Which of the following statements is most likely to be true of pseudoscience?

1)Pitch is measure of the ____________ of sound.

amid [387]

Answer:

1) wave length

2) parallel reflection

7 0

2 years ago

A horizontal force of 140 N is needed to pull a 60kg box across the horizontal floor at constant speed . Find the coefficient of

____ [38]

Explanation:

formula for force is:

force=mass × acceleration

but in case of friction

force =coefficient of friction × Normal Reaction

F. = u × R

U = F/R

but when placed horizontally

R= M×G

M=mass=60kg

G=Gravity(10m/s or 9.8m/s)

F=140N

U=140/60×10

U=140/600

U=0.2333333333

approximately to 3 significant figures

U=0.233

if i am correct rate it 5 star

4 0

1 year ago

It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil wo

vovikov84 [41]

Answer:

L = 1.11 x $10^{6}$ m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x $10^{5}$ m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x $10^{-2}$ m

Length = ?

So,

we know that,

U = 1/2 C Δ $v^{2}$

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δ $v^{2}$

52000 J = (0.5) x (C) x ( $610^{2}$ )

C = 0.28 F

And we also know that,

C = $\frac{K*E*A}{d}$

E = 8.85 x $10 ^{-12}$

K = 3.7

A = 0.20 x L

d = 2.6 x $10^{5}$ m

Plugging in the values into the formula, we get:

0.28 = $\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }$

Solving for L, we get:

L = 1.11 x $10^{6}$ m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

7 0

3 years ago

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio

Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

$d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at$

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.

Treat them separately.

At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.

Horizontal movement is not influenced by gravity.
acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

$d_x=10.0m\\d_y=3.0m$

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ? Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

$d_y=V_i_yt+\dfrac{1}{2}at^{2}$

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

$d_y=0+\dfrac{1}{2}at^{2}$

$d_y=\dfrac{1}{2}at^{2}$

From here we can derive the formula of time:

$t=\sqrt{\dfrac{2d_y}{a}}$

Now we just plug in what we know:

$t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s$

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

$vf_y=vi_y+at$

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

$0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s$

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

$d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt$

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

T = 1.564s

So we use this in our formula:

$d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x$

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

$Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s$

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

$Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17$