When is an object magnetic?

The manufacturer of a 6V dry-cell flashlight battery says that the battery will deliver 15 mA for 60 continuous hours. During th

Vaselesa [24]

Answer:

16200 J

Explanation:

t = Time the battery is on = 60 hours

I = Current = $15\times 10^{-3}\ A$

Average voltage

$V=\dfrac{6+4}{2}=5\ V$

Energy is given by

$E=V\times I\times t$

$\\\Rightarrow E=5\times 15\times 10^{-3}\times 60\times 3600$

$\\\Rightarrow E=16200\ J$

The energy delivered in the given time is 16200 J

6 0

3 years ago

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

3 years ago

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

3 0

3 years ago

Match the major muscle group with its functional role.

Naily [24]

Answer:

Abdominal

Sitting up, postural alignment

Biceps

Lifting, pulling

Deltoids

Overhead lifting

Erector Spinae

Postural alignment

Gastronemius & Soleus

Push off for walking, standing on tiptoes

Gluteus

Climbing stairs, walking, standing up

Hamstrings

Walking

Latissimus Dorsi & Rhomboids

Postural alignment, pulling open a door

Obliques

Rotation and side flexion of body

Pectoralis

Push up, pull up, bench press

Quadriceps

Climbing stairs, walking, standing up

Trapezius

Moves head sideways

Triceps

Pushing

God bless you. Because my soul almost left my body when i had to do this.

7 0

3 years ago

It would really help if anyone could answers please and thanks

mojhsa [17]

well it would be A because 55 degrees is going strait well 75 is going literally straight up

4 0

3 years ago

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