what is the value of gravitational constant​

Answer:

Microeconomics is a part of economics and the study of decisions made by people and businesses regarding the allocation of resources, and prices at which they trade goods and services.

Microeconomics helps business planning i.e. helps the business community to plan their costs, production, etc. in anticipation of demand in order to maximize profits. Microeconomics is useful in explaining how market mechanism determines the price in a free market economy.

Answer:

Solar radiation

Explanation:

Visible light, ultraviolet light, infrared, radio waves, X-rays, and gamma rays.

==>  Energy

==>  Radio noise, heat, visible light, ultraviolet radiation, X-rays, gamma rays

==>  They carry all these kinds of energy wherever they go.  Not only to the Earth.

Answer:

The total distance at 7 s is:

[tex]x_{tot}=27\: m[/tex]

Explanation:

Distance due to the force

We can use second Newton's law to find the acceleration.

[tex]F=ma[/tex]

[tex]a=\frac{F}{m}=\frac{9}{5}=1.8\: m/s^{2}[/tex]

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

[tex]x_{1}=0.5at_{1}^{2}[/tex]

[tex]x_{1}=0.5(1.8)(3)^{2}[/tex]

[tex]x_{1}=8.1\: m[/tex]

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

[tex]v=v_{i}+at_{1}=0+(1.8)3=5.4\: m/s[/tex]

So the second distance will be:

[tex]x_{2}=vt_{2}=5.4*4=21.6\: m[/tex]

Therefore, the total distance is:

[tex]x_{tot}=x_{1}+x_{2}=5.4+21.6=27\: m[/tex]

I hope it helps you!

The total distance at 7 s is: [tex]x_{total}=27\ m[/tex]

Force is an external agent applied on any object to displace it from its position. Force is a vector quantity, so with magnitude it also requires direction. Direction is necessary to examine the effect of the force and to find the equilibrium of the force.

Distance due to the force

We can use second Newton's law to find the acceleration.

[tex]F=ma[/tex]

[tex]a=\dfrac{F}{m}=\dfrac{9}{5}=1.8\ \frac{m}{s^2}[/tex]

Now, using the kinematic equation we will find the distance during this interval of time. Let's recall that the initial velocity is zero.

[tex]x_1=0.5at_1^2[/tex]

[tex]x_1=0.5(1.8)(3^2)[/tex]

[tex]x_1=8.1\ m[/tex]

In the second part of the motion, the object moves at a constant velocity, as long as there is no friction between the object and the floor.

First, we need to find the final velocity of the first interval

[tex]v=v_i+at_1=0+(1.8)(3)=5.4 \frac{m}{s}[/tex]

So the second distance will be:

[tex]x_2=vt_2=5.4\times 4=21.6\ m[/tex]

Therefore, the total distance is: [tex]x_{total}=27\ m[/tex]

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Explanation:

Given: a = -3v^2

By definition, the acceleration is the time derivative of velocity v:

[tex]a = \frac{dv}{dt} = - 3 {v}^{2} [/tex]

Re-arranging the expression above, we get

[tex] \frac{dv}{ {v}^{2} } = - 3dt[/tex]

Integrating this expression, we get

[tex] \int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt[/tex]

[tex] - \frac{1}{v} = - 3t + k[/tex]

Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as

[tex] - \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )[/tex]

or

[tex]v = \frac{10}{30t +1 } [/tex]

We also know that

[tex]v = \frac{ds}{dt} [/tex]

or

[tex]ds = vdt = \frac{10 \: dt}{30t + 1} [/tex]

We can integrate this to get s:

[tex]s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1} [/tex]

Let u = 30t +1

du = 30dt

so

[tex] \int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k[/tex]

[tex]= \frac{1}{30}\ln |30t + 1| + k[/tex]

So we can now write s as

[tex]s = \frac{1}{3}\ln |30t + 1| + k[/tex]

We know that when t = 0, s = 8 m, therefore k = 8 m.

[tex]s = \frac{1}{3}\ln |30t + 1| + 8[/tex]

Next, we need to find the position and velocity at t = 3 s. At t = 3 s,

[tex]v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s} [/tex]

[tex]s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m[/tex]

Note: velocity approaches zero as t --> [tex]\infty [/tex]

Answer:

Emissions per second = 0.36

Explanation:

Please find the attached question

Solution

Given  

Let X be the rate of background emission.

X = B/t  

Where B = 36

And t = 100

X = 36/100 = 0.36

Answer:

No,it isn't concave. The correct answer is convex lens.

Explanation:

A lens is a piece of transparent material bound by two surfaces of which at least one is curved. A lens bound by two spherical surfaces bulging outwards is called a bi-convex lens or simply a convex lens. A single piece of glass that curves outward and converges the light incident on it is also called a convex lens.

Convex lens is the answer.

See the attached diagram.

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

[tex]Mv+mu=Mv_1+mv_2[/tex]

(900 x 47) + (200 x -30)  = (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

36300 =  (900 x [tex]v_1[/tex]) + (200 x [tex]v_2[/tex])

[tex]9v_1 + 2v_2 = 363[/tex] ..............(i)

[tex]9v_1 = 363 - 2v_2[/tex]

[tex]v_1=\frac{363 - 2v_2}{9}[/tex]

The mathematical expression for the conservation of kinetic energy is

[tex]\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2[/tex]

[tex]\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2[/tex]    ................(ii)

[tex]$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$[/tex]  

[tex]21681 = 9v_1^2+2v_2^2[/tex]

Substituting (i) in (ii)

[tex]21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2[/tex]

[tex](363-2v_2)^2+18v_2^2=195129[/tex]

[tex](363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0[/tex]

[tex]22v_2^2-145v_2-63360=0[/tex]

Solving the equation, we get

[tex]v_2=96 \ m/s, -30 \ m/s[/tex]

The negative velocity is neglected.

Therefore, substituting 96 m/s for [tex]v_2[/tex] in (i), we get

[tex]v_1=\frac{363-(2 \times 96)}{9}[/tex]

     = 19

Thus, only impulse of importance is used to find final velocity.

Answer:

[tex]a=2401m/s^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=0.13kg[/tex]

Angular speed [tex]\omega=57rads/s[/tex]

Radius [tex]r=0.25m[/tex]

Time [tex]t=3s[/tex]

[tex]F=312.13N[/tex]

Generally the equation for centripetal acceleration is mathematically given by

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

[tex]a=\frac{312.13}{0.13}[/tex]

[tex]a=2401m/s^2[/tex]

Answer:

The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Explanation:

Given the data in the question;

since the train starts from rest,

Initial velocity; u = 0 m/s

final velocity; v = 42 m/s

distance covered S = 5.6 km = ( 5.6 × 1000 )m = 5600 m

acceleration a = ?

From the third equation of motion;

v² = u² + 2as

we substitute in our values

( 42 )² = ( 0 )² + [ 2 × a × 5600 ]

1764 = 0 + [ 11200 × a ]

1764 = 11200 × a

a = 1764 / 11200

a = 0.1575 ≈ 0.16 m/s²          { two decimal place }

Therefore, The acceleration during the first 5.6 km of travel is closest to 0.16 m/s²    

Option c) 0.16 m/s² is the correct answer.

Answer:

convex lens because it makes parallel light rays passing through it bends inwards and meets convex at a spot just beyond the lens known as the focal point

Answer:

Explanation:

Answer:

[tex]a=2.8*10^{-9}m/s[/tex]

Explanation:

From the question we are told that:

First Mass [tex]m=8.50kg[/tex]

2nd Mass [tex]m=14.5kg[/tex]

Distance

[tex]d_1=50=>0.50m\\\\d_2=>12cm=>0.12m[/tex]

Generally the Newtons equation for Gravitational force is mathematically given by

[tex]F_n=\frac{Gm_nm}{(r_n)^2}[/tex]

Therefore

Initial force on m

[tex]F_1=\frac{Gm_1m}{(r_1)^2}[/tex]

Final force on m

[tex]F_2=\frac{Gm_2m}{(r_2)^2}\\\\F=\frac{Gm_1m}{(r_1)^2}-\frac{Gm_2m}{(r_2)^2}[/tex]

Acceleration of m

[tex]a=\frac{F}{m}\\\\a=\frac{Gm_1}{r_1^2}-\frac{Gm_2}{r_2^2}[/tex]

[tex]a=6,67*10^{-11}{\frac{8.5}{0.12}}-\frac{14.5}{0.50}[/tex]

[tex]a=2.8*10^{-9}m/s[/tex]

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

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#SPJ2

Answer:

3.6 KJ

Explanation: Given that a 70-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.60 m, and ends with a speed of 8.5 m/s. How much nonconservative work (in kJ) was done on the boy

The workdone = the energy.

There are two different energies in the scenario - the potential energy (P.E ) and the kinetic energy ( K.E )

P.E = mgh

P.E = 70 × 9.8 × 1.6

P.E = 1097.6 J

P.E = 1.098 KJ

K.E = 1/2mv^2

K.E = 1/2 × 70 × 8.5^2

K.E = 2528.75 J

K.E = 2.529 KJ

The non conservative workdone = K.E + P.E

Work done = 1.098 + 2.529

Work done = 3.63 KJ

Therefore, the non conservative workdone is 3.6 KJ approximately

Answer:

a1 =  v1^2 / R

a2 =  v2^2 / R

a2 = (v2 / v1)^2 = (3 / 2)^2 = 9/4

F2 = 9/4 * F1 = 9/4 * 146 = N     329 N      since F = m * a

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

Explanation:

The centripetal force [tex]F_c[/tex] on the car must equal the frictional force f in order to avoid slipping off the road. Let's apply Newton's 2nd law to the y- and x-axes.

[tex]y:\:\:\:\:N - mg = 0[/tex]

[tex]x:\:\:F_c = f \Rightarrow \:\:\:m \dfrac{v^2}{r} = \mu N[/tex]

or

[tex]m \dfrac{v^2}{r} = \mu mg[/tex]

Solving for [tex]\mu[/tex],

[tex]\mu = \dfrac{v^2}{gr} = \dfrac{(12.0\:\frac{m}{s})^2}{(9.8\:\frac{m}{s^2})(52\:m)} = 0.28[/tex]

Answer:

2.55 m/s

Explanation:

A 3.00-kg crate slides down a ramp. the ramp is 1.00 m in length and inclined at an angle of 30° as shown in the figure. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.

Solution:

The work done by friction is given as:

[tex]W_f=F_f\Delta S\\\\Where\ F_f\ is\ the \ frictional\ force=-5N(the\ negative \ sign\ because\ it\\acts\ opposite\ to \ direction\ of\ motion),\Delta S=slope\ length=1\ m\\\\W_f=F_f\Delta S=-5\ N*1\ m=-5J[/tex]

The work done by gravity is:

[tex]W_g=F_g*s*cos(\theta)\\\\F_g=force\ due\ to\ gravity=mass*acceleration\ due\ to\ gravity=3\ kg*9.81\\m/s^2, s=1\ m, \theta=angle\ between\ force\ and\ displacement=90-30=60^o\\\\W_g=3\ kg*9.81\ m/s^2*1\ m*cos(60)=14.72\ J\\\\The\ Kinetic\ energy(KE)=W_f+W_g=14.72\ J-5\ J=9.72\ J\\\\Also, KE=\frac{1}{2} mv^2\\\\9.72=\frac{1}{2} (3)v^2\\\\v=\sqrt{\frac{2*9.72}{3} } =2.55\ m/s[/tex]

Answer:

B

Explanation:

Answer:

B

Explanation:

Bc i say

answer 8N 8N

la tensión del ladrillo inferior es 8N ya que los ladrillos son idénticos, decido 24 por 3 para darme 8

la tensión entre el ladrillo superior y el medio también es 8N

Complete question:

a metal of work function 1.8eV is illuminated by light of wavelength 3.0x*10-7 m,calculate the maximum kinetic energy of the emitted photons.

Answer:

the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J

Explanation:

Given;

work function, Ф = 1.8 eV

wavelength of the light, λ = 3 x 10⁻⁷ m

The maximum kinetic energy of the emitted photons is calculated from photoelectric equation.

[tex]E = K.E_{max} + \phi\\\\KE_{max} = E- \phi\\\\where;\\\\E \ is \ the \ energy \ of \ the \ incident \ light\\\\E = hf = h \frac{c}{\lambda} \\\\where;\\\\c \ is \ speed \ of \ light = 3 \times 10^8 \ m/s\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\E = \frac{ (6.626 \times 10^{-34})\times ( 3 \times 10^8)}{3\times 10^{-7}} \\\\E = 6.626 \times 10^{-19} \ J[/tex]

[tex]K.E_{max} = E - \phi\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ (1.8 \times 1.602 \times 10^{-19} \ J)\\\\K.E_{max} = 6.626\times 10^{-19} \ J \ - \ 2.884 \times 10^{-19} \ J\\\\K.E_{max} =3.742 \times 10^{-19} \ J[/tex]

Therefore, the maximum kinetic energy of the emitted photons is 3.742 x 10⁻¹⁹ J

[tex]\frac{dx}{dt} = 2.5 \: \frac{cm}{sec} [/tex]

Explanation:

The volume of a cube is V = x^3. Taking the time derivative of this expression, we get

[tex] \frac{dV}{dt} = 3 {x}^{2} \frac{dx}{dt} [/tex]

or

[tex]\frac{dx}{dt} = \frac{1}{3 {x}^{2}} \frac{dV}{dt} [/tex]

We know that dV/dt = 30 cm^3/sec so the value of dx/dt when x = 2 cm is

[tex]\frac{dx}{dt} = \frac{1}{3 {(2 \: cm)}^{2}}(30 \: \frac{ {cm}^{3} }{sec} ) = 2.5 \: \frac{cm}{sec} [/tex]

Answer:

Explanation:

[tex]V=x^3\\\\\frac{dV}{dt}=3x^2\frac{dx}{dt}\\\\30\frac{cm^3}{s}=3x^2\frac{dx}{dt}\\\\\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3x^2}~at~x=2cm,~\frac{dx}{dt}=\frac{30\frac{cm^3}{s}}{3*(2cm)^2}=\frac{5}{2}\frac{cm}{s}[/tex]

Answer:

by using p = mv equation we can find v,

6 = 1.3 v

4.615 = v

Answer:

t1= 8.40/10 =.84 s

t2 = 5.60/10 = .56s

t3= 1.4/10 = .14s

total time = 1.54 sec

Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

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