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solong [7]
2 years ago
8

At what condition does a body become weightless at the equator?

Physics
1 answer:
vesna_86 [32]2 years ago
3 0
Weightlessness is experienced when there is no net force acting on a body in the vertical direction. Whenever an object is rotating, such as the earth is, it applies an outward centrifugal force on the objects on its surface. At the equator, this force is in the direction exactly opposite to the force attracting us towards the earth, gravity.
If the earth were to spin faster by almost 17 times, the centrifugal force would become equal to the force of gravity and an object would experience weightlessness. 
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A long-distance swimmer is able to swim through still water at 4.0 km/h. She wishes to try to swim from Port Angeles, Washington
Roman55 [17]

Let \theta be the direction the swimmer must swim relative to east. Then her velocity relative to the water is

\vec v_{S/W}=\left(4.0\dfrac{\rm km}{\rm h}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

The current has velocity vector (relative to the Earth)

\vec v_{W/E}=\left(3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath

The swimmer's resultant velocity (her velocity relative to the Earth) is then

\vec v_{S/E}=\vec v_{S/W}+\vec v_{W/E}

\vec v_{S/E}=\left(\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}\right)\,\vec\imath+\left(4.0\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath

We want the resultant vector to be pointing straight north, which means its horizontal component must be 0:

\left(4.0\dfrac{\rm km}{\rm h}\right)\cos\theta+3.0\dfrac{\rm km}{\rm h}=0\implies\cos\theta=-\dfrac{3.0}{4.0}\implies\theta\approx138.59^\circ

which is approximately 41º west of north.

6 0
2 years ago
Technician A says that some ABS wheel speed sensors are used as part of the tire pressure monitoring system (TPMS) . Technician
erica [24]

Answer:

The correct answer is C) Technician A and B are both correct.

Explanation:

An anti-lock braking system (ABS) is a vehicle safety system that allows the wheels of a car to maintain tractive contact with the road surface while braking, preventing the wheels from locking up (ceasing rotation) and avoiding uncontrolled skidding. It is an automated system that uses the principles of cadence braking.

Anti-lock braking systems since their invention and introduction, have been improved remarkably in a bid to further improve driver safety and comfort.  <em>Recent technology not only prevents wheel lock up under braking, but can also provide data for the on board navigation system, traction control system, emergency brake assist, </em><u><em>hill start assist</em></u><em>, electronic stability control and the front-to-rear brake bias</em>. None of the above would be possible without wheel speed sensors.

The Tire Pressure Monitoring System (TMPS) is an electronic system in your vehicle that monitors your tire air pressure and alerts you when it falls dangerously low.

Indirect TPMS works with your car’s Antilock Braking System’s (ABS) wheel speed sensors. If a tire’s pressure is low, it will roll at a different wheel speed than the other tires. This information is detected by your car’s computer system, which triggers the dashboard indicator light.

Cheers!

3 0
2 years ago
Read 2 more answers
Which of the following exists around every object that has mass?
ICE Princess25 [194]

Answer:

A is the correct answer.

Explanation:

6 0
2 years ago
Read 2 more answers
A box of mass 8 kg slides across a frictionless surface at an initial speed 1.5 m/s into a relaxed spring of spring constant 69
Sav [38]

Answer:

1.1 sec

Explanation:

m = mass of the box = 8 kg

k = spring constant of the spring = 69 N/m

v = initial speed of the box = 1.5 m/s

t = time period of oscillation of box in contact with the spring

Time period is given as

t = \pi \sqrt{\frac{m}{k}}

Inserting the values

t = (3.14) \sqrt{\frac{8}{69}}

t = 1.1 sec

5 0
2 years ago
A charged box (m=445 g, ????=+2.50 μC) is placed on a frictionless incline plane. Another charged box (????=+75.0 μC) is fixed i
victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
2 years ago
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