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Ghella [55]
2 years ago
15

If you stood on mars and lifted a 15kg pack, you would be pulling with a force greater than...?

Physics
1 answer:
DIA [1.3K]2 years ago
7 0

Answer:

See the answers below

Explanation:

In this problem, we must be clear about the concept of weight. Weight is defined as the product of mass by gravitational acceleration.

We must be clear that the mass is always preserved, that is, the mass of 15 [kg] will always be the same regardless of the planet where they are.

W=m*g

where:

W = weight [N] (units of Newtons)

m = mass = 15 [kg]

g = gravity acceleration [m/s²]

Since we have 9 places with different gravitational acceleration, then we calculate the weight in each of these nine places.

<u>Mercury</u>

<u />w_{mercury} = 15*3.8\\w_{mercury}= 57 [N]\\<u />

<u>Venus</u>

<u />w_{venus}=15*8.8\\w_{venus}= 132 [N]<u />

<u>Moon</u>

<u />w_{moon}=15*1.6\\w_{moon}=24[N]<u />

<u>Mars</u>

w_{mars}=15*3.7\\w_{mars}=55.5 [N]

<u>Jupiter</u>

<u />w_{jupiter}=15*23.1\\w_{jupiter}= 346.5[N]<u />

<u>Saturn</u>

<u />w_{saturn}=15*9\\w_{saturn}=135[N]<u />

<u>Uranus</u>

<u />w_{uranus}=15*8.7\\w_{uranus}=130.5[N]<u />

<u>Neptune</u>

<u />w_{neptune}=15*11\\w_{neptune}=165[N]<u />

<u>Pluto</u>

<u />w_{pluto}=15*0.6\\w_{pluto}=9[N]<u />

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Answer:

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the base of a rectangular vessel measure 10m by 18cm. water is poured into a depth of 4cm. (a) what is the pressure on the base?
Alex787 [66]

Answer:

a) P =392.4[Pa]; b) F = 706.32[N]

Explanation:

With the input data of the problem we can calculate the area of the tank base

L = length = 10[m]

W = width = 18[cm] = 0.18[m]

A = W * L = 0.18*10

A = 1.8[m^2]

a)

Pressure can be calculated by knowing the density of the water and the height of the water column within the tank which is equal to h:

P = density * g *h

where:

density = 1000[kg/m^3]

g = gravity = 9.81[m/s^2]

h = heigth = 4[cm] = 0.04[m]

P = 1000*9.81*0.04

P = 392.4[Pa]

The force can be easily calculated knowing the relationship between pressure and force:

P = F/A

F = P*A

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A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p
DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

t_{total}=t_{up}+t_{down}

Time upriver is calculated with the net speed of student and 0.500 km:

t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s

(Becareful with units 0.5 km= 500m) Similarly of downriver:

t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s

So the sum is:

t_{total}=1111.0s

b) Still water does not affect student speed, so total time would be simply:

t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.  

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