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3 years ago

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]3 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.

Mercury

$w_{mercury} = 15*3.8\\w_{mercury}= 57 [N]\\$

Venus

$w_{venus}=15*8.8\\w_{venus}= 132 [N]$

Moon

$w_{moon}=15*1.6\\w_{moon}=24[N]$

Mars

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

Jupiter

$w_{jupiter}=15*23.1\\w_{jupiter}= 346.5[N]$

Saturn

$w_{saturn}=15*9\\w_{saturn}=135[N]$

Uranus

$w_{uranus}=15*8.7\\w_{uranus}=130.5[N]$

Neptune

$w_{neptune}=15*11\\w_{neptune}=165[N]$

Pluto

$w_{pluto}=15*0.6\\w_{pluto}=9[N]$

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2 years ago

An 30-turn coil has square loops measuring 0.341 m along a side and a resistance of 3.61 Ω. It is placed in a magnetic field tha

Verdich [7]

Answer:

Thus, the induced current in the coil at $t =1.73s$ is 9.98 A.

Explanation:

Faraday's law says

$\varepsilon = N \frac{d \Phi_B}{dt}$

where $N$ is the number of turns and $\Phi_B$ is the magnetic flux through the square coil:

Now,

$N = 30$

$\theta = 37.5^o$

$A = (0.341m)^2= 0.11623m^2$

$B = 1.45t^3$ ;

therefore,

$\varepsilon = N \frac{d \Phi_B}{dt} = N\frac{d ( BA\:cos(\theta))}{dt} = 30*\frac{d ( (1.45t^2)(0.1163)\:cos(37.5^o))}{dt}$

$=30*(0.1163)\:cos(37.5^o)*1.45*\dfrac{d ( t^3)}{dt} = 12.04t^2$

$\boxed{\varepsilon = 12.04t^2}$

is the emf induced in the coil.

Now, the loop is connected to $R = 3.61\Omega$ resistance; therefore, at $t = 1.73s$

$\varepsilon = RI = 12.04t^2$

$RI = 12.04(1.73)^2$

$RI = 36.03$

$I = \dfrac{36.03}{3.61\Omega }$

$\boxed{I = 9.98A }$

Thus, the current in the coil at $t =1.73s$ is 9.98 A.

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4 years ago

Project managers and their teams must keep in mind the effects of any project on the interests and needs of the entire system or

amm1812

Answer:True

Explanation:

True

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3 years ago

What is the difference between current electricity and static electricity

Allisa [31]

Answer:

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3 years ago

Read 2 more answers

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

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3 years ago