if two displacement vectors add to yeild a totsl displacement of zero, what do you know about the two displacements

Usimov [2.4K]

Answer:heat brings it up then down

Explanation:

4 0

3 years ago

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

In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so

svetlana [45]

Answer:

(a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

Explanation:

Given that,

Radius of the circle = 0.681 m

Angular acceleration = 67.7 rad/s²

Angular speed =18.6 rad/s

We need to calculate the centripetal acceleration of the ball

Using formula of centripetal acceleration

$a_{c}=\omega^2\times r$

Put the value into the formula

$a_{c}=(18.6)^2\times0.681$

$a_{c}=235.5\ m/s^2$

We need to calculate the tangential acceleration of the ball

Using formula of tangential acceleration

$a_{t}=r\alpha$

Put the value into the formula

$a_{t}=0.681\times67.7$

$a_{t}=46.104\ m/s^2$

(a). We need to calculate the magnitude of the total acceleration of the ball

Using formula of total acceleration

$a=\sqrt{a_{c}^2+a_{t}^2}$

Put the value into the formula

$a=\sqrt{(235.5)^2+(46.104)^2}$

$a=239.97\ m/s^2$

(b). We need to calculate the angle of the total acceleration relative to the radial direction

Using formula of the direction

$\theat=\tan^{-1}(\dfrac{a_{t}}{a_{c}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{46.104}{235.5})$

$\theta=11.0^{\circ}$

Hence, (a). The magnitude of the total acceleration of the ball is 239.97 m/s².

(b). The angle of the total acceleration relative to the radial direction is 11.0°

5 0

3 years ago

A radio station's channel, such as 100.7 fm or 92.3 fm, is actually its frequency in megahertz (mhz), where 1mhz=106hz. calculat

lakkis [162]

When it comes to wave behavior, there are parameters called wavelength and frequency. These two are related by speed of the radiowave. Radiowaves are electromagnetic waves which travels as fast as light. The wavelength is the distance while frequency is the reciprocal of time. When you multiply them both, you get the electromagnetic wave's speed. The equation is c = wavelength*frequency, where c is the speed of light equal to 3 x 10^8 m/s.

3 x10^8 m/s = wavelength/104.9 x 10^6 Hz (Hertz is 1/s)
wavelength = 2.86 meters

6 0

3 years ago

What is a consequence of the second law of thermodynamics?

marshall27 [118]

Answer:Increase in the entropy of the universe are the result of chemical reactions

Explanation:

The second law tells about the quality of energy such that any isolated system tends to become more disordered. Any natural Process occurring on its own is irreversible and entropy is increasing for that process.

The entropy of the Universe is increasing which is the result of chemical reactions. Any chemical reaction which increases the number of gas molecules will generate more entropy. Entropy in simple meaning is the order of randomness.

4 0

3 years ago

if two displacement vectors add to yeild a totsl displacement of zero, what do you know about the two displacements​

if two displacement vectors add to yeild a totsl displacement of zero, what do you know about the two displacements