A group of ecosystems with similar climates and organisms is called a(n)

Which of the following classifications of star temperature is coolest?

Oliga [24]

The hottest would be the O type and the coolest is M

3 0

3 years ago

Read 2 more answers

Need help with this science!!

noname [10]

Lol, I remember I helped my sister with this question but anyways, the card flew because of inertia and the quarter fell in the glass because of gravity :).

3 0

3 years ago

Can someone help me wit hthis science question?

klemol [59]

Answer:

see explanation

Explanation:

There is an increasing demand for materials and natural resources from a growing global population, especially those in more economically developed countries. The world's resources are being used up more quickly. The consumption of resources is spread unequally between MEDCs (more economically developed countries), who use more resources, and LEDCs (less economically developed countries), who use less.

The gap between the rich and poor is more evident when the resources are shared so unevenly and unfairly and natural resources like materials and natural energy cannot reach the demand of the people which can have consequences and be very difficult to manage. Having a lack of these materials in a country can result in prices going up for them, and the industry could be harder to work in because of a lack of materials.

6 0

3 years ago

A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc

guajiro [1.7K]

Answer:

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

$a_o=a\times e^{-kt}$

where,

k = rate constant

t = age of sample

$a_o$ = let initial amount of the reactant

a = amount left after decay process

We have :

$a_o=x$

$a=58\%\times x=0.58x$

t = 95 s

$0.58x=x\times e^{-k\times 95 s}$

$\k= 0.005734 s^{-1}$

Half life is given by for first order kinetics::

$t_{1/2}=\frac{0.693}{k}$

$=\frac{0.693}{0.005734 s^{-1}}=120.86 s$

$0.005734 s^{-1}$ and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0

3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

3 years ago