Hello, What is a four letter word the second letter is a O, and the hint is, “A period or measurement of time”

dsp73

°C = (5/9) · (°F-32)

The "wet" thermometer is the upper one ... you can see the wet cloth wrapped around the bulb at the end. It's reading 70° F.

°C = (5/9) · (38) = 21.1° C

The "dry" thermometer is the lower one. It's reading 80° F.

°C = (5/9) · (48) = 26.7° C

So it looks like choice-A is your answer.

6 0

3 years ago

A scientist pours 0.120 L of solution A into an Erlenmeyer flask. She adds 2.345 L of solution B. How many significant figures a

nika2105 [10]

0.120L + 2.345L = 2.465L = 4 significant figures in the answer

6 0

3 years ago

An electric generator transforms mechanical energy into electrical energy. This process could be done by which of these?

Dimas [21]

Answer:

its d

Explanation:

6 0

3 years ago

Read 2 more answers

A ball is thrown up into the air with an initial velocity of 18 m/s. A) How high does the ball go? B) Calculate the time needed

kaheart [24]

Answer:

B) t = 1.83 [s]

A) y = 16.51 [m]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f} =v_{o} -g*t$

where:

Vf = final velocity = 0

Vo = initial velocity = 18 [m/s]

g = gravity acceleration = 9.81 [m/s²]

t = time [s]

Note: the negative sign in the above equation means that the acceleration of gravity is acting in the opposite direction to the motion.

A) The maximum height is reached when the final velocity of the ball is zero.

0 = 18 - (9.81*t)

9.81*t = 18

t = 18/9.81

t = 1.83 [s], we found the answer for B.

Now using the following equation.

$y = y_{o} + v_{o}*t - 0.5*g*t^{2}\\$

where:

y = elevation [m]

Yo = initial elevation = 0

y = 18*(1.83) - 0.5*9.81*(1.83)²

y = 16.51 [m]

7 0

3 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

3 years ago