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3 years ago

13

You are traveling through the forest and encounter an ancient well. You want to know how deep the well is. If you drop a coin an

d hear it hit the bottom of the well in 3 seconds, how deep is the well?

2 answers:

eimsori [14]3 years ago

4 0

Answer:

About 30 ft

Explanation:

horrorfan [7]3 years ago

3 0

Depth = 1/2 x 9,8 x 9 = 44.1 m

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10. Solve the following numerical problems

frosja888 [35]

Answer:

$\boxed {\boxed {\sf 120 \ Joules}}$

Explanation:

Work is equal to the product of force and distance.

$W=F*d$

The force is 8 Newtons and the distance is 15 meters.

$F= 8 \ N \\d= 15 \ m$

Substitute the values into the formula.

$W= 8 \ N * 15 \ m$

Multiply.

$W= 120 \ N*m$

1 Newton meter is equal to 1 Joule
Our answer of 120 N*m equals 120 J

$W= 120 \ J$

The work done is <u>120 Joules</u>

3 0

2 years ago

Two identical loudspeakers are placed on a wall 1.00 m apart. A listener stands 4.00 m from the wall directly in front of one of

natita [175]

Answer:

The phase difference is 0.659 rad.

Explanation:

Given that,

Distance between two identical loudspeakers d= 1.00 m

Distance between speakers and listener r= 4.00 m

Frequency = 300 Hz

Suppose we need to find the phase difference in radian between the waves from the speakers when they reach the observer

We need to calculate the r'

Using Pythagorean theorem

$r'=\sqrt{d^2+r^2}$

Where, d = distance between two identical loudspeakers

r = distance between speakers and listener

Put the value into the formula

$r'=\sqrt{(1.00)^2+(4.00)^2}$

$r'=\sqrt{1.00+16.00}$

$r'=4.12\ m$

We need to calculate the path difference

Using formula of path difference

$|r'-r|=4.12-4.00$

$|r'-r|=0.12\ m$

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{v}{f}$

Where, v = speed of sound

f = frequency

Put the value into the formula

$\lambda=\dfrac{343}{300}$

$\lambda=1.143\ m$

We need to calculate the phase difference

Using formula of phase difference

$\phi=\dfrac{2\pi\times|r'-r| }{\lambda}$

$\phi=\dfrac{2\pi\times0.12}{1.143}$

$\phi=0.659\ rad$

Hence, The phase difference is 0.659 rad.

7 0

3 years ago

Assuming there are no accidents or delays, the distance that a car travels down the interstate

Ganezh [65]

Explanation:

The distance that a car travels down the interstate can be calculated with the following formula:

Distance = Speed x Time

(A) Speed of the car, v = 70 miles per hour = 31.29 m/s

Time, d = 6 hours = 21600 s

Distance = Speed x Time

D = 31.29 m/s × 21600 s

D = 675864 meters

or

$D=6.75\times 10^5\ m$

(b) Time, d = 10 hours = 36000 s

Distance = Speed x Time

D = 31.29 m/s × 36000 s

D = 1126440 meters

or

$D=1.12\times 10^6\ m$

(c) Time, d = 15 hours = 54000 s

Distance = Speed x Time

D = 31.29 m/s × 54000 s

D = 1689660 meters

or

$D=1.68\times 10^6\ m$

Hence, this is the required solution.

7 0

2 years ago

The dot or scalar product of two (3d) vec- tors ⃗a = ⟨a1,a2,a3⟩ and ⃗b = ⟨b1,b2,b3⟩ is defined as

Neko [114]

Yes, yes, we know all of that. It certainly took you long enough to
get around to asking your question.

If
   a = (14, 10.5, 0)
and
   b = (4.62, 9.45, 0) ,

then, to begin with, neither vector has a z-component, and they
both lie in the x-y plane.

Their dot-product a · b = (14 x 4.62) + (10.5 x 9.45) =

         (64.68)   +   (99.225) = 163.905 (scalar)

I feel I earned your generous 5 points just reading your treatise and
finding your question (in the last line). I shall cherish every one of them.

7 0

3 years ago

A car is strapped to a rocket (combined mass = 661 kg), and its kinetic energy is 66,120 J.

aliina [53]

Answer:

9.43 m/s

Explanation:

First of all, we calculate the final kinetic energy of the car.

According to the work-energy theorem, the work done on the car is equal to its change in kinetic energy:

$W=K_f - K_i$

where

W = -36.733 J is the work done on the car (negative because the car is slowing down, so the work is done in the direction opposite to the motion of the car)

$K_f$ is the final kinetic energy

$K_i = 66,120 J$ is the initial kinetic energy

Solving,

$K_f = K_i + W = 66,120 + (-36,733)=29,387 J$

Now we can find the final speed of the car by using the formula for kinetic energy

$K_f = \frac{1}{2}mv^2$

where

m = 661 kg is the mass of the car

v is its final speed

Solving for v, we find

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(29,387)}{661}}=9.43 m/s$

3 0

3 years ago