Please help with this question​

Answer:

The system will tend to person A

Explanation:

A force is defined as either a pull or a push, in this scenario person A weighs lass that person B, so the resultant effect of the 10N interactive force will tend towards person A.

This is solely because person A is less than person B in weigh

Answer:

67.5

Explanation:

The plane accelerates at 2.7m/s,^2

Time is 25 seconds

The velocity can be calculated as follows

= 25×2.7

= 67.5

Hence the speed f the plane is 67.5

Answer:

Waves. Refraction is an effect that occurs when a light wave, incident at an angle away from the normal, passes a boundary from one medium into another in which there is a change in velocity of the light. ... The wavelength decreases as the light enters the medium and the light wave changes direction.

Explanation:

As a wavelength increases in size, its frequency and energy (E) decrease. From these equations you may realize that as the frequency increases, the wavelength gets shorter. ... Mechanical and electromagnetic waves with long wavelengths contain less energy than waves with short wavelengths.

Answer:

0.011

Explanation:

Translate to English please

Answer:

T = 0.01 s

Explanation:

Given that,

The frequency of the beats of a hummingbird, f = 100 Hz

We need to find the period of the hummingbirds flaps. Let the time is t. We know that the relation between frequency and time period is given by :

T = 1/f

Put all the values,

T = 1/100 = 0.01 s

So, the time period of the humming bird is 0.01 s.

Answer:

Aerobic Fitness. Aerobic fitness improves overall health and well-being. ...

Muscular Fitness. Strength training improves your muscle and bone health and helps with weight loss. ...

Flexibility. Flexibility allows you to move your body freely. ...

Stability and Balance

Explanation:

UP

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = [tex]\sqrt{\frac{2y_o}{g} }[/tex]

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

Answer:

[tex] \large{ \tt{☄ EXPLANATION}} : [/tex]

[tex] \large{ \tt{♡ \: LET'S \: GET \: STARTED}} : [/tex]

[tex] \large{ \boxed{ \tt{❃ \: WORK ( \: W \: ) = FORCE\: ( \: F) \times DISPLACEMENT \: ( \: D \: )}}}[/tex]

- Plug the values and then simplify !

[tex] \large{ \bf{↦ \: 150 = F \times 10}}[/tex]

[tex] \large{ \bf{↦F\times 10 = 150}}[/tex]

[tex] \large{ \bf{↦ \: F = \frac{150}{10}}} [/tex]

[tex] \large {\bf{↦F = \boxed{ \bf{15} \: N}}}[/tex]

[tex] \boxed{ \boxed{ \large{ \tt{✤ \: OUR \: FINAL \: ANSWER : \boxed{ \bf{15 \: n}}}}}}[/tex]

ツ Hope I helped ! ۵

☪ Have a wonderful day / evening ! ☼

# StayInAndExplore ! ☂

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Answer:

[tex]D_T=18.567m[/tex]

Explanation:

From the question we are told that:

Acceleration [tex]a=8.0 m/s^2[/tex]

Displacement [tex]d=1.05 m[/tex]

Initial time [tex]t_1=6.0s[/tex]

Final Time [tex]t_2=2.5s[/tex]

Generally the equation for Velocity of 1.05 travel is mathematically given by

Using Newton's Law of Motion

 [tex]V^2=2as[/tex]

 [tex]V=\sqrt{2*6*1.05}[/tex]

 [tex]V=4.1m/s[/tex]

Generally the equation for Distance traveled before stop is mathematically given by

 [tex]d_2=v*t_1[/tex]

 [tex]d_2=3.098*4[/tex]

 [tex]d_2=12.392[/tex]

Generally the equation for Distance to stop is mathematically given by

Since For this Final section

Final velocity [tex]v_3=0 m/s[/tex]

Initial velocity [tex]u_3=4.1 m/s[/tex]

Therefore

Using Newton's Law of Motion

 [tex]-a_3=(4.1)/(2.5)[/tex]

 [tex]-a_3=1.64m/s^2[/tex]

Giving

 [tex]v_3^2=u^2-2ad_3[/tex]

Therefore

 [tex]d_3=\frac{u_3^2}{2ad_3}[/tex]

 [tex]d_3=\frac{4.1^2}{2*1.64}[/tex]

 [tex]d_3=5.125m[/tex]

Generally the Total Distance Traveled is mathematically given by

 [tex]D_T=d_1+d_2+d_3[/tex]

 [tex]D_T=5.125m+12.392+1.05 m[/tex]

 [tex]D_T=18.567m[/tex]

Answer:

Each parent has a set of genes, and a combination of the genes from the mom and the dad could come together to make the child look totally different.

A child looks different to both parents because children usually look like a "combination" of both their parents. Each parent gives their child their genes.  So a child wouldn't look exactly like one parent.

[tex]\huge{ \mathrm{  \underline{ Answer }\:  \:  ✓ }}[/tex]

Total force acting on right side = 800 N

Total force acting on left side :

Now, equivalent force acting on the plane is :

And the direction of equivalent force will be the direction of greater force, that is right direction.

Hence, Correct option is :

_____________________________

[tex]\mathrm{ \:TeeNForeveR\:}[/tex]

Force is a push or pull that changes or attempts to change the shape or size of an object's position.

F = m × a

Here

F= Force

m= mass of object

a= Acceleration

Answer:

Sound is transmitted by longitudinal waves thru air - the observer cannot hear the explosion because no air is present between the moon and the ship

Light is transmitted by electromagnetic waves so the light from the explosion can travel between the  moon and the observer.

Explanation:

given Time Period = 5 Sec

Amplitude = 1m, Wave length = 15 m

using time period = 1/ frequency , Frequency = 1/5 cycles per sec = 0.2 sec = 12 per min

wave speed= Frequency * Wave length

speed = 0.2*15 = 3m/s

The answer for this question is negative externality

Answer:

a)  [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

b)  [tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]

c) [tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]

Explanation:

From the question we are told that:

Kinetic energy of Proton[tex]K.E_p= 6.0 MeV[/tex]

Radius [tex]r=0.750[/tex]

Energy [tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

a)

[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

b)

Generally the equation for acceleration of proton is mathematically given by

[tex]a_p=\frac{v^2}{r}[/tex]

Where

Speed of Proton particle is

[tex]V_p=2.12 *10^3 m/s[/tex]

[tex]a_p=\frac{(2.12 *10^3)^2}{0.750}\\a_p=1.27*10^{10}m/s^2[/tex]

Therefore

[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(1.27*10^{10})^2)}{(3*10^8)^3}[/tex]

[tex]\frac{dE}{dt}=6.881810^{-51}J/s[/tex]

Energy radiated per sec

[tex]mev=\frac{6.881810^{-51}}{1.6*10^{-19}}\\mev=4.3*10^{-32}ev[/tex]

Therefore the Fraction of its energy  it radiates per second is given as

[tex]x=\frac{ K.E_p}{mev}\\[/tex]

[tex]x=\frac{6.0}{4.3*10^{-32}}[/tex]

c)

Generally the equation for acceleration of proton is mathematically given by

[tex]a_p=\frac{v^2}{r}[/tex]

Where

Speed of Proton particle is

[tex]V_p=2.5 *10^7 m/s[/tex]

[tex]a_p=\frac{(2.5 *10^7)^2}{0.750}\\a_p=0.33*10^{14}m/s^2[/tex]

Therefore

[tex]\frac{dE}{dt}=\frac{q^2a^2}{4\pi E_0C^3}[/tex]

[tex]\frac{dE}{dt}=\frac{9*10^9*(1.6*10^{-19}^2*(0.33*10^{14})^2)}{(3*10^8)^3}[/tex]

[tex]\frac{dE}{dt}=59.26^{-25}J/s[/tex]

Energy radiated per sec

[tex]mev=\frac{59.26^{-25}}{1.6*10^{-19}}\\mev=2.22*10^{-3}ev[/tex]

Therefore the Fraction of its energy  it radiates per second is given as

[tex]x'=\frac{ K.E_p}{mev}\\[/tex]

[tex]x'=\frac{6.0}{2.22*10^{-3}}[/tex]

Answer:

 v_f = 20 m / s

Explanation:

For this exercise let's use the relationship between momentum and moment

        I = Δp

        F t = m v_f - m v₀

as the body starts from rest v₀ = 0

        F t = m v_f

        v_f = [tex]\frac{F t}{m}[/tex]

let's calculate

        v_f = 4 2 / 0.4

         v_f = 20 m / s

Answer:

C.

A factor of something is an ingredient, or it makes up that thing. For example, in math, we always try to find the factors of a number. We are basically trying to find the "ingredients" of a number.

Answer:

P = 113.4 N

Explanation:

We know that if it is on the Moon it has another gravity. Therefore, it is known through the International System that the gravity of the Moon is 1.62 m/s², so..

Data:

Use formula:

Replace and solve:

The man's weight is 113.4 Newtons.

Greetings.

Answer: [tex]3920\ J[/tex]

Explanation:

Given

mass of ball m=10 kg

It is placed at a height of 150 m

It is dropped from the height and allowed to free fall for 40 m

Velocity acquired by the ball during this fall is given by [tex]v^2-u^2=2as[/tex]

Insert u=0, a=g

[tex]\Rightarrow v^2-0=2\times 9.8\times 40\\\Rightarrow v=\sqrt{784}\\\Rightarrow v=28\ m/s[/tex]

Kinetic energy at this instant

[tex]K.E.=\dfrac{1}{2}\times 10\times 28^2\\\\\Rightarrow K.E.=3920\ J[/tex]

Answer:

The path difference must be equal to an integral (1 * lambda, 2 * lambda, -------n * lambda) number of wavelengths for constructive interference to occur.

Answer:

Explanation:

Angular momentum is L = mvr:

L = (5.2)(1.8)(.86) so

L = 8.05 kg*m/s

Answer:

Initial y-component of speed

Vy = 28 * sin 30 = 14 m/sec vertically

1/2 m Vy^2 = 2 m g h    conservation of energy of y-component

h = Vy^2 / (2 * g) = 14^2 / (2 * 9.8) = 10 m

Answer:

Coefficient of kinetic friction = 0.38 (Approx.)

Explanation:

Given:

Applied force = 45 N

Mass of wooden crate = 12 kg

Find:

Coefficient of kinetic friction

Computation:

Coefficient of kinetic friction = Applied force / (Mass)(Acceleration due to gravity)

Coefficient of kinetic friction = 45 / (12)(9.8)

Coefficient of kinetic friction = 45 / 117.6

Coefficient of kinetic friction = 0.3826

Coefficient of kinetic friction = 0.38 (Approx.)