All categories

3 years ago

3. How does Earth’s rotation affect our view of stars

Physics

2 answers:

Elis [28]3 years ago

5 0

Since we ride along with the Earth while it's doing whatever it does,
the Earth's rotation causes our eyes to constantly point in a different
direction.

If we try to keep watching one star, we have to keep changing the
direction of our eyes to keep looking at the same star.

We can't feel the Earth rotating, so our brains say that the star ... and
the sun and the moon too ... is actually moving across the sky.

fomenos3 years ago

4 0

This is because when the Earth spins on it's axis the view of our sky shifts as well.

You might be interested in

The wave frequency is where 1 hertz equals 1 wave passed a fixed point in 1 second.

ahrayia [7]

   (10 waves) / (5 sec)

   = (10/5) (wave/sec)

   = 2 per sec

   = 2 Hz .

3 0

3 years ago

A 13561 N car traveling at 51.1 km/h rounds

Minchanka [31]

Answer:

a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

3 0

3 years ago

Explain why all variables but one should be controlled in a well designed experiment.

zlopas [31]

All variables should be controlled except the one being tested because it would allow to ascertain the researcher that the results is truly an effect of that one and tested variable. This variable is called the independent variable. It is what is being changed and manipulated in a research to effect a change to the dependent variable which is the variable being studied.

3 0

3 years ago

Calculate the acceleration of a car (in m/s^2) that accelerates from 0 to 30 m/s in 6 s along a straight road.

tester [92]

Answer: 5 m/s^2

Explanation: In order to solve this question we have to use the kinematic equation given by:

Vf= Vo+a*t where V0 is zero.

we know that it takes Vf( 30 m/s) in 6 seconds

a=(30 m/s)/6 s= 5 m/s^2

8 0

3 years ago

If a bar magnet is suspended at the center on a string and allowed to swing freely, in what direction will its south pole point

nika2105 [10]

Explanation : If a bar magnet is suspended at the center on a string and allow to swing freely, then the magnet will rotate so that it will line with the pole of the earth .

So, we can say that when the magnet suspended freely by the string, then the magnet will rotate and stop in north and south direction . The north pole of the magnet will stop in the south direction of the earth and the south pole of the magnet will stop in the north direction of the earth.

8 0

3 years ago