A machine lifts a load of 100N through a vertical distance of 2m in seconds. What is the work done by machine?​

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

Answer:

a battery, wires, and a switch.

Explanation:

All circuits include?

Answer:

Explanation:

s = s₀ + v₀t + ½at²

if the throw point is origin and UP the positive direction and ignoring air resistance.

s = 0 + (-5)(7) + ½(-10)(7²)

s = 0 - 35 - 245

s = - 280 m

Explanation:

The range R of a projectile is given the equation

[tex]R = \dfrac{v_0^2}{g}\sin{2\theta}[/tex]

The maximum range is achieved when [tex]\theta = 45°[/tex] so our equation reduces to

[tex]R_{max} = \dfrac{v_0^2}{g}[/tex]

We can solve for the initial velocity [tex]v_0[/tex] as follows:

[tex]v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}[/tex]

or

[tex]v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}[/tex]

[tex]\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}[/tex]

To find the maximum altitude H reached by the missile, we can use the equation

[tex]v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy[/tex]

At its maximum height H, [tex]v_y = 0[/tex] so we can write

[tex]0 = (v_0\sin{45°})^2 - 2gH[/tex]

or

[tex]H = \dfrac{(v_0\sin{45°})^2}{2g}[/tex]

[tex]\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}[/tex]

[tex]\:\:\:\:\:\:= 2.4×10^6\:\text{m}[/tex]

Answer:

a thing or event that evokes a specific functional reaction in an organ or tissue.

or

a thing that rouses activity or energy in someone or something; a spur or incentive.

Answer:

Answer: At its lowest point, the kinetic energy of a watermelon just as it touches the ground is zero if it does not touch anything on its way down.

Explanation:

This is because that upon having been dropped from a height, an object no longer has any kinetic energy at all. Kinetic energy transforms to gravitational potential energy during the fall and there's nothing left over for kinetic once you stop accelerating anymore. Fortunately, things don't stay still until they land very often! For example, if a person catches the fruit with his hands after some air resistance slows him down - making him more similar in speed to the lag of the trajectory - then he'll be able to share some of his saved up gravitational potential with that watermelon and do some

Km values are standardized because half the Vmax (Vmax/2) is more informative than Vmax. This value (Km) can be used to calculate the affinity of the enzyme by a given substrate.

The Km (Michaelis constant) of the enzyme refers to the value in which the concentration of substrate is equal to half of its maximum velocity (Vmax/2).

This value (Km) is inversely proportional to the affinity of an enzyme by a given substrate.

An enzyme showing a high Km also exhibits a low affinity for its specific substrate, and thereby this enzyme requires a high concentration of the substrate to reach its maximum velocity (Vmax).

In consequence, the Km value is a more informative value than the maximum velocity (Vmax), which only indicates the concentration of an enzyme catalyzing a reaction under ideal conditions.

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Answer:

You can jump 1.5 feet on Earth.

Explanation:

Because the moons gravity is weaker that earth so it would be easier to jump further on the moon.

Answer:

Explanation:

Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.

In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.

That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.

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Acceleration due to gravity on that planet is half than that of om earth ~

that is ~

And it's weight is equal to ~

now, As we know ~

here ,

let find the mass of stone ~

Mass of the stone is 100 kg ~

I hope it helped

Hi there!

We can begin by finding the period of the pendulum.

[tex]T = \text{ # of complete swings / seconds} = 12 / 8 = \boxed{\text{1.5 sec}}[/tex]

The frequency is simply the reciprocal of the period, so:

[tex]f = \frac{1}{T} = \frac{1}{1.5} = \frac{2}{3}Hz \text{ or } \boxed{0.67 Hz}[/tex]

Answer:

A) -31.2 rad/s

B) 28.1 rad/s^2

Explanation:

Answer:

Explanation:

I guess we are ASSUMING that this is a rear wheel drive car as a front wheel drive car will never get the front wheel normal force to zero

If we consider it as a statics problem and choose our moment center carefully...say 0.7 m above the rear wheel to ground contact point.

Call the traction force at the rear wheels F

The normal force on the front wheels will be zero, so no moment generated by the front wheels.

Summing moments about our chosen point to zero

1600(9.8)[2.6 / 2] - F[0.7] = 0

F = 291,200

this force will create an acceleration of

a = F/m

a = 291200/1600

a = 182 m/s²

which is about 18.6 times gravity acceleration

Answer:

Sample Response: No image will be formed because the rays will not converge to or diverge from a common point.

Explanation:

Explanation:

a. (i) When the variable resistor is set at zero, the only resistance in the circuit is due to the lamp. So the current flowing through the circuit is

[tex]I = \dfrac{V}{R} = \dfrac{220\:\text{V}}{440\:Ω} = 0.5\:\text{A}[/tex]

(ii) The power output P of the lamp is given by

[tex]P = I^2R = (0.5\:\text{A})^2(440\:Ω) = 110\:\text{W}[/tex]

b. (i) The variable resistor is in a series connection to the lamp so when its value is set to its maximum value of 660 Ω, the total resistance of the circuit is simply the sum of the two resistances:

[tex]R_T = R_{vr} + R_L = 660\:Ω + 440\:Ω = 1100\:Ω[/tex]

Therefore, the current through the circuit is

[tex]I = \dfrac{V}{R_T} = \dfrac{220\:\text{V}}{1100\:Ω} = 0.20\:\text{A}[/tex]

(ii) Using the result in Part (ii), we can solve for the potential difference across the lamp as follows:

[tex]V_L = IR_L = (0.20\:\text{A})(440\:Ω) = 88\:\text{V}[/tex]

(iii) The power output of the lamp is

[tex]P = I^2R_L = (0.20\:\text{A})^2(440\:Ω) = 17.6\:\text{W}[/tex]

(iv) The rate at which electrical energy is supplied, i.e., the power output of the circuit is equal to the square of the current multiplied by the total resistance of the circuit:

[tex]P = I^2R_T = (0.20\:\text{A})^2(1100\:Ω) = 44\:\text{W}[/tex]

Explanation:

Fgravity = G*(mass1*mass2)/D²

G is the gravitational constant throughout the universe.

D is the distance between the 2 objects.

the distance is now quadrupled.

Fgravitynew = G*(mass1*mass2)/(4D)² =

= G*(mass1*mass2)/(16D²) =

= (G*(mass1*mass2)/D²) / 16 = Fgravity/16

the new gravitational force will be 179/16 = 11.1875 units

Hi there!

With a Force/Displacement curve, we must take the integral (area underneath the curve) to calculate the work done.

We know that:

W = ΔKE

Calculate the work by finding the area underneath the force curve from

x = 0 to 3 m:

We can use a trapezoid:

A = 1/2(3 + 2)(3) = 7.5 J

This is the amount of work done, and since the object starts from rest:

7.5J = KEf - KEi (0 J)

7.5J = KEf = 7.5 J

Answer:

A. Determine the equation of constraints

Explanation:

Fgravity = G*(mass1*mass2)/D².

G is the gravitational constant, which has the same value throughout our universe.

D is the distance between the objects.

so, if you triple one of the masses, what does that do to our equation ?

Fgravitynew = G*(3*mass1*mass2)/D²

due to the commutative property of multiplication

Fgravitynew = 3* G*(mass1*mass2)/D² = 3* Fgravity

so, the right answer is 3×12 = 36 units.

9514 1404 393

Answer:

  about 5.89 seconds

Explanation:

The penny will hit the ground at the same time it would if it were simply dropped. The equation for the vertical motion is ...

  h(t) = -4.9t^2 +170 . . . . . where 170 is the initial height in meters

h(t) = 0 when ...

  4.9t^2 = 170

  t = √(170/4.9) ≈ 5.89

The penny will hit the ground in about 5.89 seconds.

Answer:

Explanation:

s 0.12 + 1.20(1.80) = 2.28 m from the top.

Answer:

Explanation:

At the top of the arc, 3/4 of the acceleration of gravity is use to supply the necessary centripetal acceleration.

0.75g = ω²R

R = 0.75g/ω²

R = 0.75(9.81) / (0.15 rev/s)(2π rad/rev)²

R = 8.283006...

R = 8.28 m

The characteristics of the expression of the simple harmonic motion allows to find the results for the expression of the mass- block system are:

     A) The constant Ces: C = xinit

     B) The ocsntna S is: S = 0

     C) The equation of the system is: x = xinit cos wt

     D) If the reference system is at some extreme, the equation is:

              [tex]L - x_{init} = x_{init} \ cos \ wt[/tex]  

The simple harmonic movement is an oscillatory movement where the restoring force is proportional to the displacement, the general equation that describes this movement is indicated.

          x = C cos wt + S sin wt

Where x is the displacement C and S are constants. W the angular velocity and t the time.

A) The initial position of the body occurs when the time is zero, t = 0

We substitute.

           x = C cos 0 + S sin 0

           [tex]x_{init}[/tex] = C

B) The velocity of the particle is defined.

          [tex]v= \frac{dx}{dt} \\ v= C w \ sin \ wt - Sw \ cos \ wt[/tex]  

The initial velocity occurred for time zero t = 0

          v = - S w

It indicates that the initial velocity is zero, since the angular velocity must be different from zero, it implies that the constant is valid.

          S = 0

C) The equation for the block remains.

           x (t) = [tex]x_{init} \ cos \ wt[/tex]  

D) In ​​some cases it is measured with respect to another reference system, the most common are:

In these cases, the change that must be made is

          x =  [tex]L - x_{min}[/tex] t

we substitute

          [tex]L - x_{init} = x_{init} \ cos \ wt[/tex]  

          L = [tex]x_{init}[/tex]  (1 + cos wt)

In conclusion, using the characteristics of the expression of the simple harmonic motion we can find the results for the expression of the mass- block system are:

    A) The constant Ces: C = xinit

     B) The ocsntna S is: S = 0

     C) The equation of the system is: x = xinit cos wt

     D) If the reference system is at some extreme, the equation is:

              [tex]L - x_{init} = x_{init} \ cos \ wt[/tex]  

Learn more about simple harmonic motion here:  brainly.com/question/17315536

a.

The object's speed at 2.20 m below balcony level is 8.74 m/s

Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.

Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and

So, E = E'

U + K + f = U' + K' + f'

where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).

So,

U + K + f = U' + K' + f'

mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh

mgh = mgh' + 1/2mv² + 0.10mgh

Dividing through by m, we have

gh = gh' + 1/2v² + 0.10gh

So, gh -  0.10gh = gh' + 1/2v²

0.90gh = gh' + 1/2v²

1/2v² = 0.90gh - gh'

1/2v² = g(0.90h - h')

v² = 2g(0.90h - h')

Taking square-root of both sides, we have

v = √[2g(0.90h - h')]

where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and  g = acceleration due to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

v = √[2g(0.90h - h')]

v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]

v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]

v = √[2 × 9.8 m/s²(3.901 m)]

v = √[76.4596 m²/s²]

v = 8.74 m/s

So, the object's speed at 2.20 m below balcony level is 8.74 m/s

b.

Yes it does matter when we apply 10% loss before V calculations

We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.

So, yes it does matter when we apply 10% loss before V calculations

Learn more about speed of a falling object here:

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Hi there!

Since the collision is elastic, we must also satisfy the following condition:

Ei = Ef, or:

KEi = KEf

Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.

mv1 + mv2 = mvf1 + mvf2

0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)

0.2882/0.220 = vf1 + vf2

1.31 = vf1 + vf2

Now, we can express this as a conservation of energy:

1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²

Plug in values and simplify:

0.403315 = 1/2m(vf1² + vf2²)

Simplify further:

3.6665 = vf1² + vf2²

Use the equation derived from momentum above and solve for one variable:

vf2 = 1.31 - vf1

Plug in this expression for vf2:

3.6665 = vf1² + (1.31 - vf1)²

Expand:

3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²

Simplify:

1.9504 = -2.62vf1 + 2vf1²

Solve for vf1 using a graphing calculator:

vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.

Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.

Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):

vf2 = 1.31 - (-0.53) = 1.84 m/s.

Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.

Answer:

78

Explanation:

x=xi+vi(t)+1/2a(t)^2

x=10+16(4)+1/2(0.50)(4)^2

x=74+4

x=78 m

Answer:

Explanation:

s = s₀ + v₀t + ½at²

assuming that the acceleration is in the direction of initial velocity.

(it would not have to be so)

s = 0 + 10(8) + ½(3.5)(8²)

s = 192 m

Hi there!

We know that:

I = Δp = m(vf - vi)

Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.

I = 1.1(5 - (-23)) = 30.8 Ns