When a 1 is on the input of an inverter, what is the output?​

The relationship between the potential and the electric field allows to find the results for the value of the electric field as a function of the distance is:

Electric potential is defined by the change in potential energy of a test charge between two points, between the value of the test charge.

          dV = - E . ds

          E = [tex]- \frac{dV}{ds} \ \hat s[/tex]  

Where the bold letters indicate vectors, V is the potential difference, E the electric field and s the path.

Let's apply this expression for each section of the given graph:

1) section from x₀ = 0 to x_f = 2 m, the potential is V₀ = 2 V is constant.

  The derivative of a constant is zero.

        E = 0

2) Section between x₀ = 2 and x_f = 4 m, the potential varies linearly from V₀ = 2 v to V_f = -2 V.

We look for the equation of the line.

       V-V₀ = m (x- x₀)

We carry out the derivative.

      E = - m i ^

The slope (m) is:

       [tex]m= \frac{V_f - V_o}{x_f- x_o}[/tex]  

Let's calculate.

       [tex]m= \frac{-2 -2}{4-2} = \ -2 \ V/m[/tex]  

Let's substitute.

       E =  [tex]2 \hat i \ V/m[/tex]  

3) From x₀ = 4 to x_f = 4.5 m, the potential varies from V₀ = -2 to V_f = 0.

We look for the equation of the line and we derive.

      E = - m i ^

Let's  substitute.

      [tex]m = \frac{0-(-2)}{4.5-4} = \ 4 V/m[/tex]  

    E = - 4 [tex]\hat i[/tex] V / m

4) From x₀ = 4.5 m to x_f = 6m.  The potential is constant and the derivative of a constant is zero.

      E = 0

5) From x₀ = 6m to x_f = 8 m, the potential changes linearly from v₀ = 0 to V_f = 1 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{1-0}{8-6} = \ 0.5 \ V/m[/tex]  

      E = - 0.5 [tex]\hat i[/tex] V/m

6) From x₀ = 8m to x_f = 9m, the potential changes linearly from V₀ = 1 V to V_f = -1.

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-1-1}{9-8} = \ -2 \ V/m[/tex]

Let's substitute.

       E = 2 [tex]\hat i[/tex] V/m

7) From x₀ = 9m to x_f = 10 m, the potential changes linearly from V₀ = -1 V to V_f = -2 V

We look for the equation of the line and we derive.

       E = - m i ^

       [tex]m = \frac{-2+1}{10-9} = \ -1 \ V/m[/tex]

Let's substitute.

       E = 1 [tex]\hat i[/tex]  V/m

In the attachment we can see these Electric fields as a function of distance.

In conclusion, the relationship between the potential and the electric field we can find the results for the value of the electric field as a function of the distance is:

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Explanation:

We use the Theorem of conservation of mechanical energy for finding the velocity when it strikes the ground:

Ei = Ef

Ki + Ui = Kf + Uf

Ui = Kf

m g h = 1/2 m v^2

v = sqrt(2gh)

So the momentum will be:

p = mv = m * sqrt(2gh)

Answer:

FA = 2FB

Force on spring A is twice the Force on spring B

Explanation:

F = kx

FB = (kB)x

FA = (kA)x

FA= (2kB)x

FA = 2(kB)x

FA = 2FB

The force [tex]F_A[/tex] needed to stretch spring A is going to be twice as much as the force [tex]F_B[/tex] needed to stretch spring B.

Explanation:

We know that the spring constants are related as

[tex]k_A = 2k_B[/tex]

The force [tex]F_A[/tex] needed to stretch spring A is given by

[tex]F_A = -k_Ax[/tex]

Also, the force [tex]F_B[/tex] needed to stretch spring is

[tex]F_B = -k_Bx[/tex]

Taking the ratio of the forces, we get

[tex]\dfrac{F_A}{F_B} = \dfrac{-k_Ax}{-k_Bx} = \dfrac{k_A}{k_B}[/tex]

Since [tex]k_A = 2k_B,[/tex] the equation above becomes

[tex]\dfrac{F_A}{F_B} = \dfrac{2k_B}{k_B} = 2[/tex]

or

[tex]F_A = 2F_B[/tex]

This shows that since the spring constant of spring A is twice as large as that of spring B, the force needed is going to be twice as large.

Answer:What is the half life of the element Lokium?

The half-live of the element Lokium is 4.

Explanation:

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Answer:

The half-life of the element Lokium is 4

What is half-life?

Whether or not a given isotope is radioactive is a characteristic of that particular isotope. Some isotopes are stable indefinitely, while others are radioactive and decay through a characteristic form of emission. As time passes, less and less of the radioactive isotope will be present, and the level of radioactivity decreases. An interesting and useful aspect of radioactive decay is the half-life. The half-life of a radioactive isotope is the amount of time it takes for one half of the radioactive isotope to decay. The half-life of a specific radioactive isotope is constant; it is unaffected by conditions and is independent of the initial amount of that isotope.

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Answer:

Explanation:

a) F = ma

a = F/m

a = 9(800) / 1 x 10⁹ = 7.2 x 10⁻⁶ m/s

b) t = v/a

t = 200 / 7.2 x 10⁻⁶

t = 2.8 x 10⁷ s       about 10½ months

c) v² = u² + 2as

s = (v² - u²) / 2a

s = (200² - 0²) / (2( 7.2 x 10⁻⁶))

s = 2.8 x 10⁹ m    nearly 7 times around the earth

And all this assumes NO FRICTION.

Answer:

Explanation:

A wonderful (NOT) mixed unit problem

convert mph to m/s

v = 10 mi/hr(5280 ft/mi)(12 in/ft)(2.54 cm/in) / (100 cm/m) / (3600 s/hr)

v = 4.4704 m/s

KE = ½(0.006)4.4704²

KE = 0.05995342848 J

KE = 0.06 J

Question

what is the force exerted on a selectively permeable membrane beacuse water has moved from an area of higher concentration to an area of Lower concentration of water?

Answer

OSMOSIS

Explanation

Osmosis: This is the hydrostatic force acting to equalize the concentration of water on both sides of the membrane that is impermeable to substances dissolved in that water. Water will move along its concentration gradient.

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None of the given options is correct based on the relationship for the components of the crate's weight.

The given parameter:

The vertical component of the force on the crate is calculated as follows;

[tex]F_y = W \times cos(q)\\\\F_y = F_gcos(q)[/tex]

The horizontal component of the crate is calculated as follows;

[tex]F_x = F_g \times sin(q)\\\\F_x = F_g sin(q)[/tex]

Thus, we can conclude that none of the given options is correct based on the relationship for the components of the crate's weight.

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Answer:

The main condition necessary for a controlled thermonuclear fusion:

The ions must be held together in close proximity at high temperature with a confinement time long enough to avoid cooling.

Answer:

Explanation:

Ignoring friction, the initial kinetic energy will convert to maximum potential energy at its highest point.

PE = KE

mgh = ½mv²

    h = v²/2g

    h = 36.4²/ (2(9.81))

    h = 67.53109...

    h = 67.53 m

Answer:

Explanation:

We can subtract directly the corresponding components and check using the parallelogram rule.

Explanation:

Have a look:

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Where, graphically, I used the fact that:

→

A

−

→

B

=

→

A

+

(

−

→

B

)

For the magnitude we use Pythagoras (with the components) to get:

∣

∣

∣

→

A

−

→

B

∣

∣

∣

=

√

(

−

1

)

2

+

(

5

)

2

=

√

1

+

25

=

√

26

≈

5.1

For the direction I can see that will be

90

∘

from the

x

axis up to the

y

axis, plus the little bit passed the

y

axis given as:

θ

=

arctan

(

1

5

)

=

11.3

∘

giving in total: angle

=

90

∘

+

11.3

∘

=

101.3

∘

Answer:

Explanation:

Objects want to continue doing what they're already doing because they are "lazy." This is called inertia.

Answer:

fifth harmonic

Explanation:

Answer:

The removal of heat energy slows the speed of particles

Explanation:

When you add heat to a substance, the heat energy gets transferred to kinetic energy, and the molecules began to move a greater distance at a greater speed. When you remove heat, the opposite happens.

Answer:

35m/s

Explanation:

Answer:

35m/s

Explanation:

Simply multiply 5 and 7.

Answer:

I believe The Knowledge that we apply everyday or based on these methods and discoveries may represent us

Explanation:

Answer:

Decant it.

Explanation:

Pour the water/sugar solution off the sand. When the sand wants to start coming out as well, Stop and add fresh water to the beaker, stir to rinse the remaining solution into a less concentrated solution and decant again.

Repeat the dilution process until the mix is essentially sand and water, then drive the remaining water from the sand by drying.

Answer:

lithosphere

Explanation:

hope this helps you!!

If it takes 4.4 Newton to lift 1 pound, then to lift 2 pounds the force required will be equal to 8.8 Newton.

A force in physics is an input that has the power to change an object's motion. A mass-containing object's velocity can vary, or accelerate, as a result of a force. Intuitively, a push or a pull can also be used to describe forces.

Being such a vector quantity, a force does have magnitude and direction. The SI unit metric newton is used to measure it (N). The letter F stands for force.

According to Newton's second law's original formulation, an object's net force is equal to the speed that its momentum is changing over time.

As per the given data in the question,

It takes 4.4 N to lift 1 pound,

Let the total force required to lift 2 pounds is x.

x = (4.4 × 2)/1

x = 8.8 N

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The work done if the forklift is operating at full capacity is 11,900 J.

We have to recall that power is defined as the rate of doing work. The rate of doing work is defined as;

Power = Work done/time taken

When;

Power = 4950 W

Time taken = 2.40 s

Work done = Power × time taken

Work done = 4950 W × 2.40 s

Work done = 11,900 J

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Answer:

D.

correct me if im wrong

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Answer:

a = ω^2 A      formula for max acceleration (ignoring sign)

V = ω A         formula for max velocity

V^2 = ω^2 A^2 = a A   from first equation

E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J

(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule

Answer:

(a) 91 kg (2 s.f.)    (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

                                                   [tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}[/tex]

     Subsequently,

                                                  [tex]F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})[/tex]

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

                                                    [tex]v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}[/tex]

     According to Newton's First Law which states objects will remain at rest

     or in uniform motion (moving at constant velocity) unless acted upon by

     an external force. Hence, the block of ice by the end of the first 5

     seconds, experiences no acceleration (a = 0) but travels with a constant

     velocity of 4.4 [tex]m \ s^{-1}[/tex].

                                                    [tex]s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}[/tex]

      Therefore, the ice block traveled 22 m in the next 5 seconds after the

      worker stops pushing it.

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

[tex]x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;[/tex]

[tex]\Rightarrow mg\sin{\theta} = F[/tex]

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at [tex]\theta.[/tex] Solving for the angle, we get

[tex]\sin{\theta} = \dfrac{F}{mg}[/tex]

or

[tex]\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)[/tex]

[tex]\;\;\;= \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right][/tex]

[tex]\;\;\;=42.9°[/tex]

Answer:

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Hi there!

1.

Hooke's law states that:

F = -kx

k = Spring constant (N/m)

x = DISPLACEMENT from equilibrium (m)

Essentially, the force of a spring is PROPORTIONAL to its spring constant and its displacement from its equilibrium point.

2.

The force of the spring (T) is not proportional to the spring's length (l), but rather its DISPLACEMENT from its equilibrium length. (Δl)

3.

The equilibrium length is where the force of the spring (T) = 0N. Looking at the graph, the line intersects this value at l = 30cm.

4.

We can begin by looking at the given graph.

When the spring force = 4N, the total length of the spring is 35 cm.

Now, the EQUILIBRIUM length is 30 cm, so the total elongation is:

35 - 30 = 5 cm.

5.1.

If the spring elongates by 10 cm, the total length of the spring is:

30 + 10 = 40 cm

According to the graph, a length of 40 cm corresponds to a force of 8N.

5.2.

We can solve for the weight of the ball using the following:

W (weight) = m (mass) · acceleration due to gravity (10N/kg)

Using a summation of forces:

∑F = T - W

The elongation that we are solving for occurs at the equilibrium point (net force = 0 N), so:

0 = T - W

T = W = 8 N

5.3.

0 = T - Mg

T = Mg

Use the prior value of T and gravity to solve:

8 = 10M

m = 0.8 kg

Answer:

when the ball is at its highest point

Explanation:

Provided the ball returns to where it was thrown. The velocity, and therefore kinetic energy, will be momentarily zero at the highest point of the throw.

The Impulse delivered to the baseball is 89 kgm/s.

To solve the problem above, we use the formula of impulse.

⇒ Formula:

Where:

From the question,

⇒ Given:

⇒ Substitute these values into equation 1

Note: The negative tells that the impulse is in the same direction as the final velocity and therefore can be ignored.

Hence, The Impulse delivered to the baseball is 89 kgm/s.

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