A 115 kg hockey player, Adam, is skating east when he tackles a stationary 133 kg player, Bob. Afterward, they move at 1.35 m/s east. What was Adam's velocity before the collision? (Unit = m/s) ​

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

Answer:

A) -31.2 rad/s

B) 28.1 rad/s^2

Explanation:

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.

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:

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

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

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

Explanation:

here the file has everything

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.

Explanation:

In this chapter, we will study the important concepts of kinetic energy and the

closely related concept of work and power.

A- Kinetic Energy

Kinetic energy is a physical quantity, which is associated with the moving objects

and defined as:

K = ½ mv2

If the body is stationary (v=0), its kinetic energy is zero. The SI unit of kinetic

energy is kg.m2

/s2

or Joule (J), where 1 J=1 kg.m2

/s2

. Kinetic energy is a scalar

quantity.

B- Work

The work is defined as the ability to perform a force along a certain displacement.

There are different types of work as follows:

1- Work done by a constant force

The work done by the constant force F is given by the scalar product of the force F

and the displacement d.

WF = F.d = Fd cosθ

where θ is the angle between the force and displacement. The above equation means

that the work is the product of the displacement magnitude by component of the

force parallel to the displacement. Therefore, work is a scalar quantity (only

magnitude, no direction) and can be positive, negative, or zero. The SI unit of work

is (N.m) or joule (J) where 1 N.m = 1 J.

Special cases and remarks:

• If the angle between the force and displacement is zero (parallel), the work is

WF = F d (maximum work)

Addition of a proton to germanium  will convert it to arsenic (element 33) having different properties.

The atomic number of an atom is the number of protons in the nucleus of the atom. The atomic number serves as the identity of an atom. If the atomic number is changed by adding or removing protons, the identity of that atom changes.

Hence, when a proton is added to germanium, the  atom would become arsenic (element 33) and have different properties.

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Answer: Its B and D

I had the same question on usatestprep

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

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

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.

Answer:

the one in which the fluid has the lowest density

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]

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 battery, wires, and a switch.

Explanation:

All circuits include?

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.

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:

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:

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

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]  

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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. Determine the equation of constraints

Answer:

Explanation:

s = s₀ + v₀t + ½at²

ASSUMING the acceleration is in the direction of initial motion.

s = 0 + 3.3(10) = ½(3.7)(10²)

s = 218 m

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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