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3 years ago

Which would be a common-sense practice in a lab environment?

Physics

1 answer:

algol133 years ago

6 0

Using safe research practices would be a very good common-sense practice in a lab environment.

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A 4-kg hammer is lifted to a height of 10 m and dropped from rest. What was the velocity (in m/s) of the hammer when it was at a

ipn [44]

Answer:

v = 10.84 m/s

Explanation:

using the equation of motion:

v^2 = (v0)^2 + 2×a(r - r0)

<em>due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.</em>

v^2 = 2×g(r - r0)

v = \sqrt{2×(-9.8)×(4 - 10)}

= 10.84 m/s

therefore, the velocity at r = 4 meters is 10.84 m/s

4 0

2 years ago

In many cartoon shows, a character runs of a cliff, realizes his predicament and lets out a scream. He continues to scream as he

Rzqust [24]

Answer:

Increasing until terminal velocity is reached

Explanation:

Provided the scream is a constant pitch at the source, Doppler effect will make the pitch increase as the velocity of the source towards the listener increases.

5 0

2 years ago

The electromagnetic wave that CT scans are based on is called a

kherson [118]

X-ray)
because Electromagnetic waves are in act

6 0

2 years ago

If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb

Anna [14]

Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

P1V1 = P2V2 ( at temperature is constant )

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg(torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987*(X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L

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3 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$