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Marina CMI [18]

2 years ago

7

A fish appears to be 2.00 m below the surface of a pond (nwater = 1.33) when viewed almost directly above by a fisherman. What i

s the actual depth of the fish?

2 answers:

Alja [10]2 years ago

8 0

Explanation:

The given data is as follows.

Apparent depth = 2.00 m, refractive index = 1.33

It is known that formula to calculate real depth is as follows.

The refractive index = $\frac{\text{real depth}}{\text{apparent depth}}$

or, real depth = refractive index × apparent depth

Now, putting the given values into the above formula as follows.

real depth = 1.33 × 2

= 2.66 m

Thus, we can conclude that the actual (real) depth of the fish is 2.66 m.

Olegator [25]2 years ago

6 0

Answer:

Actual depth = 2.66 m

Explanation:

Given:

Apparent depth = Depth of fish as seen below the surface of water = $D_{a}$ = 2 m

Refractive index of water = n = 1.33

Actual depth is related to the apparent depth as follows:

n = $D_{r}$ / $D_{a}$

⇒ Actual depth = $D_{r}$ = n × $D_{a}$

= (1.33) ( 2 ) = 2.66 m

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 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme

Artyom0805 [142]

Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider $T_{1}\ and\ T_{2}$ are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

$Let\ Q_{1}\ and\ Q_{2}$ are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

$W\ =\ Q_{1}-Q_{2}$

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

$Efficiency\ [\eta]=\frac{W}{Q_{1}}$

$\eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}$

$=\ 1-\frac{Q_{2}} {Q_{1}}$

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

6 0

2 years ago

Read 2 more answers

Which of the following are scalar quantities? select all that apply

dolphi86 [110]

Answer:

1, 4, 5, see the explanation below

Explanation:

We must remember that scalar magnitudes are distinguished by having only a physical quantity, that is, they have no sense or direction as an example of scalar quantities, we find mass, temperature, energy, specific heat, power among others.

1 . 150 [grams] , because is a mass = scalar

4. 5 kilometer [race], is an amount = scalar

5. 34 steps, is an amount = scalar

Number 2, and 3 are vectors because they have amount and direction.

3 0

2 years ago

A 6.0-ohm resistor that obeys Ohm’s Law is connected to a source of variable potential difference. When the applied voltage is d

leva [86]

Is halved. A 6Ω resistor connected to a voltage source which voltage is decreased from 12V to 6V the current passing through the resistor is halved.

The key to solve this problem is applying Ohm's Law V = R I, clearing I from the equation, we obtain I = V/R. Then, the current is directly proportional to the voltage and inversely proportional to the resistance.

V = 12V and R = 6Ω

I = 12V/6Ω = 2A

V = 6V and R = 6Ω

V = 6V/6Ω = 1A

As we can see the current is halved if the voltage descreased from 12V to 6V

5 0

2 years ago

The moon’s orbital speed around Earth is 3.680 × 10^3 km/h. Suppose the moon suffers a perfectly elastic collision with a comet

Naily [24]

Answer:

Speed of comet before collision is

$v_{2_{i}}=-2.5\times10^{3}\quad km/h$

Explanation:

Correction: (As stated after collision comet moves away from moon so velocity of moon and moon and comet must be opposite in direction. as spped of moon after collision is −4.40 × 10^2km/h so that comet's must be 5.740 × 10^3km/h instead of -5.740 × 10^3km/h)

Solution:

$mass \quad of\quad moon = m_{1}\\\\mass\quad of \quad comet = m_{2} = 0.5 m_{1}\\\\speed\quad of\quad moon\quad before\quad collision = v_{1_{i}}=3.680\times 10^3 km/h\\\\speed \quad of\quad moon\quad after\quad collision=v_{1_{f}} = -4.40 \times 10^2 km/h\\\\speed\quad of\quad comet\quad after\quad collision =v_{2_{f}} =5.740 \times 10^3 km/h$

Case is considered as partially inelastic collision, by conservation of momentum

$m_{1}v_{1_{i}}+m_{2}v_{2_{i}}=m_{1}v_{1_{f}}+m_{2}v_{2_{f}}\\\\m_{1}v_{1_{i}}+0.5m_{1}v_{2_{i}}=m_{1}v_{1_{f}}+0.5m_{1}v_{2_{f}}\\\\v_{1_{i}}+0.5v_{2_{i}}=v_{1_{f}}+0.5v_{2_{f}}\\\\v_{2_{i}}=2(v_{1_{f}}+0.5v_{2_{f}}-v_{1_{i}})\\\\v_{2_{i}}=2(-4.40 \times 10^2+0.5(5.740 \times 10^3)-3.680 \times 10^3 )\\\\v_{2_{i}}=-2.5\times10^{3}\quad km/h$

8 0

2 years ago