A bowling ball with a mass of 7.0 kg, generates 56.0 kgᐧm/s units of momentum. What

What is the voltage in a circuit if the current is 6.2 A and the resistance is 18 ohms? A) 2.9 V B) 11.8 V C) 24.2 V D) 111.6 V

maks197457 [2]

V = I · R = (6.2 A) · (18 Ω) = 111.6 volts (D)

7 0

2 years ago

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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

3 0

3 years ago

Which of the following CANNOT be

Nezavi [6.7K]

Answer:

Well, there is a kind of magnet to pick up a coin.. I think you can pick up a needle with one too.. I think safety pins. depending on what its made of though.

Explanation:

7 0

2 years ago

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Someone help

AURORKA [14]

Answer:

Explanation:

No 1 is linked to the option provided at no 5.

A direct object is the noun that follows the verb and answers the question.

No 2 is linked to the option provided at no 8.

A verb is a word that expresses action.

No 3 is linked to the option provided at no 5.

The subject is what or whom the sentence is about.

No 5 is linked to the option provided at no 3.

A predicate noun follows a linking verb, and renames the subject.

No 6 is linked to the option provided at no 4.

A predicate adjective follows a linking verb, and describes a subject.

No 7 is linked to the option provided at no 1.

A linking verb joins a subject and a predicate.

No 8 is linked to the option provided at no 2.

A sentence expresses a complete thought....

5 0

2 years ago

David moves a 350 N bucket with 175 N of force using a pulley. What mechanical advantage does he have when using the pulley?

NikAS [45]

Answer:

the mechanical advantage David had when using the pulley is 2.

Explanation:

Given;

load moved by David, L = 350 N

effort applied by David, E = 175 N

The mechanical advantage David had when using the pulley is calculated as;

$M.A = \frac{Load}{Effort} \\\\M.A = \frac{350}{175} \\\\M.A = 2$

Therefore, the mechanical advantage David had when using the pulley is 2.

7 0

2 years ago