Which phrase describes what a scientist does?

How do lines of latitude affect how direct or indirect the Sun’s rays are on the Earth?

IgorLugansk [536]

The technical definition of latitude is the angular distance north or south from the earth's equator measured through 90 degrees. ... Locations at lower latitudes receive stronger and more direct sunlight than locations near the poles. Energy input from the sun is the main driving force in the atmosphere.

The Seasons at Different Latitudes
The seasonal effects are different at different latitudes on Earth. Near the equator, for instance, all seasons are much the same. Every day of the year, the Sun is up half the time, so there are approximately 12 hours of sunshine and 12 hours of night.

When we consider Latitude alone as a control, we know that the low latitudes (say from the Equator to approximately 30 degrees N/S) are the warmest across the year (on an annual basis).

8 0

3 years ago

A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant k=450N/m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

For a simple harmonic motion, the magnitude of the maximum acceleration is given by
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

we have already calculated it at point b), and it is given by
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

What is your acceleration while sitting in your chair. the latitude of corvallis is 44.4˚.?

marta [7]

You can start with the equations you know

a=v^2/r = (2pi*r/T)^2/r = 4pi^2r/T^2

Radius of earth (R) = 6378.1 km
Time in one day (T) = 86400 seconds
Latitude = 44.4 degrees

If you draw a circle and have the radius going out at a 44.4 degree angle above the center you can then find the r.

r=Rcos(44.4)
r=6378.1cos(44.4)
r= 4556.978198 km or 4556978 m

Now you can plug this value into the acceleration equation from above...

a= 1.8*10^8/7.47*10^9
a= .0241 m/s^2

8 0

3 years ago

2 ways to change frictional force between 2 objects

Zanzabum

In order to change the frictional force between two solid surfaces, it can be changed by shorter distances and by the amount of weight it has or the amount of force that is pushing that object to go however distance it can.

5 0

3 years ago

Read 2 more answers

Breaking a pencil in half is an example of: options:

Licemer1 [7]

Answer:

b

Explanation:

8 0

3 years ago

Read 2 more answers