The most useful unit of measurement for measuring the thickness of a side of a lab beaker would be a(n)

A 3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

anyanavicka [17]

Answer:

I think the answer is

Explanation:

140N/m/20N =

7m is the answer

8 0

2 years ago

Read 2 more answers

Kate is working on a project in her tech education class. She plans to assemble a fan motor. Which form of energy does the motor

andreyandreev [35.5K]

Motion Energy

I am writing this so it can be more than 20 letters

7 0

3 years ago

The Type K thermocouple has a sensitivity of about 41 uV /°C, i.e. for each degree difference in the junction temperature, the o

Novosadov [1.4K]

Answer:

ΔTmin = 3.72 °C

Explanation:

With a 16-bit ADC, you get a resolution of $2^{16}=65536$ steps. This means that the ADC will divide the maximum 10V input into 65536 steps:

ΔVmin = 10V / 65536 = 152.59μV

Using the thermocouple sensitiviy we can calculate the smallest temperature change that 152.59μV represents on the ADC:

$\Delta Tmin = \frac{\Delta Vmin}{41 \mu V/C}= 3.72 C$

3 0

2 years ago

A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.

Nataliya [291]

Answer:

c

Explanation:

4 0

3 years ago

After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter

Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, $\Delta L=3.1\ cm$

(a) The change in length per unit original length is called strain. So,

$\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193$

(b) The force acting per unit area is called stress.

$\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa$

(c) The ratio of stress to the strain is called Young's modulus. So,

$Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2$

Hence, this is the required solution.

8 0

2 years ago