Two carts, cart AA (mass 4.00 kgkg) and cart BB (mass 6.50 kgkg) move on a frictionless, horizontal track. Initially, cart BB is at rest and cart AA is moving toward it at 4.00 m/sm/s. The carts are equipped with ideal spring bumpers, meaning they conserve energy. The collision is head-on, so all motion before and after the collision is along a straight line. Let x x be the direction of the initial motion of cart AA.

Answer: [tex]0.0353\ s^{-1}[/tex]

Explanation:

Given

Radioactive material is found to decrease 40% of its original value in [tex]2.59\times 10\ s[/tex]

Sample at any time is given by

[tex]N=N_oe^{-\lambda t}[/tex]

where, [tex]\lambda=\text{decay constant}[/tex]

Put values

[tex]\Rightarrow 0.4N_o=N_oe^{-\lambda\cdot 2.59\times 10}\\\Rightarrow 0.4=e^{-\lambda\cdot 2.59\times 10[/tex]

Taking natural logarithm both side

[tex]\Rightarrow \lambda=\dfrac{\ln 2.5}{25.9}\\\\\Rightarrow \lambda =0.0353\ s^{-1}[/tex]

Answer:

Jill will move down first

Explanation:

The nonmechanical energy or the loss of energy after the collision will be equal to  [tex]E=6\ J[/tex]

The conservation of the momentum is defined as when two bodies collide with each other then the total energy of the masses will remain constant.

It is given in the question that

Mass of the body X is  [tex]M_X=4\ kg[/tex]

The velocity of the body X is  =[tex]V_x=3 \ \frac{m}{s}[/tex]

Mass of the body Y is   [tex]M_Y= 2\ kg[/tex]

The velocity of body Y is [tex]V_y= 0[/tex]

Now to find out the energy converted after the collision we will first find the final velocity of both the bodies.

Now from the conservation of the momentum

[tex]M_X V_X +M_YV_Y=(M_X+M_Y)V_F[/tex]

[tex](4\times 3)+(2\times0)=(4+2)\times V_F[/tex]

[tex]V_F= \dfrac{12}{6}[/tex]

[tex]V_F = 2 \frac{m}{s}[/tex]

Now to find the Change in the energy of the body

[tex]\Delta E=E_i-E_f[/tex]

[tex]\Delta E=\dfrac{1}{2} M_XV_X^2+\dfrac{1}{2} M_YV_Y^2-\dfrac{1}{2} (M_X+M_Y)V_F^2[/tex]

[tex]\Delta E=\dfrac{1}{2} (4)(3^2)+\dfrac{1}{2} (2)(0^2)-\dfrac{1}{2} ( 4+2) 2^2[/tex]

[tex]\Delta E= 6\ J[/tex]

Thus the nonmechanical energy or the loss of energy after the collision will be equal to  [tex]E=6\ J[/tex]

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Answer:

0.5 m/s 2 is the acceleration

Explanation:

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Answer:The three main types are reflecting telescopes, refracting telescopes, and catadioptric telescopes. Radio telescopes collect and focus radio waves from distant objects. Space telescopes orbit Earth, collecting wavelengths of light that are normally blocked by the atmosphere.

Answer:

oii me manda mensagem fofo vc tem namorada fofo

Answer: See explanation

Explanation:

The reason why astronomers think that the rates of impact for the Moon must have been higher earlier than 3.8 billion years ago is because on the older highlands, there are ten times more craters than on the younger maria.

It is believed that the impact rate was higher earlier and thus can be seen when the numbers of the craters that can be seen on the lunar highlands is being compared to that on the maria. It should be noted that there are about 10 times more craters that can be found on the highlands than those on the maria.

If there was a constant rate of impact throughout the history of the Moon, then the highlands be about 10 times older and therefore will have been formed about 38 billion years ago.

Answer:

E = 1.38 x 10⁸ eV = 138 MeV

Explanation:

The energy associated with the given wavelength can be found from the following formula:

[tex]E = \frac{hc}{\lambda}[/tex]

where,

E = Energy of electron = ?

h = Plank's Constant = 6.625 x 10⁻³⁴ J.s

c = Speed of Light = 3 x 10⁸ m/s

λ = wavelength = 9 fm = 9 x 10⁻¹⁵ m

Therefore,

[tex]E = \frac{(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{9\ x\ 10^{-15}\ m}\\\\E = (2.21\ x\ 10^{-11}\ J)(\frac{1\ eV}{1.6\ x\ 10^{-19}\ J})[/tex]

E = 1.38 x 10⁸ eV = 138 MeV

Answer:

C) 26.6

Explanation:

I don't know how to calculate vector

The angle between the two vectors is 117⁰

The given parameters;

vector A = 3.00i + 1.00j

vector B = -3.00i + 3.00j

The  angle between the two vectors is calculated as follows;

[tex]cos \ \theta = \frac{A\ . \ B}{|A| \ . \ |B|}[/tex]

The dot product of vector A and B is calculated as;

[tex]A \ . \ B = (3i \ + j) \ . \ (-3i \ + 3j) = (3\times -3) + (1 \times 3) = -9 + 3 =- 6[/tex]

The magnitude of vector A and B is calculated as;

[tex]|A| = \sqrt{3^2 + 1^2} = \sqrt{10} \\\\|B| = \sqrt{(-3)^2 + (3)^2} = \sqrt{18}[/tex]

The  angle between the two vectors is calculated as;

[tex]cos \ \theta = \frac{-6}{\sqrt{10} \ . \sqrt{18} } \\\\cos \ \theta = \frac{-6}{\sqrt{180} } \\\\cos \ \theta = -0.4472\\\\\theta = cos \ ^{-1} (-0.4472) \\\\\theta = 116.6^0 \approx 117^0[/tex]

Thus, the angle between the two vectors is 117⁰

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Answer:

B

Explanation:

Answer:

B

Explanation:

Answer:

a. When the magnetic field is constant, there is only a magnetic field in the region of space. When the magnetic field is time dependent, there will be both a magnetic field and an electric field in the region of space.

Explanation:

This is true because according to Maxwell's equation, a time dependent magnetic field produces an electric field.

When the magnetic field is constant, there is no electric field in that region of space. But, when it begins to vary, an electric field is then generated by Maxwell's equation which is given by

CurlE = -dB/dt

So, there is initially no electric field when the magnetic field is constant, but it exists when the magnetic field becomes time dependent.

So when dB/dt = 0

CurlE = 0 ⇒ E = 0

So when dB/dt ≠ 0

CurlE ≠ 0 ⇒ E ≠ 0

Answer:

[tex]B_{int}=-0.015T[/tex]

Explanation:

From the question we are told that:

RF source oscillation speed [tex]\sigma= 34 MHz[/tex]

The external field [tex]Bext =0.78 T[/tex].

Pro- ton magnetic moment component [tex]\mu=1.41 X 10-26 J/T[/tex]

Generally the equation for magnitude of [tex]B_{int}[/tex] is mathematically given by

[tex]B_{int}=B_{ext}-\frac{h\triangle \sigma}{2 \mu}[/tex]

[tex]B_{int}=0.78-\frac{6.6*10^{-34}*34*10^6}{2*1.41*10^{26}}[/tex]

[tex]B_{int}=0.78-0.7957[/tex]

[tex]B_{int}=-0.015T[/tex]

Answer:

It makes it lighter when its closer and heavier when its farther way.

Explanation:

Answer:

0.3625

Explanation:

From the given information:

Consider the equilibrium conditions;

On the ladder, net torque= 0

Thus,

[tex]\tau_{net} = 0[/tex]; and

[tex]-fL \s in \theta +m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta = 0[/tex]

However, by rearrangement;

[tex]fL \s in \theta =m_L g \dfrac{L}{2} cos \theta + mg (7.46\ m) cos \theta \\ \\ \mu(m_L + m) gL \ sin \theta = (323 \ N) ( 10.8 \ meters) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0[/tex]

[tex]\mu= \dfrac{ (323 \ N) ( 10.8 \ m) \ cos 56^0 + (734 \ N) (7.46 \ m) \ cos \ 66.46^0}{\Big [(323 \ N)+(734 \ N) \Big] (10.8 \ m)}[/tex]

[tex]\mathbf{\mu= 0.3625 }[/tex]

Answer:

89 N downward.

Explanation:

Both forces are working in the same direction -- down.

Therefore you can add them together.

F = F1 + F2

F is the total force

F1 is the gravitational force

F2 is the force the person provides

F1 = 19

F2 = 70

F = ?

F = 19 + 70

F = 89 Newtons downward.

Answer:

wE dONt taLk aBt brunooooOoooOoOo ;')

Explanation:

120 works for acellus

Answer: 119.9

Explanation:

F = (mv^2)/r

Here we know m (0.309kg), v (12.9m/s) and r (0.429m)

So F = (0.309*12.9^2)/0.429 = 119.861748

The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.

The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.

In this case, we need to determine that specific heat for solid and liquid states of matter. By Heat Physics, we understand that specific heat is contained in the slopes of the two sensible phases in the following form:

[tex]\frac{\Delta T}{\Delta Q} = \frac{1}{m\cdot c}[/tex] (1)

Where:

Solid phase

If we know that [tex]m = 9.80\,g[/tex], [tex]T_{1} = 40\,^{\circ}C[/tex], [tex]T_{2} = 235\,^{\circ}C[/tex], [tex]Q_{1} = 183\,J[/tex] and [tex]Q_{2} = 819\,J[/tex], then the specific heat of the solid phase is:

[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]

[tex]c = \frac{819\,J-183\,J}{(9.80\,g)\cdot (235\,^{\circ}C - 40\,^{\circ}C)}[/tex]

[tex]c = 0.333\,\frac{J}{g\cdot ^{\circ}C}[/tex]

The specific heat of the solid phase is 0.333 joules per gram-degree Celsius.

Liquid phase

If we know that [tex]m = 9.80\,g[/tex], [tex]T_{3} = 230\,^{\circ}C[/tex], [tex]T_{4} = 471\,^{\circ}C[/tex], [tex]Q_{3} = 1470\,J[/tex] and [tex]Q_{4} = 2870\,J[/tex], then the specific heat of the liquid phase is:

[tex]c = \frac{\Delta Q}{m\cdot \Delta T}[/tex]

[tex]c = \frac{2870\,J - 1470\,J}{(9.80\,g)\cdot (471\,^{\circ}C - 230\,^{\circ}C)}[/tex]

[tex]c = 0.593\,\frac{J}{g\cdot ^{\circ}C}[/tex]

The specific heat of the liquid phase is 0.593 joules per gram-degree Celsius.

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                  least two STDs that are caused by different microorganisms (such as bacteria, viruses, and so on). Propose at least two ethical strategies for helping to prevent these STDs OTHER THAN abstinence and male and female condoms.

Answer:

Explanation:

Chlamydia is a bacterial infection that affects the reproductive system in a myriad of nasty and invasive ways, including pain during micturition, genital discharge, and spreading to the rectum and eyes. Other than safe sex practices and abstinence, ways to prevent and treat Chlamydia are antibiotics for both partners (if they're both infected) and surgical intervention in severe cases.

Trichomoniasis is a parasitic sexually transmitted disease that infects women but can affect men during intercourse, women who are infected are more likely to suffer from pain in the vaginal area during moments including micturition and intercourse as foul-smelling discharge. The more concerning aspect of this disease is how women afflicted are more likely to give premature birth and spread the infection to the baby upon birth. Much like Chlamydia, Trichomoniasis can be prevented via safe-sex practices and abstinence, but it can also be treated by antibiotics for both sexual partners to prevent spreading it between them.

Answer:

a) false

b) True

c) True

d) False

e) False

Explanation:

a) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

b) True

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

c) True

d) False

For a diverging lens, the focal length is negative while it is positive for a converging lens

e) False

Image distances  for virtual images are always negative and it also forms on the the same side of the lens as the object and is enlarged

Answer:

moving

Explanation:

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Answer:

B

Explanation:

Answer:

Part A:

Distance=864000 m=864 km

Part B:

Energy Used=ΔE=8638000 Joules

Part C:

[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]

Explanation:

Given Data:

v=20m/s

Time =t=12 hours

In Secs:

Time=12*60*60=43200 secs

Solution:

Part A:

Distance = Speed**Time

Distance=v*t

Distance= 20*43200

Distance=864000 m=864 km

Part B:

Energy Used=ΔE= Energy Required-Kinetic Energy of swans

Energy Required to move= Power Required*time

Energy Required to move=200*43200=8640000 Joules

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

[tex]K.E\ of\ Swans=\frac{1}{2} *10*(20)^2=2000\ Joules[/tex]

Energy Used=ΔE=8640000 -2000

Energy Used=ΔE=8638000 Joules

Part C:

Fraction of Mass used=Δm/m

For This first calculate fraction of energy used:

Fraction of energy=ΔE/Energy required to move

ΔE is calculated in part B

Fraction of energy=8638000/8640000

Fraction of energy=0.99977

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

Now, the relation between energies ratio and masses is:

[tex]\frac{\triangle E}{E}=\frac{\triangle m}{2m}v^2[/tex]

[tex]\frac{\triangle m}{m}=\frac{2}{v^2} *\frac{\triangle E}{E}\\\frac{\triangle m}{m}=\frac{2}{20^2} *0.99977[/tex]

[tex]\frac{\triangle m}{m}=0.004998=0.49985\%[/tex]

Answer:

the Earth's crust is broken into about 12 plates that float on hotter, softer rocks in the underlying mantle

Explanation:

Answer:

[tex]0.0084575\ \text{T}[/tex]

[tex]0.272\ \text{m}[/tex]

2.2 cm easily observable

Explanation:

[tex]m_1[/tex] = Mass of 12 C = [tex]1.99\times 10^{-26}\ \text{kg}[/tex]

[tex]m_2[/tex] = Mass of 13 C = [tex]2.16\times 10^{-26}\ \text{kg}[/tex]

[tex]r_1[/tex] = Radius of 12 C = [tex]\dfrac{25}{2}=12.5\ \text{cm}[/tex]

B = Magnetic field

v = Velocity of atom = 8.5 km/s

[tex]r_2[/tex] = Radius of 13 C

The force balance of the system is

[tex]qvB=\dfrac{m_1v^2}{r}\\\Rightarrow B=\dfrac{m_1v}{rq}\\\Rightarrow B=\dfrac{1.99\times 10^{-26}\times 8500}{12.5\times 10^{-2}\times 1.6\times 10^{-19}}\\\Rightarrow B=0.0084575\ \text{T}[/tex]

The required magnetic field is [tex]0.0084575\ \text{T}[/tex]

Radius is given by

[tex]r=\dfrac{mv}{qB}[/tex]

[tex]r\propto m[/tex]

So

[tex]\dfrac{r_2}{r_1}=\dfrac{m_2}{m_1}\\\Rightarrow r_2=\dfrac{m_2}{m_1}r_1\\\Rightarrow r_2=\dfrac{2.16\times 10^{-26}}{1.99\times 10^{-26}}\times 12.5\times 10^{-2}\\\Rightarrow r_2=0.136\ \text{m}[/tex]

The required diameter is [tex]2\times 0.136=0.272\ \text{m}[/tex]

Separation is given by

[tex]2(r_2-r_1)=2(0.136-0.125)=0.022\ \text{m}[/tex]

The distance of separation is 2.2 cm which is easily observable.

Answer:

1482.35J

Explanation:

Efficency=(output energy/input energy) x100

17.0 =(252/input energy) x100

input energy =100x252/17=1482.35J

wasted energy=1482.35-252=1230.35J