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3 years ago

6

Sonar question: A person's yells into a deep well, sending a sound wave that travels at a speed of 50 m/s. The echo is heard in

6 seconds.
How deep is the well? Remember, speed distance/time

1 answer:

velikii [3]3 years ago

3 0

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a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte

IgorLugansk [536]

<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
m1 = 5.5g

$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

m2 = 22.6g

$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
M1 = .0055 kg
v1i = ?
M2 = 0.0226 kg
v2i = 0 m/s
dx = 2.5 m
dy = -1.5 m

Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf.

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x.
</span>

dy = -1.5 m

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

<span>We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

$dy = v_{iy}t + \frac{1}{2} at^{2}$ </span>

$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

<span> $1.5 = \frac{1}{2} (9.8)t^{2}$

$\frac{(2)(1.5)}{9.8}=t^{2}$

$\frac{(3)}{9.8}=t^{2}$

$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
$v_{x} =d_{x}t$
$v_{x} =(2.5m)(0.553s)$
$v_{x} =4.52m/s$

Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it.

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

Now let's put in all our given and what we solved:

$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

The initial speed of the dart is 23.09 m/s or 23.10 m/s.</span>

7 0

4 years ago

Compute VO when Vin = 0.5 V, in two different ways: a) using the equation VO = G (V+-V-) with G = 106; and b) using the Golden R

Mrrafil [7]

Is b because yes thanks

7 0

3 years ago

A man pulls on his dogs leash to keep him from running after the bicycle. Which term best describes this example?

Luda [366]

What term do you mean? like what he did to the dog is he stopped the dog

5 0

3 years ago

Read 2 more answers

f a single circular loop of wire carries a current of 45 A and produces a magnetic field at its center with a magnitude of 1.50

Lelu [443]

Answer:

Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

Current flowing through the wire, I = 45 A

Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T

Number of turns in circular wire, N = 1

Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

$B=\frac{N\mu_{0} I}{2R}$

Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.

Substitute the suitable values in the above equation.

$1.50\times10^{-4} =\frac{4\pi \times10^{-7}\times45 }{2R}$

R = 0.18 m

4 0

3 years ago

A woman was recently given the opportunity to ride in a porsche race car on their test in Hapeville, Georgia. Below is a graph o

Zinaida [17]

Answer:

During the segments B - C and D - E, the car stopped since the y axis is the distance and the distance stayed the same in between those segments.

For a simpler answer, the flat horizontal lines on the graph are the times when the car was stopped.

6 0

3 years ago