which materials are good for constructing bridges and tall buildings​

Explanation:

The buoyant force [tex]F_B[/tex] is defined as

[tex]F_B = \rho_wgV[/tex]

where [tex]\rho_w[/tex] is the density of the displaced fluid (freshwater), g is the acceleration due to gravity and V is the volume of the submerged object. In the case of freshwater, its density is [tex]997\:\text{kg/m}^3.[/tex] Since the buoyant force is 20 N, we can solve for the volume of the displaced fluid:

[tex]F_B = \rho_wgV \Rightarrow V = \dfrac{F_B}{\rho_wg}[/tex]

Plugging in the values, we get

[tex]V = \dfrac{20\:\text{N}}{(997\:\text{kg/m}^3)(9.8\:\text{m/s}^2)}[/tex]

[tex]\:\:\:\:\:= 2.05×10^{-3}\:\text{m}^3[/tex]

Recall that the weight of an object in terms of its density and volume is given by

[tex]W = \rho gV[/tex]

Using the value for the volume above, we can solve for the density of the object as follows:

[tex]\rho = \dfrac{W}{gV} = \dfrac{900\:\text{N}}{(9.8\:\text{m/s}^2)(2.05×10^{-3}\:\text{m}^3)}[/tex]

[tex]\:\:\:\:\:= 44,798\:\text{kg/m}^3[/tex]

Answer:

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

Explanation:

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

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

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.

A Torque is a twisting force, or turning moment, it is a vector quantity with both magnitude and direction e.g Turning the handle of a cork-screw clockwise and then counterclockwise will advance the screw first inward and then outward By convention, counterclockwise torques are positive and clockwise torques are negative.

The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation.

Counterclockwise is the positive rotation direction and clockwise is the negative direction.

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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: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".

Explanation:

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

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:

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:

The importance of signal processing science as has been made clear. Since we now know that it help to convert discrete signal from time domain to frequency domain. Moreover, we are now aware of the difference in DF5 and FFT algorithms as well as the implementation in Matlab.

Answer:

Explanation:

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

Answer:

a battery, wires, and a switch.

Explanation:

All circuits include?

Answer:

3

Explanation:

state of matter are solid

liquid and

gases

Answer:

3

Explanation:

state of a matter are solid liquid and gas

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:

A) -31.2 rad/s

B) 28.1 rad/s^2

Explanation:

Hi there!

We can begin by finding the total time taken for the ball to reach the net using the equation:

dₓ = vₓt

12.9 = 44.7t

12.9/44.7 = t = 0.289 s

Now, we can use the following equation to solve for displacement in the Y direction:

d = y₀ + vit + 1/2at²

There is no initial vertical velocity, so:

d = y₀ + 1/2at²

Plug in known values:

d = 1.28 + 1/2(-9.8)(0.289²)

d = 0.87m

Thus, since 0.87 m < 0.914 m, the tennis ball does NOT make it over the net.

Answer:

A. Determine the equation of constraints

Answer:

Thermal Conductivity Easily Transmits Heat Among Fine Ceramics

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

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

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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:

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.

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:

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:

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