What is the Y-component of a vector A, which is of magnitude
16-12 and at a 45° angle to the horizontal?

Answer:

a) RADIATION, b) CONVECTION, c)  CONDUCTION, d) CONVECTION

Explanation:

In the heating processes there can be three types: conduction, convention and radiation.

The conduction process occurs when the movement of atoms or thermal agitation of molecules creates the transfer of thermal energy.

The convention process occurs when there is a movement of matter creating the transfer of energy

The process of Radiation an electromagnetic wave indexes on a material and is absorbed, creating the process of energy transfer.

Now let's examine each situation>

a)Warming of the Earth's surface on a sunny day

in this case the sunlight heats the earth as it is absorbed, which is why it is a RADIATION process

b) On a sunny afternoon, you watch cumulus clouds forming

in this case the nines rise from the surface, since it is a moving mass, the process is CONVECTION

c) A very shallow layer of air in contact with the ground is warmed

in this case the thermal movement of the layer molecules heat the earth, for which the process of CONDUCTION

d) A south wind carries warm air into the central United States

As we have a movement of a mass of matter, the process is CONVECTION.

Answer:

Rahul's weight

Explanation:

In fact, the force Rahul exerts on Earth corresponds to the force of gravity. But Rahul's weight is, in fact, the force of gravity exerted by the Earth on Rahul, and these two forces correspond to the action-reaction pair of Newton's third law, which states that the two forces are equal.

Using formulas, Rahul's weight is equal to

W=mg

where m is Rahul mass and g is the gravitational acceleration (g=9.81 m/s^2).

Answer:

0.444atm

Explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (

P2 = final pressure (

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question,

P1 = 101.3 kPa = 101.3 × 0.00987 = 0.999atm

P2 = ?

V1 = 80L

V2 = 160L (double of V1)

T1 = 34°C = 34 + 273 = 307K

T2 = 0°C = 0 + 273 = 273K

Using P1V1/T1 = P2V2/T2

0.999 × 80/307 = P2 × 160/273

79.92/307 = 160P2/273

Cross multiply

307 × 160P2 = 79.92 × 273

49120P2 = 21818.16

P2 = 21818.16 ÷ 49120

P2 = 0.444

P2 = 0.444atm

Answer:

Economists, I guess.

Explanation:

Answer:

the law of conservation of energy.

Explanation:

An electric circuit can be defined as an interconnection of electrical components which creates a path for the flow of electric charge (electrons) due to a driving voltage.

Generally, an electric circuit consists of electrical components such as resistors, capacitors, battery, transistors, switches, inductors, etc.

In Physics, a point where at least three circuit paths (wires) meet is referred to as a junction.

The Kirchhoff’s circuit laws are two(2) equations first published by Gustav Kirchhoff in 1845.

Fundamentally, they address the conservation of energy and charge in the context of electrical circuits.

One of the laws known as Kirchoff's Current Law (KCL) deals with the principle of application of conserved energy in electrical circuits.

Kirchoff's Current Law (KCL) states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction.

This simply means that the algebraic sum of currents in a network of conductors(wires) meeting at a point is equal to zero.

The Law of Conservation of Energy states that energy cannot be destroyed but can only be transformed or converted from one form to another.

This ultimately implies that, Kirchhoff's junction rule is a statement of the law of conservation of energy.

Answer:

1.5 m is ur answer

Explanation:

Solution :

The distance between the starting point and the end point, [tex]L_0[/tex] = 10 light years

But due to the relativistic motion of Bob and Charlie, the distance will be reduced following the Lorentz contraction. The contracted length will be different since they are moving with different speeds.

For Bob,

Speed of Bob's rocket with respect to Alice, [tex]L_b = 0.7 \ c[/tex]

So the distance appeared to Bob due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_b^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.49} \ Ly$[/tex]

    [tex]$=7.1 \ Ly$[/tex]

Therefore, the time required to finish the race by Bob is

[tex]$t_b = \frac{L_b}{V_b}$[/tex]

  [tex]$=\frac{7.1 \ c}{0.7 \ c}$[/tex]

  = 10.143 year

For Charlie,

Speed of Charlie's rocket with respect to Alice, [tex]L_c = 0.866 \ c[/tex]

So the distance appeared to Charlie due to the length contraction,

[tex]$L_b=L_0\sqrt{1-\frac{V_c^2}{c^2}$[/tex]

[tex]$L_b=10\times \sqrt{1-0.75} \ Ly$[/tex]

    [tex]$=5 \ Ly$[/tex]

The time required to finish the race by Charlie is

[tex]$t_b = \frac{L_c}{V_c}$[/tex]

  [tex]$=\frac{5 \ c}{0.866 \ c}$[/tex]

  = 5.77 year

Answer:

4 ohm

Explanation:

The equivalent resistance (Re) of three resistors in parallel is given by;

1/Re = 1/R1 + 1/R2 + 1/R3

Where; R1 = 16 ohm, R2 = 16 ohm, R3 = 8 ohm

1/Re= 1/16 + 1/16 + 1/8

1/Re= (0.0625) + (0.0625) + (0.125)

Re= 4 ohm

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

Answer:

venus - 2

earth - 3

mars - 4

mercury - 1

Explanation:

bosu

here is your answer

Answer:

The right solution is "24.39 per sec".

Explanation:

According to the question,

⇒ [tex]v=\frac{502.1}{\sqrt{3} }[/tex]

      [tex]=289.9 \ m/s[/tex]

The time will be:

⇒ [tex]t=\frac{d}{v}[/tex]

      [tex]=\frac{2\times 6}{289.9}[/tex]

      [tex]=\frac{12}{289.9}[/tex]

      [tex]=0.041 \ sec[/tex]

hence,

⇒ [tex]N=\frac{1}{t}[/tex]

        [tex]=\frac{1}{0.041}[/tex]

        [tex]=24.39 \ per \ sec[/tex]

Answer:

40479.6 J

Explanation:

Applying,

q = cm(t₂-t₁).................... Equation 1

Where q = change in heat content of the system, c = specific heat capacity of the system, m = mass of the system, t₁ = initial temperature, t₂ = final temperature.

From the question,

Given: m = 79 g = 0.079 kg, t₁ = 21°C, t₂ = 143°C

Constant: c = 4200 J/kg.°C

Substitute these values into equation 1

q = 4200(0.079)(143-21)

q = 331.8(122)

q = 40479.6 J

Answer:

[tex]V_2=3.3m/s[/tex]

Explanation:

From the question we are told that:

Distance [tex]d_1=1.4m[/tex]

Tangential speed [tex]V=2.2m/s[/tex]

Distance 2 [tex]d_2=2.1m[/tex]

Generally the equation for Angular velocity is mathematically given by

 [tex]w=\frac{v}{r}[/tex]

Therefore

 [tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]

 [tex]V_2=\frac{2.2*2.1}{1.4}[/tex]

 [tex]V_2=3.3m/s[/tex]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =[tex]F^>[/tex] × [tex]x^>[/tex]

W = [tex]Fxcos\theta[/tex]    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = [tex]Fxcos([/tex] 0° [tex])[/tex]

W = [tex]Fx[/tex] ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

[tex]W_{net[/tex] = ΔKE

= [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

[tex]Fx[/tex] = [tex]\frac{1}{2}[/tex]mv² -  [tex]\frac{1}{2}[/tex]mu²

we make F, the subject of formula

F = [tex]\frac{m}{2x}[/tex]( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = [tex]\frac{830}{(2)(0.255)}[/tex]( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

Answer:

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Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

[tex] f = f_{0}\frac{v + v_{o}}{v - v_{s}} [/tex]        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

[tex] v_{o}[/tex]: is the speed of the observer = 0 (it is heard in the town)

[tex] v_{s}[/tex]: is the speed of the source =?

The frequency of the train before slowing down is given by:

[tex] f_{b} = f_{0}\frac{v}{v - v_{s_{b}}} [/tex]  (1)                  

Now, the frequency of the train after slowing down is:

[tex] f_{a} = f_{0}\frac{v}{v - v_{s_{a}}} [/tex]   (2)  

Dividing equation (1) by (2) we have:

[tex] \frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}} [/tex]

[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}} [/tex]   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

[tex] v_{s_{b}} = 2v_{s_{a}} [/tex]     (4)

Now, by entering equation (4) into (3) we have:

[tex] \frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}} [/tex]  

[tex] \frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}} [/tex]

By solving the above equation for [tex]v_{s_{a}}[/tex] we can find the speed of the train after slowing down:

[tex] v_{s_{a}} = 11.06 m/s [/tex]

Finally, the speed of the train before slowing down is:

[tex] v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s [/tex]

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

Answer:

α = 13.7 rad / s²

Explanation:

Let's use Newton's second law for rotational motion

         ∑ τ = I α

we will assume that the counterclockwise turns are positive

         F₁  0 + F₂ R₂ - F₃ R₃ = I α

give us the cylinder moment of inertia

        I = ½ M R₂²

        α = (F₂ R₂ - F₃ R₃)  [tex]\frac{2}{M R_2^2}[/tex]

let's calculate

        α = (24  0.22 - 13  0.10) [tex]\frac{2}{12 \ 0.22^2}[/tex]2/12 0.22²

        α = 13.7 rad / s²

The boy pushed the sled for 16 seconds.

We have a boy who pushes his little brother on a sled.

We have to determine for how long time does boy push the sled.

The Work - Energy theorem states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

According to the question -

The sled is initially at rest → initial velocity (u) = 0.

Final velocity (v) = 4 m/s

Mass of boy and sled (M) = 40 kg

Power developed (P) = 20 W = 20 Joules/sec

According to work - energy theorem -

Work done (W) = Δ E(K) = E(f) - E(i)

Therefore -

W = ([tex]\frac{1}{2} \times 40 \times 4 \times 4 - \frac{1}{2}[/tex] x 40 x 0) = 320 Joule

Now, Power is defined as the rate of doing work -

P = [tex]\frac{dW}{dt}[/tex] = [tex]\frac{W}{t}[/tex]

20 = [tex]\frac{320}{t}[/tex]

t = 16 seconds

Hence, the boy pushed the sled for 16 seconds.

To solve more questions on Work, Energy and Power, visit the link below -

https://brainly.com/question/208670

#SPJ2

Answer:

[tex]I=0.902kg*m^2[/tex]

Explanation:

From the question we are told that:

Mass [tex]m=92.0kg[/tex]

Radius [tex]r=0.140m[/tex]

Generally the equation for moment of Inertia is mathematically given by

 [tex]I = 0.5*m*r^2[/tex]

 [tex]I=0.5*(92)*0.14^2[/tex]

 [tex]I=0.902kg*m^2[/tex]

Answer:

559.5 N

Explanation:

Applying,

v² = u²+2gs............. Equation 1

Where v = final velocity,

From the question,

Given: s = 5.10 m, u = 0 m/s ( from rest)

Constant: 9.8 m/s²

Therefore,

v² = 0²+2×9.8×5.1

v² = 99.96

v = √(99.96)

v = 9.99 m/s

As the diver eneters the water,

u = 9.99 m/s, v = 0 m/s

Given: t = 1.34 s

Apply

a = (v-u)/t

a = 9.99/1.34

a = -7.46 m/s²

F = ma.............. Equation 2

Where F = force, m = mass

Given: m = 75 kg, a = -7.46 m/s²,

F = 75(-7.46)

F = -559.5 N

Hence the average force exerted on the diver is 559.5 N

Answer:

See explanation below

Explanation:

The length of the arrow represents the magnitude/size of the force. The longer it is, the higher the force's magnitude is.

Answer:

Explanation:wave=wavelength×frequency,

Answer:

The electric field in the center is  [tex]\frac{2k}{\pi R^{2}}\left ( Q_{2}-Q_{1 } \right )[/tex].

Explanation:

because they work on the same object

Explanation:

A push and pull factor

Answer:

Explanation:

From the given information:

total mass = 25 kg

distance d = 100 m

height = 4.0 m

Force F = 9.0 N

The speed at (bottom) u = 5.0 m/s

Using the concept of energy conservation;

[tex]\dfrac{1}{2}mu^2 + W = \dfrac{1}{2}mv^2 + mgh[/tex]

divide both sides by m

[tex]\dfrac{1}{2}u^2 + \dfrac{W }{m}= \dfrac{1}{2}v^2 + gh[/tex]

multiply both sides by 2

[tex]\dfrac{1}{2}u^2\times 2 + \dfrac{W }{m}\times 2= \dfrac{1}{2}v^2\times 2 + gh\times 2[/tex]

[tex]u^2 +2 \dfrac{W }{m}=v^2 + 2gh[/tex]

[tex]v^2 =u^2 - 2gh+ 2 \dfrac{W }{m}[/tex]

Recall that:

W = Fd

∴

[tex]v^2 =u^2 - 2gh+ 2 \dfrac{Fd }{m}[/tex]

[tex]v^2 =(5.0 \ m/s)^2 - 2(9.81 \ m/s)(4.0 \ m)+ 2 \dfrac{( 9.0 \ N \times 100 \ m) }{25 \ kg}[/tex]

[tex]v^2 =(25.0 ) - 78.48 +72[/tex]

[tex]v^2 = 18.52 \ m^2/s^2[/tex]

[tex]v = \sqrt{18.52 \ m^2/s^2}[/tex]

v = 4.30 m/s