PHYSICS COLLEGE

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Answers

Answer 1

Answer:

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

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A screw can be considered a type of

Answers

Fastener because a fastener is something that connects to objects and usually can come apart but can also be permanent

Much of our knowledge of the interior of the Earth comes from the study of planetary vibrations, which is the science of

Answers

Answer:

Seismology.

Explanation:

Seismology is the beach of physical science that deals with the study of vibrations that comes out from the interior of the earth onto the surface and these vibrations are in the form of seismic waves that are primary, secondary, and surface waves.

The science of seismology tells about the magnitude and intensity of these waves that lead to planetary vibrations. These waves trigger earthquakes, floods, and even landslides.

How many nanoseconds are in one hour? How do you write the following in scientific notation?2,560,000m

Answers

Answer:

3.6 × 10¹² nanoseconds

Explanation:

Hour is the unit of time. Seconds is the SI unit of time.

Hour and seconds are related as:

1 hour = 60 minutes

1 minute = 60 seconds

So,

1 hour = 60 ×60 seconds = 3600 seconds

Thus,

3600 seconds are in one hour

Also,

1 sec = 10⁹ nanoseconds

Thus,

3600 sec = 3600 × 10⁹ nanoseconds = 3.6 × 10¹² nanoseconds

Thus,

3.6 × 10¹² nanoseconds are in one hour.

What is effort arm
don't say the answer of gogle

Answers

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

A 1.5v battery stores 4.5KJ of energy. How long can it light a flashlight bulb that draws 0.60A

Answers

Answer:

The 1.5V battery can power the flashlight bulb drawing 0.60A for 83.33 minutes before it is depleted.

Explanation:

To determine how long a 1.5V battery can power a flashlight bulb drawing 0.60A, you can use the formula for calculating the energy (in joules) consumed by an electrical device over time:

Energy (Joules) = Power (Watts) × Time (Seconds)

In this case, the power (P) is given by the product of the voltage (V) and current (I):

Power (Watts) = Voltage (Volts) × Current (Amperes)

So, first, calculate the power consumption of the flashlight bulb:

Power (Watts) = 1.5V × 0.60A = 0.90 Watts

Now, you want to find out how long the battery can power the bulb, so rearrange the energy formula to solve for time:

Time (Seconds) = Energy (Joules) / Power (Watts)

Given that the battery stores 4.5 kJ (kilojoules), which is equivalent to 4,500 joules, and the power consumption is 0.90 watts:

Time (Seconds) = 4,500 J / 0.90 W = 5,000 seconds

Now, to express the time in more practical units, convert seconds to minutes:

Time (Minutes) = 5,000 seconds / 60 seconds/minute ≈ 83.33 minutes

So, the 1.5V battery can power the flashlight bulb drawing 0.60A for approximately 83.33 minutes before it is depleted.

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A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction (a) If the applied stress if 0.45 MPa, what will be the resolved shear stress, tRss, along the [101] direction within the (11 T) plane? (b) What tensile stress is required to produce a critical resolved shear stress, TcRss of 0.242 MPa

Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)