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3 years ago

10

You are stranded in a blizzard. You need water to drink to drink,and you're trying to stay warm.should the melt the snow and dri

nk it or just eat the snow? Explain

1 answer:

Crazy boy [7]3 years ago

7 0

It would be a really bad idea to eat the snow because you obviously are trying to stay warm right? Well, the best thing to do is melt the snow. However, the process of melting the snow would have a few complications as well. But yes, the latter idea (drinking the snow) is a better idea (not the best).

You might be interested in

Calculate the speed of magnetic field rotation in an ac machine with 4 poles operating at the frequency of 60 hz.

geniusboy [140]

I think the answer is 15hz

4 0

3 years ago

The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU

Ipatiy [6.2K]

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

8 0

3 years ago

In a nuclear station if the temperature of the superheated water starts to decrease what should the engineer in charge of the re

Sophie [7]

Answer:

They sh0uld g0 t0 the reactor and then, see what the issue is...

Explanation:

then, see if they can fix the problem, im sorry if its wr0ng.

8 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is

Andreas93 [3]

Answer:

$v=1.08\times 10^7\ m/s$

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, $\dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2$

Separation between the plates, $d=1.75\times 10^{-2}\ m$

An electron is released from rest, u = 0

Using equation of kinematics,

$v^2-u^2=2ad$ ..........(1)

Acceleration of the electron in electric field, $a=\dfrac{qE}{m}$ ............(2)

Electric field, $E=\dfrac{\sigma}{\epsilon_o}$ ............(3)

From equation (1), (2) and (3) :

$v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}$

v = 10840393.1799 m/s

or

$v=1.08\times 10^7\ m/s$

So, the electron is moving with a speed of $1.08\times 10^7\ m/s$ before it reaches the positive plate. Hence, this is the required solution.

3 0

3 years ago

A spinning wheel has a rotational inertia of 2 kg•m². It has an angular velocity of 6.0 rad/s. An average counterclockwise torqu

ira [324]

Answer:

$-20.0 kg m^2/s$

Explanation:

The angular momentum of an object in rotation is given by

$L=I \omega$

where

I is the moment of inertia

$\omega$ is the angular speed

In this problem, initially we have

$I=2 kg m^2$ is the moment of inertia of the wheel

$\omega_i = 6.0 rad/s$ is the initial angular speed

So the initial angular momentum is

$L_i = I\omega_i = (2)(6.0)=12 kg m^2/s$

Later, a counterclockwise torque of

$\tau=-5.0 Nm$ is applied

So the angular acceleration of the wheel is:

$\alpha = \frac{\tau}{I}=\frac{-5.0}{2}=-2.5 rad/s^2$ in the direction opposite to the initial rotation.

As a result, the final angular velocity of the wheel will be:

$\omega_f = \omega_i + \alpha t$

where

t = 4.0 is the time interval

Solving,

$\omega_f = +6.0 +(-2.5)(4.0)=-4 rad/s$

which means that now the wheel is rotating in the counterclockwise direction.

Therefore, the new angular momentum of the wheel is:

$L_f = I\omega_f =(2)(-4.0)=-8.0 kg m^2/s$

So, the change in angular momentum is:

$\Delta L=L_f - L_i = (-8.0-(12))=-20.0 kg m/s^2$

7 0

3 years ago