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jek_recluse [69]
3 years ago
11

A swimmer swims north at 0.10 m/s relative to still water across a river that flows at 0.25 m/s from east to west. Calculate the

swimmer's velocity relative to the riverbank.
Please no guessing.
Showing work is optional but appreciated
Physics
1 answer:
ValentinkaMS [17]3 years ago
7 0
0.15 m/s East
If you follow the equation a=vf-vi/t, you'll discover that subtracting the final velocity, 0.25 m/s, by the initial velocity, 0.10 m/s, and divide by zero, (bc there was no given time) the answer is 0.15 m/s East
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He back window of this car contains a heating element.The heating element is part of an electrical circuit connected to the batt
Strike441 [17]
Um okay then????????????
7 0
3 years ago
I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove
Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
<span><span><span><span><span>69,816,900 km
0. 466 697 AU</span>

</span> </span> </span> <span> Distance at Perihelion (</span></span><span>point in it's orbit that's closest to the sun):</span>
<span><span><span><span>46,001,200 km
0.307 499 AU</span> </span>

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is  1/2  of the orbital period.

</span><span>Mercury's Orbital period = <span><span>87.9691 Earth days</span></span></span></span>

1/2 (50%) of that is  43.9845  Earth days

The average of the aphelion and perihelion distances is

     1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
     1/2 ( 0.466697 + 0.307499) = 0.387 098  AU
 
This also happens to be 1/2 of the major axis of the elliptical orbit.


3 0
3 years ago
The temperature inside my refrigerator is about 4 degrees C. If I place a balloon in my fridge that initially has a temperature
maksim [4K]

v

Convert the given temperatures from celsius to kelvin since we are dealing with gas.

To convert to kelvin, add 273.15 to the temperature in celsius.

T1 = 22 + 273.15 = 295.15 k

T2 = 4 + 273.15 = 277.15 k

V1 = 0.5 L

Let's find the final volume (V2).

To solve for V2 apply Charles Law formula below:

\frac{V_1}{T_1}=\frac{V_2}{T_2}

5 0
1 year ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
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jek_recluse [69]

Answer:

Electrons are located in specific orbit corresponding to discrete energy levels

Explanation:

In Bohr's model of the atom, electron orbit the nucleus in specific levels, each of them corresponding to a specific energy. The electrons cannot be located in the space between two levels: this means that only some values of energy are possible for the electrons, so the energy levels are quantized.

A confirmation of Bohr's model is found in the spectrum of emission of gases. In fact, when an electron jumps from a higher energy level to a lower energy level, it emits a photon whose energy is exactly equal to the difference in energy between the two levels: since the energy levels are discrete, this means that the emitted photons cannot have any value of wavelength, but also their wavelength will appear as a discrete spectrum. This is exactly what it is observed in the spectrum of emission of gases.

3 0
3 years ago
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