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3 years ago

What do we use as a reference point to prove that we are always moving even when we are sitting down

2 answers:

4 0

Im not quite sure what you mean but do you mean us or the earth? If you mean us a reference is that our hearts are beating, we are breathing, we (may be) blinking, we are using our brains to do everything, etc.

mafiozo [28]3 years ago

3 0

This is a elementary question? in my opinion we could use stars. as the earth is in constant rotation, therefor we are technically still moving if we are sitting, just like being in a car, if that makes since

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Bats are capable of navigating using the earth's field-a plus for an animal that may fly great distances from its roost at night

Nataliya [291]

Answer:

The correct magnitude of the coil's magnetic field= =50μT

Explanation :

The magnetic field takes place as a result of movement of charge I e current, can also occurs from magnetised material, magnetic field as a result of charge movement can be deducted using right hand grip rule.

At the equator magnetic field lines are parallel towards the earth's surface and the angle of inclination of the magnetic lines of force at the horizontal position is referred to as the angle of dip at the point.

As the current is produced then the varying magnetic field is opposed ,then there is induced current when the coil is positioned at varying magnetic field.

Given from the question,

angle below horizontal θ=60-degree

The Earth's magnetic field B=50μT

The horizontal magnetic field can be expressed in terms of the formula below;

BH=Bcos⁡θ

B(H) = earth's horizontal component of magnetic field

Ø is the angle between coil's field

B=the magnetic field in Tesla

Then,

BH=50μT×cos60∘

=50μT× 0.5

As the current is passed through the coil to produce a field , when combined with the earth's field, which creates a net field with the same strength and dip angle (60 degrees below horizontal) as the earth's field.

It can be deducted that B has the same magnitude and angle which makes the those vertical component to cancel each other since they are the same.

For the magnetic field to be pointed out at North direction, we can calculate the corrected magnetic field using the formula below

Bc=2BH

Bc=2×50μT× 0.5=50μT

The correct magnitude of the coil's magnetic field= =50μT

4 0

4 years ago

What should a sailboat operator do when approaching a PWC head-on

nekit [7.7K]

In accordance with the International Regulation for the prevention of collisions at sea:

1.- A sailing boat has a passing preference over a motorized boat, except when the motor boat is limited by its draft.

2.- The sailboat must maintain its course and speed.

3.- If it is evident that the PWC does not respond, the sailboat must sound the warning signal, and change its course to starboard.

4.- All actions must be taken as soon as possible.

5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.

8 0

3 years ago

A weather balloon is inflated to a volume of 26.8 LL at a pressure of 744 mmHgmmHg and a temperature of 31.2 ∘C∘C. The balloon r

elena-s [515]

Answer:

43.96 L

Explanation:

We are given that

$V_1=26.8 L$

$P_1=744mm Hg$

$T_1=31.2^{\circ} C=31.2+273=304.2K$

$P_2=385mmHg$

$T_2=-14.8^{\circ}=273-14.8=258.2K$

We know that

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1T_2}{P_2T_1}$

Substitute the values

$V_2=\frac{744\times 26.8\times 258.2}{385\times 304.2}$

$V_2=43.96 L$

Hence, the volume of balloon at -14.8 degree Celsius=43.96 L

5 0

3 years ago

A 4400 W motor is used to do work. If the motor is used for 200 s, how much work could it do? (Power: P = W/t)

guajiro [1.7K]

So if p=w/t
then 4400=(w)(200)
so you would multiply 4440•200 and get 880,000

3 0

4 years ago

When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.6

Bezzdna [24]

Answer:

Distance between the charges, r = 0.8 meters

Explanation:

Given that,

Charge 1, $q_1=+8.4\ \mu C=+8.4\times 10^{-6}\ C$

Charge 2, $q_2=+5.6\ \mu C=+5.6\times 10^{-6}\ C$

Repulsive force between charges, F = 0.66 N

Let r is the distance between charges. The formula for the electrostatic force is given by :

$F=k\dfrac{q_1q_2}{r^2}$

$r=\sqrt{\dfrac{kq_1q_2}{F}}$

$r=\sqrt{\dfrac{9\times 10^9\times 8.4\times 10^{-6}\times 5.6\times 10^{-6}}{0.66}}$

r = 0.8009 meters

r = 0.8 meters

So, the distance between the charges i 0.8 meters. Hence, this is the required solution.

4 0

3 years ago