A change in temperature of 25K is equivalent to a change in temperature on the celsius scale of:

An object exerts a reaction force when it

r-ruslan [8.4K]

Answer:

all the forces occur in pairs that if one object exerts a force on another object, then the second object exerts an equal and opposite reaction force on the first.

5 0

3 years ago

How would life on earth be different if earth’s axis were not tilted with respect to its orbit?

Vladimir [108]

If Earth's axis was "straight up and down" instead of tilted, then ...

<span>-- There would be no seasons.

-- The climate at any one place would be the same all year around.

-- The days would be the same length, everywhere,
    and all year around.

-- So would the nights.

-- The sun would be up a little more than 12 hours every day.
    It would be down a little less than 12 hours every day.

-- At the middle of the day, the sun would be at the same height
   in the sky all year around, not higher in some months and lower
   in others.

-- The equator would be the only place on Earth where the sun
    could ever be directly over your head.

-- If you were at the north pole or the south pole, the sun would be
   down on the horizon, and it would just go around and around you
   every day. It would never rise or set, and it would never get any
   higher or lower.

</span>

4 0

3 years ago

Read 2 more answers

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

Bezzdna [24]

Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$ , the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q = -3.923\,C$ and $r = 100\,m$ , then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0

3 years ago

Suppose that the half-life of an element is 1000 years. How many half-lives will it take before one-eighth of the original sampl

34kurt

Answer:

3

Explanation:

The half-life of a radioactive isotope is the time it takes for the mass of the sample to halve.

This can be rewritten as follows:

$m(t) = m_0 (\frac{1}{2})^n$