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4 years ago

If a spear is thrown at a fish swimming in a lake, it will often miss the fish completely. Why does this happen?

1 answer:

5 0

The light that makes up the image of the fish is refracted when it leaves the water and enters the air. Your brain thinks the light traveled straight, so the fish is not located in the direction from you that your brain thinks it is.

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A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Oduvanchick [21]

To solve this problem it is necessary to apply the concepts based on Newton's second law and the Centripetal Force.

That is to say,

$F_c = F_w$

Where,

$F_c =$ Centripetal Force

$F_w =$ Weight Force

Expanding the terms we have to,

$mg = \frac{mv^2}{r}$

$gr = v^2$

$v = \sqrt{gr}$

Where,

r = Radius

g = Gravity

v = Velocity

Replacing with our values we have

$v = \sqrt{(9.8)(11.8)}$

$v = 10.75m/s$

Therefore the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top is 10.75m/s

5 0

3 years ago

On top of Mount Everest, the temperature is -19 ∘C in July. Being a physicist, you determine by how many degrees Celsius one nee

kondaur [170]

Answer:

254 °C

Explanation:

The average kinetic energy of gas molecules K = 3RT/2N where R = gas constant = 8.314 J/mol-K, N = avogadro's constant = 6.022 × 10²³ atoms/mol

T = temperature in Kelvin.

Let K be its average kinetic energy at t = -19°C = 273 + (-19) = 273 - 19 = 254 K = T. K = 3RT/2N = 3 × 8.314 J/mol-K × 254 K/(2 × 6.022 × 10²³ atoms/mol) = 5.26 × 10⁻²¹ J

When its average kinetic energy doubles, it becomes K₁ = 2K = 2 × 5.26 × 10⁻²¹ = 10.52 × 10⁻²¹ J at temperature T₂. So,

K₁ = 3RT₁/2N

T₁ = 2NK₁/3R

T₁ = 2 × 6.022 × 10²³ atoms/mol × 10.52 × 10⁻²¹ J/3 × 8.314 J/mol-K = 508 K

The temperature difference is thus ΔT = T₁ - T = 508 K - 254 K = 254 K.

Since temperature change in kelvin scale equals temperature change in Celsius scale ΔT = 254 °C

So, we need to change the temperature of the air by 254 °C to double its average kinetic energy.

3 0

3 years ago

A balloon is rising vertically upwards at a velocity of 10m/s. When it is at a height of 45m from the ground, a parachute bails

harina [27]

(a) 30.9 m

Let's analyze the motion of the parachutist. Its vertical position above the ground is given by

$y=h+ut+\frac{1}{2}gt^2$

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t=3 s , we find the height of the parachutist when it opens the parachute:

$y=45 m+(10 m/s)(3 s)+\frac{1}{2}(-9.8 m/s^2)(3 s)^2=30.9 m$

(b) 44.1 m

Here we have to find first the height of the balloon 3 seconds after the parachutist has jumped off from it. The vertical position of the balloon is given by

y = h + ut

where

h = 45 m is the initial height

u = 10 m/s is the initial velocity (upward)

t is the time

Substituting t = 3 s, we find

y = 45 m + (10 m/s)(3 s) = 75 m

So the distance between the balloon and the parachutist after 3 s is

d = 75 m - 30.9 m = 44.1 m

(c) 8.2 m/s downward

The velocity of the parachutist at the moment he opens the parachute is:

v = u +gt

where

u = 10 m/s is the initial velocity (upward)

t is the time

g = -9.8 m/s^2 is the acceleration of gravity (downward)

Substituting t = 3 s,

v = 10 m/s + (-9.8 m/s^2)(3 s)= -19.4 m/s

where the negative sign means it is downward

After t=3 s, the parachutist open the parachute and it starts moving with a deceleration of

a =+5 m/s^2

where we put a positive sign since this time the acceleration is upward.

The total distance he still has to cover till the ground is

d = 30.9 m

So we can find the final velocity by using

$v^2-u^2 = 2ad$

where this time we have u = 19.4 m/s as initial velocity. Taking the downward direction as positive, the deceleration must be considered as negative:

a = -5 m/s^2

Solving for v,

$v=\sqrt{u^2 +2ad}=\sqrt{(19.4 m/s)^2+2(-5 m/s^2)(30.9 m)}=8.2 m/s$

(d) 5.24 s

We can find the duration of the second part of the motion of the parachutist (after he has opened the parachute) by using

$a=\frac{v-u}{t}$

where

a = -5 m/s^2 is the deceleration

v = 8.2 m/s is the final velocity

u = 19.4 m/s is the initial velocity

t is the time

Solving for t, we find

$t=\frac{v-u}{a}=\frac{8.2 m/s-19.4 m/s}{-5 m/s^2}=2.24 s$

And added to the 3 seconds between the instant of the jump and the moment he opens the parachute, the total time is

t = 3 s + 2.24 s = 5.24 s

8 0

3 years ago

A photon in a laboratory experiment has an energy of 5 eV. What is the frequency of this photon? (using the idea of the electron

andreyandreev [35.5K]

Answer and working shown on photo

7 0

3 years ago

It took a crew 3 h 20 min to row 5 km upstream and back again. if the rate of flow of the stream was 2 km/h, what was the rowing

Basile [38]

Suppose rowing speed is V km/s.

So, upstream speed = V-2

Downstream speed = V+ 2

Total time is taken 3 h 20 min to cover up and down journey of 5 km.

So, total time = upstream time + downstream time

$3. 3 \ hours = \frac{5 km}{ V- 2} + \frac{5 km}{V+2} \\\\\ 3.3 (V^2-2^2) = 10V\\\\V^2- 4-3V= 0$

$V^2 -4V+V -4=0 \\\\\ V( V-4) +1 (V-4)=0$

$V= 4 km/h$ and $V= -1 \ km/h$ , we take only positive velocity.

Thus, the the rowing speed of the crew in still water is 1 km/s.

8 0

3 years ago