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3 years ago

The specific gravity of a fluid with a weight density of 31.2 lb/ft is a. 2.00 b. 0.969 c. 0.500 d. 1.03

Engineering

1 answer:

kykrilka [37]3 years ago

7 0

Answer:

Answer is c 0.500

Explanation:

$SpecificGravity=\frac{\rho _{fluid}*g}{\rho _{water}*g}$

We know that $\rho_{water}=62.42lb/ft^{3}$

Applying values we get

$SpecificGravity=\frac{31.2}{62.4}=0.5$

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The domain of discourse is the members of a chess club. The predicate B(x, y) means that person x has beaten person y at some po

satela [25.4K]

Answer:A. No one has ever beat Nancy.

Explanation:

The dormain of discourse in a simple language is the set of entities upon which our discussions are based when discussing about something.

The dormain of discourse is also known simply as universe, can also be said to be a set of entities o

upon which certain variables of interest in some formal treatment may range.

The dormain of discourse is generally attributed to Augustus De Morgan, it was also extensively used by George Boole in his Laws of Thought.

THE LOGICAL UNDERSTANDING OF THE THE QUESTION IS THAT NO ONE HAS EVER BEAT NANCY.

8 0

3 years ago

Race car is accelerating and has a velocity of 10 m/s @ t=0. It completes a lap on a circular track of 400 m in 14 seconds. Calc

wariber [46]

Answer:

component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²

magnitude of acceleration is 22.98 m/s²

Explanation:

given data

velocity = 10 m/s

initial time to = 0

distance s = 400 m

time t = 14 s

to find out

components and magnitude of acceleration after the car has travelled 200 m

solution

first we find the radius of circular track that is

we know distance S = 2πR

400 = 2πR

R = 63.66 m

and tangential acceleration is

S = ut + 0.5 ×at²

here u is initial speed and t is time and S is distance

400 = 10 × 14 + 0.5 ×a (14)²

a = 3.37 m/s²

and here tangential acceleration is constant

so velocity at distance 200 m

v² - u² = 2 a S

v² = 10² + 2 ( 3.37) 200

v = 38.05 m/s

so radial acceleration at distance 200 m

ar = $\frac{v^2}{R}$

ar = $\frac{38.05^2}{63.66}$

ar = 22.74 m/s²

so magnitude of total acceleration is

A = $\sqrt{a^2 + ar^2}$

A = $\sqrt{3.37^2 + 22.74^2}$

A = 22.98 m/s²

so magnitude of acceleration is 22.98 m/s²

8 0

3 years ago

Because there is a one-to-many relationship between sales reps and customers in the TAL Distributors database, one sales rep can

stiv31 [10]

Answer:

True

Explanation:

The relationship between sales reps and customers is an example of a one-to-many relationship. This is because one sales rep can be associated with many customers but a customer must have one sales rep. It is impossible for a customer to have zero sales rep but this is quite possible for a sales rep to have either zero, one or even more customers. Therefore the statement is True.

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3 years ago

A length of pipe will weigh the most when

FinnZ [79.3K]

Answer:

Mark me please as brainliest

Explanation:

A length of pipe will weigh the most when O A. part dimensions are such that the pipe is in the MMC. O B. the OD and ID are at the minimum allowable limit. OC. the OD and ID are at the maximum allowable limit. OD.part dimensions are such that the pipe is in the LMC

4 0

3 years ago

Read 2 more answers

In Millikan's oil drop experiment, if the electricfield between the plates was of just the right magnitude, it wouldexactly bala

Vilka [71]

Answer:

The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb

Explanation:

In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, $|F_{electrostatic}| = |F_{gravitational}|$

where

$F_{electrostatic}=q.E$

where q is the charge of the droplet and E is the electric field. On the other hand

$F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g$

where m,V ,d and r are the mass, volume, density and radius of the oil droplet respectively and g is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,

$q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5} (N/C)$

(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))

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4 years ago