how do I calculate equilibrant and fx and fy. I don't understand what they are asking

Answer:

eˣ-8.

Explanation:

(eˣ-8x+7)'=eˣ-8.

Answer:

Discrimination is the worst thing in the world you can't even do a thing so you will do such a physical thing or do a mathematics problem ok done it's ok

Answer:

Since 1 cal = 4.19 J     the heat Q received by the metal is

Q = 13,500 J * 1 C / 4.19 J = 3,222 calories

S = ΔQ / (ΔT * ΔM)  = 3222 cal / (45 deg C / 750 gm)

S = .095 cal / gm deg C

Note that specific heat capacity for Cu is .093 cal / gm deg C

The speed at which the medicine leaves the needle is  2.462 m/s

When an incompressible, ideal fluid is flowing through a tube or pipe, the total energy remains constant.

p₁ /ρg + v₁²/2g +z₁ = p₂ /ρg + v₂²/2g +z₂

Where, p/ρg = pressure energy

            v²/2g = kinetic energy

                   z = potential energy

Given is during an injection, pressure in the barrel of syringe is 1.03  atm while pressure in the needle section is 1.00 atm. Assuming the syringe lays horizontally and mass density of the liquid medicine ρ =1000 kg/m³.

The fluid is initially at rest.

Using the Bernoulli's equation, we have

v₁² -  v₂²= 2 x (p₂ -p₁) / ρ

Substituting the values, we get

0 -  v₂² = 2 x (1.00 -1.03) x 1.01 x 10⁵ /1000

v₂² = 6.06

v₂ = 2.462 m/s

Thus, the speed at which the medicine leaves the needle is 2.462 m/s

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Answer:

28

Explanation:

Or from this equation, when we can write the magnification of talents is F times F plus the object distance knowledge, substitute the value to find out the magnification and that equals minus 54 centimeters Over -54 minus 26 centimeters. And we got the magnification of the image produced by The lens is 0.675.

The magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

Electric field at a given distance is calculated as follows;

E = kq/r²

E₂ = (9 x 10⁹ x q)/(2²)

E₂ = 2.25 x 10⁹q

E₂ + E₀ = 0

2.25 x 10⁹q + 4 = 0

2.25 x 10⁹q = - 4

q = -4 / (2.25 x 10⁹)

q = -1.78 x 10⁻⁹

E₄ = (9 x 10⁹ x (-1.78 x 10⁻⁹) ) / (4²)

E₄ = - 1 N/C

|E₄| = 1 N/C

Thus, the magnitude of the electric field at x =4m on the x axis at this time 1 N/C.

The complete question is below:

Suppose a uniform electric field of 4N/C is in the positive x direction. When a charge is placed and at a fixed to the origin, the resulting electric field on the x axis at x =2m becomes zero. What is the magnitude of the electric field at x =4m on the x axis at this time?

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[tex]q = -21 * 10^{-6} C[/tex]

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

[tex]1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\[/tex]

and its sign must be negative so that it will have upward electric force

so it is

[tex]q = -21 * 10^{-6} C[/tex]

The charge of a particle of mass is [tex]-21 * 10^{-6} C[/tex]

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The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.

Unit conversion;

1 km/sec = 1000 m/sec

Given data;

Spaceprobe speed  = 1.795 km/s = 1795 m /sec

Probe mass = 635.0 kg

Fuel mass = 4092.0 kg

Expelled propellent velocity = 4.161 km/s = 41461 m/sec

From the momentum conservation principle;

[tex]\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times 41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec[/tex]

Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

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At 11.3°C the rod will just come into contact

Coefficient of linear expansion of aluminium [tex]\alpha_{Al}[/tex] = [tex]23*10^-^6[/tex] °[tex]C[/tex]

Coefficient of linear expansion of Brass [tex]\alpha_{B} =19*10^-^6[/tex] °[tex]C[/tex]

[tex]\alpha = \Delta{L}/Lo(T2-T1)[/tex]

For aluminium

[tex]\alpha_{Al} = \Delta{L}/Lo(T-0)[/tex]

[tex]\Delta{L_{1}} = (23*10^-^6*50*T)[/tex]

For Brass

[tex]\alphaB = \Delta{L_{2}/Lo(T-0)[/tex]

[tex]\Delta{L_{1} +\Delta{L_{2} =0.024cm(23*10^-^6*50*T)+(19*10-6*50*T) =0.024[/tex]

T =11.43 °C

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The direction of this vector is 170° from the positive x-axis.

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The field exerts 6.5 x 10⁸ Newton of force on EACH COULOMB of charge in the field.

We're putting 1.6 x 10⁻¹⁹ Coulomb of charge into the field.

The force on it will be

      (1.6 x 10⁻¹⁹ Coulomb) x  (6.5 x 10⁻⁸ Newton/Coulomb).

That's 1.04 x 10⁻¹⁰ Newton.

The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.

During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:

Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.

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Statement B  is true about the energy level of gas, Gases have more energy than liquids and solids.

A gas is a sample of matter that adopts the shape of the container in which it is housed and develops a uniform density inside the container.

Liquid molecules are more energetic than solid molecules. Further heating will cause the molecules to move so quickly that they won't stick together at all. The gas molecules have the highest energy content.

Gases have more energy than liquids and solids.

Hence statement B  is true about the energy level of gas.

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According to the pavlovian perspective, people who suffer from PTSD are classically conditioned.

This is referred to as post traumatic stress disorder which is a mental disorder from a terrifying incident.

The pavlovian perspective however view this disorder as being a type of classical conditioned one.

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Answer:

Explanation:

missing info .....

d. is wrong...the temperature cannot be measured due to physical changes

Answer:

coplanar When all forces are acting in the same 

resultunt the single force and associated torque obtained by combining a system of forces and torques acting on a rigid body via vector addition

(a) The average velocity of the particle in the time interval t₁=2sec and t₂=3sec is 10 m/s.

(b) The velocity and acceleration at any time t is v =  (4ti + j) m/s and a = a = 4i m/s²

(c)  The average acceleration in the time interval given in part (a)​ is 3.98 m/s².

x = at²i + btj

x = 2t²i + tj

Δv = Δx/Δt

x(2) = 2(2)²i + 2j

x(2) = 8i + 2j

|x(2)| = √(8² + 2²) = 8.246

x(3) =  2(3)²i + 3j

x(3) = 18i + 3j

|x(3)| = √(18² + 3²) = 18.248

Δv = (18.248 - 8.246)/(3 - 2)

Δv =  10 m/s

v = dx/dt

v =  (4ti + j) m/s

a = dv/dt

a = 4i m/s²

v(2) = 4(2)i + j

v(2) = 8i + j

|v(2)| = 8.06 m/s

v(3) = 4(3)i + j

v(3) = 12i + j

|v(3)| = 12.04 m/s

a = (12.04 - 8.06)/(3 - 2)

a = 3.98 m/s²

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Answer:

a = (V2 - V1) / t      equation  for constant acceleration

if t the independent variable is placed on the abcissa or x-axis and V the dependent variable is placed on the ordinate or y-axis then the acceleration would be the slope at the point considered.

We will use the design process to meet the requirements of the competition by the given explained in the main answer.

The initiative was taken by two or more parties operating independently to win the business of a third party by competing to hire contractors on the best terms possible.

The complete question is

"A local businessperson placed an advert in the local newspaper for a competition for girls only from different schools to participate in the competition. The competition is about drawing ideas for a shopping complex for the local community. Girls from the different schools are excited about the competition. you are a group of boys in the local schools who saw the advert and feel it is unfair for you to be excluded from the competition.

consider yourself as one of the learners in the local schools.

Explain why the project should be given to boys also who have better skills for this project than the girls do.

1.2 think about the design process skills, investigates, design, make, evaluate and communicate."

Explain how you will use the design process to meet the requirements of the competition.

[ 1.1 ]Boys should also be given the project because some of them have more artistic talent than girls do when it comes to sketching designs for local shopping centers.

Boys can also present a clear picture of how local communities can expand their shopping options.

For instance, they can encourage local merchants to offer seasonal items because doing so will benefit their businesses.

They can also suggest that retailers encourage customers to bring a friend to participate in competitions, which will be highly regarded.

[1.2] In order to fulfill the needs of the competition, I would make sure to explore how a shopping center for the neighborhood functions, for instance, by making sure I am aware of the strategies one may use to enhance his or her firm in the neighborhood.

In order to win the competition, I would make sure that I was confident in what I was saying and that I was audible enough to be heard during the competition.

Finally, I would make sure I met the competition's conditions by staying on-topic and solely addressing the issues raised in the advertisement

Hence will use the design process to meet the requirements of the competition.

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Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

Answer:

a force of acts on a body of mass 50 kg for 10 sec when the first of acting on the body the body covers 80 and in the next 10 second

Answer:

a = 0.4 ms⁻²

Explanation:

• First calculate the resultant force on the body:

If we consider upwards to be the positive direction and downwards to be the negative direction, then the resultant force will be:

F = +100 + (-80)

  = +20 N

The resultant force is positive, so it will act upwards.

• Next calculate the acceleration:
F = ma

a = F/m

a = 20/ 50

a = 0.4 ms⁻²

As the force is acting upwards, the acceleration will also be upwards.

Answer:

8.4335 x 10^-²¹, 1.78204 x 10²⁶

Explanation:

p = mv

p = (5.05 x 10⁶)(1.67 x 10^-²⁷)

p = 8.4335 x 10^-²¹

p = mv

p = (5.98 x 10²⁴)(2.98 x 10)

p = 1.78204 x 10²⁶

(a) Let [tex]v[/tex] be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

[tex]a_{\rm rad} = \dfrac{v^2}R[/tex]

where [tex]R[/tex] is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

[tex]F = (1.500\,\mathrm{kg}) a_{\rm rad}[/tex]

so that

[tex](1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N[/tex]

Solve for [tex]v[/tex] :

[tex]v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}[/tex]

(b) The net force equation in part (a) leads us to the relation

[tex]F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}[/tex]

so that [tex]v[/tex] is directly proportional to the square root of [tex]R[/tex]. As the radius [tex]R[/tex] increases, the maximum linear speed [tex]v[/tex] will also increase, so the cord is less likely to break if we keep up the same speed.

Answer:

a (2m)

Explanation:

according to wavelenght = speed / frequency,

the wavelength is 2 m

Answer:

a

Explanation:

The magnitude of the reaction force of the air pushing the balloon would be equal and opposite.

A push or a pull that an object experiences as a result of interacting with another item is known as a  reaction force.

Newton's third law is officially expressed as follows: There is an equal and opposite reaction to every action. The implication of the statement is that there are always two forces acting on the two interacting objects. The force acting on the first object is equal in size to the force acting on the second. The force acting on the first object is acting in the opposite direction to the force acting on the second object. Force pairs—equal and opposing action-reaction force pairs—always exist in pairs.

Knowing that everything has an equal and opposite reaction according to Newton's second law. The preasured air that a balloon had to push out into the free air acts as the reaction force.

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Answer:

See below....the question is unclear....pick the answer you think they want

Explanation:

Car goes from 30 m/s    to   24 m/s   in 10 seconds

acceleration = change in velocity/change in time

                     =   -6 m/s   / 10 s   =  - 3/5   m/s^2  

The wording of the question is not very good.....perhaps they meant to say that after the 5 seconds the velocity is now 24 m/s ?

a1 *10   +   a2 *5   = -6 m/s  

            ( 6 m/s is the change in velocity from 30 to 24 m/s)    

    a1/a2 = 1.5    so   a2 = a1/1.5   (substitute this in)

10 a1   + 5 ( a1/1.5)  =-6

15 a1 + 5 a1  = -9

a1 =  - .45 m/s^2

The velocity of the car at the end of the initial ten-second interval, given that the car further slows down for the next five seconds, is 25.5 m/s

From the first statement, we have:

v₁ = u₁ + a₁t₁

v₁ = 30 + (a₁ × 10)

v₁ = 30 + 10a₁ ......(1)

From the second statement, we have:

v₂ = u₂ + a₂t₂

24 = u₂ + (a₂ × 5)

24 = u₂ + 5a₂   .......(2)

But,

u₂ = v₁ = 30 + 10a₁

Thus, we have:

24 = 30 + 10a₁ + 5a₂

But,

a₁/a₂ = 1.5

a₁ = 1.5a₂ ..... (3)

Thus, we have:

24 = 30 + 10a₁ + 5a₂

24 = 30 + 10(1.5a₂) + 5a₂

24 = 30 + 15a₂ + 5a₂

24 = 30 + 20a₂

Collect like terms

24 - 30 = 20a₂

-6 = 20a₂

Divide both sides by 20

a₂ = -6 / 20

= -0.3 m/s²

Substituting the value of a₂ into equation 3, we have

a₁ = 1.5a₂

= 1.5 × -0.3

= -0.45 m/s²

Substitute the value of a₁ into equation 1 to obtain the velocity, v₁ at the end of the initial to 10 s

v₁ = 30 + 10a₁

= 30 + (10 × -0.45)

= 30 - 4.5

= 25.5 m/s

Thus, velocity of the car at the end of the initial ten-second interval is 25.5 m/s

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Answer:

Approximately [tex]3.14\; {\rm rad \cdot s^{-1}}[/tex].

Explanation:

Fact: the angular velocity [tex]\omega[/tex] of a simple harmonic oscillator is the ratio between the maximum velocity [tex]v_{\text{max}}[/tex] and the maximum displacement [tex]x_\text{max}[/tex] of this oscillator. In other words:

[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}}\end{aligned}[/tex].

Derivation of the previous equation:

Let [tex]A[/tex] denote the amplitude of this oscillation, and let [tex]\omega[/tex] denote the angular velocity.

The displacement of the oscillator at time [tex]t[/tex] would be:

[tex]x(t) = A\, \sin(\omega\, t)[/tex].

The maximum displacement of this oscillator would be [tex]x_\text{max} = A[/tex].

The velocity of this oscillator at time [tex]t[/tex] is the derivative of displacement with respect to time:

[tex]\begin{aligned} v(t) &= \frac{d}{d t}\, [x(t)] \\ &= \frac{d}{d t} [A\, \sin(\omega\, t)] \\ &= A\, \omega\, \cos(\omega\, t)\end{aligned}[/tex].

The maximum velocity of this oscillator would be [tex]v_\text{max} = A\, \omega[/tex].

Notice that dividing [tex]v_\text{max} = A\, \omega[/tex] by [tex]x_\text{max} = A[/tex] would give:

[tex]\displaystyle \frac{v_\text{max}}{x_\text{max}} = \frac{A\, \omega}{A} = \omega[/tex].

It is given that [tex]v_\text{max} = 2.76\; {\rm m\cdot s^{-1}}[/tex] while [tex]x_\text{max} = 0.0880\; {\rm m}[/tex]. Therefore:

[tex]\begin{aligned} \omega &= \frac{v_{\text{max}}}{x_{\text{max}}} \\ &= \frac{2.76\; {\rm m\cdot s^{-1}}}{0.0880\; {\rm m}} \\ &\approx 3.14\; {\rm s^{-1}}\end{aligned}[/tex].

(Radians per second.)