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4 years ago

% = Wo/Wi x 100 Solve for Wo % = Wo/Wi x 100 Solve for Wi

1 answer:

8 0

1) % = (Wo /Wi) * 100

Solve for Wo => Wo = (% / 100) * Wi

For example, % =30% and Wi = 250 => Wo = (30 /100) * 250 = 0.30 * 250 = 75

Wo = 75

2) % = (Wo / Wi) * 100

Solve for Wi

=> Wi = Wo * (%/100)

For example, Wo = 125 and % = 40%

=> Wi = 125 * (40 / 100) = 125 * 0.40 = 50

Wi = 50

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Under specific circumstances, waves will bend around obstacles. This type of bending is called

BartSMP [9]

That type of bending is called "diffraction" of waves.

6 0

3 years ago

Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.

miss Akunina [59]

Answer:

Δt = $\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}$

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 + Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

$\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}$ = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0

4 years ago

A force of 10. N toward the right is exerted on a wooden crate initially moving to the right on a horizontal wooden floor. the c

Artyom0805 [142]

1) 5 N

The crate is initially moving, so we must calculate the force of kinetic friction, which is given by:

$F_f = \mu_k mg$

where

$\mu_k=0.2$ is the coefficient of friction between the crate (made of wood) and the floor (made of wood). The coefficient of kinetic friction between wood and wood is about 0.2.

$mg=25 N$ is the weight of the crate

Substituting the numbers into the formula, we find

$F_f=(0.2)(25 N)=5 N$

2) 5 N

There are two forces acting on the crate along the horizontal direction:

- The force that pushes the crate toward the right, of magnitude $F=10 N$

- The force of friction, which acts in the opposite direction (so, towards the left), of magnitude $F_f = 5 N$

Since the two forces are in opposite directions, the net force is given by their difference:

$F_{net}=F-F_f = 10 N-5 N=5 N$

3) Yes

The crate is accelerating. In fact, according to Second Newton's Law:

$F_{net}=ma$ (1)

where Fnet is the net force on the crate, m is its mass, a is its acceleration. We can immediately see that since Fnet is not zero, the acceleration is also non-zero, so the crate is accelerating.

We can even calculate the magnitude of the acceleration. In fact, the mass of the crate is given by:

$m=\frac{Weight}{g}=\frac{25 N}{9.8 m/s^2}=2.55 kg$