s[/tex]

Let [tex]\theta[/tex] is the angular displacement of each wheel during this time. Using the second equation of motion as :

[tex]\theta=\omega_o t+\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta=44.35\times 4.5+\dfrac{1}{2}\times 6.7\times (4.5)^2[/tex]

[tex]\theta=267.41\ rad[/tex]

So, the angular displacement of each wheel during this time is 267 radian.