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Ad libitum [116K]
3 years ago
14

HELP. A river flows South with a speed of 8m/s. From the shore, you see a boat moving South with a speed of 10 m/s. From their f

rame of reference, how fast are the rowers moving?
Physics
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:

18 m/s

Explanation:

Given that,

Speed of river is 8 m/s due South

From the shore, you see a boat moving South with a speed of 10 m/s. Both boat and river are moving in same direction. The resultant velocity will add up.

v = 10 m/s + 8 m/s

v = 18 m/s

Hence, the rowers are moving with a speed of 18 m/s.

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What type of device forms images by changing the speed at which light
Nady [450]

Answer:

Any Lens

Explanation:

I Hope it's right if not so Sorry :)

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3 years ago
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According to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the differen
Kryger [21]

Answer:

dT(t)/dt = k[T5 - T(t)]

Explanation:

Since T(t) represents the temperature of the object and T5 represents the temperature of the surroundings, according to Newton's law of cooling, the rate at which an object's temperature changes is directly proportional to the difference in temperature between the object and the surrounding medium, that is dT(t)/dt ∝ T5 - T(t)

Introducing the constant of proportionality

dT(t)/dt = k[T5 - T(t)]

which is the desired differential equation

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3 years ago
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A bird lands on a bare copper wire carrying a current of 51
nirvana33 [79]
The resistance of the piece of wire is
R= \frac{\rho L}{A}
where
\rho = 1.68 \cdot 10^{-8}\Omega m is the resistivity of the copper
L=5.1 cm=0.051 m is the length of the piece of wire
A=0.13 cm^2 = 0.13 \cdot 10^{-4} m^2 is the cross sectional area of the wire
By substituting these values, we find the value of R:
R= \frac{\rho L}{A}=6.6 \cdot 10^{-5} \Omega

Then, by using Ohm's law, we find the potential difference between the two points of the wire:
V=IR=(51 A)(6.6 \cdot 10^{-5} \Omega )=3.4 \cdot 10^{-3} V
7 0
3 years ago
1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it
spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

   PE = (2.5)(98)(14) and

   PE = 340 J

b. KE=\frac{1}{2}mv^2 so

   KE=\frac{1}{2}(2.5)(14)^2 and

   KE = 250 J

c. TE = KE + PE so

   TE = 340 + 250 and

   TE = 590 J

d. PE at 8.7 m:

   PE = (2.5)(9.8)(8.7) and

   PE = 210 J

e. The KE at the same height:

   TE = KE + PE and

   590 = KE + 210 so

   KE = 380 J

f. The velocity at that height:

   380=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(380)}{2.5} } so

   v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

   590 = KE + PE and

   PE = (2.5)(9.8)(11.6) so

   PE = 280 then

   590 = KE + 280 so

   KE = 310 then

   310=\frac{1}{2}(2.5)v^2 and

   v=\sqrt{\frac{2(310)}{2.5} } so

   v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

   v=v_0+at and

   26 = 0 + 9.8t and

   26 = 9.8t so the time at 26 m/s is

   t = 2.7 seconds. Now we use that in the equation for displacement:

   Δx = v_0t+\frac{1}{2}at^2 and filling in the time the object was at 26 m/s:

   Δx = 0t + \frac{1}{2}(-9.8)2.7)^2 so

   Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0
3 years ago
What are some actions that can be taken to remedy air pollution? Select all that apply. Run your air conditioner at a higher set
Sati [7]

Answer:  Government passing regulatory laws

Car emissions testing

Carpooling and less driving overall

Air pollution results when a mixture of particles and gases that can reach in harmful concentrations both outdoors as well as indoors. This increase in concentration can lead to rise in the temperatures and global warming it is also related with the health hazards. Soot, smoke, pollen, carbon dioxide, carbon monoxide and methane are the examples of the common pollutants.

Government passing regulatory laws: The government can ban some of the fossil fuels to prevent air pollution, and some vehicles which are likely to emit more partially combustible gases.

Car emission testing: Nowadays, cars and other vehicles are equipped with catalytic converters which converts the uncombustible gaseous products into less harmful gases which are emitted from the vehicle. The car gas emission testing should be performed to either allow a vehicle or ban the movement of the vehicle.

Carpooling and less driving overall: Carpooling is the sharing of one vehicle with more than one individual. This will prevent the need of others to drive another vehicle. The less the number of vehicles on the road the lesser will be the emission of gases and less pollution will be caused.

4 0
3 years ago
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