En 2.0 s, una particula con aceleración constante a lo largo del eje x se mueve desde x =10 m
hasta x =50 m. La rapidez al final del recorrido es de 10 m/s. ¿Cuál es la aceleración de la partícula?

Explanation:

9420 km/hr is the correct answer

Hope this helps...☺

Answer:

I don't see anything on your question?

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Answer:

Answer:

Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.

Answer:

u=8.868 m/s

Explanation:

The displacement of the ball is 4 meters

The final speed of the ball is 0.5 m/s

The initial speed of the ball is to be calculated. Using the equation of the rectilinear motion,

[tex]v^{2} =u^{2} +2as[/tex]

Plugging the values in the above expression,

[tex]\\0.5^{2} =u^{2} +2*(-9.8)*4\\\\u^{2} =78.4+0.25\\\\u^{2} =78.65\\\\u=8.868 m/s[/tex]

Answer:

Fr = 1.91 * 9*10⁹*q²/L²

Explanation:

Let´s say that the corners of the square are  A B C and D

We are going to find out the force on the charge placed on B  ( the charge placed in the upper right corner.

As all the charges are positive (the same sign), then all the three forces on the charge in B are of rejection.

Force due to charge placed in A

module   Fₓ =  K* q² / L²   in the direction of x

Force due to charge placed in C

module  Fy = K* q²/L²   in the direction of y

Force due to  the charge placed in D

That force will have the direction of the diagonal of the square, and the distance between charges placed in D and A is the length of the diagonal.

d²  =  L²  +  L²  =  2*L²

d  =  √2 * L

The module of the force due to charge place in D

F₄₅ = K*q²/ 2*L²

To get the force we need to add first  Fₓ  and  Fy  

Fx + Fy  =  F₁

module of  F₁ = √ Fx² + Fy²    the direction will be the same as the diagonal of the square then:

F₁   =   √  ( K* q²/L² )²  +   ( K* q²/L² )²

F₁  =  √ 2  *  K*q²/L²

And now we add forces F₁   and F₄₅   to get the net force Fr on charge in point B.

The direction of Fr is the direction of the diagonal and is of rejection

the module is

Fr  =  F₁  *  F₄₅

Fr  =  √ 2  *  K*q²/L²  +   K*q²/ 2*L²

Fr  = ( √ 2 + 0,5 ) * K*q² /L²

K  =  9*10⁹  Nm²C²

Fr = 1.91 * 9*10⁹*q²/L²

We don´t know units of L and q

Answer:

We want to solve the sum:

6.74*10⁴ + 8.95*10⁴

first, we take the common factor 10⁴ out, so we get:

(6.74 + 8.95)*10⁴

Now we solve the sum:

(15.66)*10⁴

Now we want to rewrite it in exponential form, wo we can rewrite it as:

(15.66)*10⁴ = (1.566*10)*10⁴ = (1.566)*10*10⁴ = (1.566)*10⁴⁺¹ = 1.566*10⁵

k = 5.

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

[tex]I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2[/tex]

The maximum magnetic field is calculated using the following intensity formula;

[tex]I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T[/tex]

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

Answer:

499.7 J

Explanation:

Since total mechanical energy is conserved,

U₁ + K₁ + W₁ = U₂ + K₂ + W₂ where U₁ = potential energy at bottom of incline = mgh₁, K₁ = kinetic energy at bottom of incline = 1/2mv₁² and W₁ = work done by friction at bottom of incline, and U₂ = potential energy at top of incline = mgh₂, K₁ = kinetic energy at top of incline = 1/2mv₂² and W₂ = work done by friction at top of incline. m = mass = 4 kg, h₁ = 0 m, v₁ = 21.3 m/s, W₁ = 0 J, h₂ = radius of circular ramp = 10 m, v₂ = 2.8 m/s, W₂ = unknown.

So, U₁ + K₁ + W₁ = U₂ + K₂ + W₂

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

Substituting the values of the variables into the equation, we have

mgh₁ + 1/2mv₁²  + W₁ = mgh₂ + 1/2mv₂²  + W₂

4 kg × 9.8 m/s²(0) + 1/2 × 4 kg × (21.3 m/s)²  + 0 = 4 kg × 9.8 m/s² × 10 m + 1/2 × 4 kg × (2.8 m/s)²  + W₂

0 + 2 kg × 453.69 m²/s² = 392 kgm²/s² + 2 kg × 7.84 m²/s²  + W₂

907.38 kgm²/s² = 392 kgm²/s² + 15.68 kgm²/s²  + W₂

907.38 kgm²/s² = 407.68 kgm²/s² + W₂

W₂ = 907.38 kgm²/s² - 407.68 kgm²/s²

W₂ = 499.7 kgm²/s²

W₂ = 499.7 J

Since friction is a non-conservative force, the work done by all the non-conservative forces is thus W₂ = 499.7 J

The reason why we are unlikely to see fluctuations in light output in extragalactic sources with a diameter of about one light day over timescales shorter than about one day is due to the size and distance of the source, as well as the speed of light.

When we observe an extragalactic source with a diameter of about one light day, we are essentially observing light that has traveled a very long distance through space to reach us. This light may have originated from a region of the source that is changing in brightness or emitting intense bursts of light, but by the time the light reaches us, these fluctuations are smeared out over a longer period of time due to the speed of light.

For example, if the source were emitting a burst of light that lasted for only a few hours, by the time that light travelled a distance of one light day (which is about 25 billion miles or 40 billion kilometres), the burst would be spread out over a longer period of time. This is because the light emitted at the beginning of the burst would have already traveled a significant distance away from the light emitted at the end of the burst by the time it reached us. As a result, we would observe the burst as a more gradual increase and decrease in light output over a period of several days, rather than a sharp increase and decrease over a few hours.

In addition, the turbulent interstellar and intergalactic media that the light passes through can also scatter and delay the light, further smearing out any short-term fluctuations in light output. This effect is known as interstellar scintillation and can make it even more difficult to observe short-term variations in the light output of extragalactic sources.

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Answer:

the person  is sitting 1.5 m from the left end of the board

Explanation:

Given the data in the question;

Wb = 125 N

Wm = 500 N

T₂ = 250 N

Now, we know that;

T₁ + T₂ = Wb + Wm

T₁ + 250 = 125 + 500

T₁ = 125 + 500 - 250

T₁ = 375 N

so tension of the left chain is 375 N.

Now, taking torque about the left end

500 × d + 125 × 2 = 250 × 4

500d + 250 = 1000

500d = 1000 - 250

500d = 750

d = 750 / 500

d = 1.5 m

Therefore, the person  is sitting 1.5 m from the left end of the board.

Explanation:

Specifically his leg muscles. As the leg muscles expand, they push down on the ground. Newton's 3rd law says that for any action, there's an opposite and equal reaction. That means a downward push into the ground will have the ground push back, more or less, and that's why the kangaroo will jump. The ground (and the earth entirely) being much more massive compared to the animal means that the ground doesn't move while the kangaroo does move. Perhaps on a very microscopic tiny level the ground/earth does move but it's so small that we practically consider it 0.

This experiment can be done with a wall as well. Go up to a wall and lean against it with your hands. Then do a pushup to move further away from the wall, but you don't necessarily need to lose contact with the wall's surface. As you push against the wall, the wall pushes back, and that causes you to move backward. If the wall was something flimsy like cardboard, then you could easily push the wall over and you wouldn't move back very much. It all depends how much mass is in the object you're pushing on.  

Answer:

0.3333

Explanation:

Acceleration = change in velocity/time

a = 20 m/s / 60 m/s

a = 0.3333 m/s^2

Answer:

Explanation:

From the question;

We will make assumptions of certain values since they are not given but the process to achieve the end result will be the same thing.

We are to calculate the following task, i.e. to determine the electric field at the distances:

a)  at 4.75 cm

b)  at 20.5 cm

c) at 125.0 cm

Given that:

the charge (q) = 33.3 nC/m

= 33.3 × 10⁻⁹ c/m

radius of rod = 5.75 cm

a) from the given information, we will realize that the distance lies inside the rod. Provided that there is no charge distribution inside the rod.

Then, the electric field will be zero.

b) The electric field formula [tex]E = \dfrac{kq }{d}[/tex]

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{0.205}[/tex]

E = 1461.95 N/C

c) The electric field E is calculated as:

[tex]E = \dfrac{9 \times 10^9 \times (33.3 \times 10^{-9}) }{1.25}[/tex]

E = 239.76 N/C

Answer:

We can also prove the conservation of mechanical energy of a freely falling body by the work-energy theorem, which states that change in kinetic energy of a body is equal to work done on it. i.e. W=ΔK. And ΔE=ΔK+ΔU. Hence the mechanical energy of the body is conserved

Explanation:

Answer:

Explanation:

There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.

Answer:

Explanation:

a)

Given that:

V = 25 mi/hr

To ft/sec, we have:

[tex]V = 25 \times \dfrac{5280}{3600} ft/s[/tex]

[tex]V = \dfrac{110}{3} ft/s[/tex]

[tex]\rho = 0.075 \ lb/ft^3[/tex]

[tex]\rho = 0.075 \times \dfrac{1 \ lbf s^2/ft}{32.174 \ lbm}[/tex]

[tex]\rho = \dfrac{0.075}{32.174 } lbf.s^2/ft^4[/tex]

[tex]C_d = 0.28[/tex]

A = 25ft²

Recall that:

The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]

[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{110}{3})^2[/tex]

[tex]F_d =10.967 \ lbf[/tex]

[tex]P = F_dV \\ \\ P = 10.97 \times (\dfrac{110}{3}} \\ \\ P = 402.3 \ hp[/tex]

For 70 miles per hour, we have:

[tex]V = 70 \times \dfrac{5280}{3600} ft/s[/tex]

[tex]V = \dfrac{308}{3} ft/s[/tex]

The drag force [tex]F_d =\dfrac{C_dA \rho V^2}{2}[/tex]

[tex]F_d =\dfrac{1}{2}\times 0.28 \times 25\times \dfrac{0.075}{32.174}\times (\dfrac{308}{3})^2[/tex]

[tex]F_d =85.99 \ lbf[/tex]

[tex]P = F_dV \\ \\ P = 85.99 \times (\dfrac{308}{3}}) \\ \\ P = 8828.2 \ hp[/tex]

k = [tex] \dfrac{ (\dfrac{h}{ \lambda} )^{2} }{2m} [/tex]

k = (6.626×10-¹⁹/590 × 10-⁹ )^{2} /2 × 1.673 × 10-²⁷

k = (1.12 × 10-³⁰)^2/3.346×10-²⁷

k = 1.25 × 10-⁶⁰ /3.346×10-²⁷

k = 0

ldk why, my answer is coming this :(

I don't know

because I don't know

Answer:

[tex]F=78.3hz[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=12[/tex]

Length [tex]l=80cm=0.8m[/tex]

Linear density [tex]\mu= 6.0g[/tex]

Generally the equation for Frequency is mathematically given by

 [tex]F=\frac{1}{2l}\sqrt{\frac{T}{K}}[/tex]

 [tex]F=\frac{1}{2(0.8)}\sqrt{\frac{12*9.8*0.8}{6*10^{-3}}}[/tex]

 [tex]F=78.3hz[/tex]

Answer:

a) the most probable speed of the molecules is 409.2 m/s

b) required temperature of xenon is 1322 K

Explanation:

Given the data in the question;

a)

Maximum probable speed of hydrogen molecule (H₂)

[tex]V_{H_2[/tex] = √( 2RT / [tex]M_{H_2[/tex] )

where R = 8.314 m³.Pa.K⁻¹.mol⁻¹ and given that T = 20.3 K

molar mass of H₂; [tex]M_{H_2[/tex] = 2.01588 g/mol

we substitute

[tex]V_{H_2[/tex] = √( (2 × 8.314 × 20.3 ) / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = √( 337.5484 / 2.01588 × 10⁻³  )

[tex]V_{H_2[/tex] = 409.2 m/s

Therefore, the most probable speed of the molecules is 409.2 m/s

b)

Temperature of xenon  = ?

Temperature of hydrogen = 20.3 K

we know that;

T = (Vxe² × Mxe) / 2R

molar mass of xenon; Mxe = 131.292 g/mol

so we substitute

T = ( (409.2)² × 131.292 × 10⁻³) / 2( 8.314  )

T = 21984.14167 / 16.628

T = 1322 K

Therefore, required temperature of xenon is 1322 K

Answer:

V = 2.58 m/s

Explanation:

Below is the calculation:

Given the weight of Erica = 37 kg

The weight of Danny = 45 kg

Danny's speed to move upward = 4.7 m/s

Use below formula to find the answer.

m1 * u1 = (m1+m2) * V

V = m1*u1 / (m1+m2)

Here, m1 = 45

u1 = 4.7

m1 = 45

m2 = 37

Now plug the values in formula:

V = m1*u1 / (m1+m2)

V = (45*4.7)/(45+37)

V = 2.58 m/s

Answer:

Josiah Spiers in 1867 was when scripture union was founded

Answer:

[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

Explanation:

The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.

The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]

From the given options, the second option is the correct answer.

[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]

Answer:

A

Explanation:

a pex

The color orange has a wavelength of 590 nm. The energy of an orange photon is approximately 0.337 eV.

The correct answer is option E.

To calculate the energy of a photon, we can use the equation:

E = (hc) / λ

where E is the energy of the photon, h is the Planck's constant (6.626 x [tex]10^-^3^4[/tex]J·s or 6.626 x[tex]10^-^1^9^[/tex] eV·s), c is the speed of light (3.00 x [tex]10^8[/tex] m/s), and λ is the wavelength of the light.

Given that the wavelength of orange light is 590 nm (or 590 x [tex]10^-^9[/tex]m), we can substitute the values into the equation:

E = [(6.626 x[tex]10^-^1^9^[/tex] eV·s) x (3.00 x [tex]10^8[/tex] m/s)] / (590 x[tex]10^-^9[/tex]m)

E = (1.9878 x [tex]10^-^1^0[/tex]eV·m) / (590 x [tex]10^-^9[/tex] m)

E = 3.3695 x [tex]10^-^1[/tex] eV

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The question probable may be:

The color orange has a wavelength of 590 nm. What is the energy of an orange photon? (h = 6.626 x [tex]10^-^1^9^[/tex], 1 eV = 1.6 x[tex]10^-^1^9^[/tex]J)

A) 2.81 eV

B) 3.89 eV

C) 2.10 eV

D) 2.78 eV

E)  0.337 eV

Answer:

because it is helpful to human beings I think