Ask question

Ask question

All categories

2 years ago

11

Which level of acceleration should you use when accelerating on a short highway entry ramp?

1 answer:

ahrayia [7]2 years ago

5 0

The lowest level of acceleration should be use when accelerating on a short highway entry ramp.

<h3>What is an acceleration?</h3>

This means the rate at which velocity changes with time in terms of both the speed and direction.

In a machine, this means rate at which a vehicle can increase its speed in terms of the time that it takes to reach a particular speed.

Because the short highway entry ramp need careful driving, then, the lowest level of acceleration should be use when accelerating on a short highway entry ramp.

Read more about acceleration

brainly.com/question/460763

#SPJ1

You might be interested in

with a digital system, if you have measured incorrectly and use too low of a kvp for adequate penetration, what do you need to d

Lubov Fominskaja [6]

The x-ray beam's penetrating power is regulated by kVp (beam quality). Every time an exposure is conducted, the x-rays need to be powerful (enough) to sufficiently penetrate through the target area.

<h3>How does kVp impact the exposure to digital receptors?</h3>

The radiation's penetration power and exposure to the image receptor both increase as the kVp value is raised.

<h3>Exposure to the image receptor is enhanced with an increase in kVp, right?</h3>

Due to an increase in photon quantity and penetrability, exposure at the image receptor rises by a factor of five of the change in kVp, doubling the intensity at the detector with a 15% change in kVp.

To know more about kVp visit:-

brainly.com/question/17095191

#SPJ4

5 0

1 year ago

It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d

denis23 [38]

The exit temperature is 586.18K and compressor input power is 14973.53kW

Data;

Mass = 50kg/s
T = 288.2K
P1 = 1atm
P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\$

Let's substitute the values into the formula

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K$

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

$P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW$

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

brainly.com/question/10121263

6 0

2 years ago

A car accelerates from rest with an acceleration of 5 m/s^2. The acceleration decreases linearly with time to zero in 15 s, afte

Tpy6a [65]

Answer: At time 18.33 seconds it will have moved 500 meters.

Explanation:

Since the acceleration of the car is a linear function of time it can be written as a function of time as

$a(t)=5(1-\frac{t}{15})$

$a=\frac{d^{2}x}{dt^{2}}\\\\\therefore \frac{d^{2}x}{dt^{2}}=5(1-\frac{t}{15})$

Integrating both sides we get

$\int \frac{d^{2}x}{dt^{2}}dt=\int 5(1-\frac{t}{15})dt\\\\\frac{dx}{dt}=v=5t-\frac{5t^{2}}{30}+c$

Now since car starts from rest thus at time t = 0 ; v=0 thus c=0

again integrating with respect to time we get

$\int \frac{dx}{dt}dt=\int (5t-\frac{5t^{2}}{30})dt\\\\x(t)=\frac{5t^{2}}{2}-\frac{5t^{3}}{90}+D$

Now let us assume that car starts from origin thus D=0

thus in the first 15 seconds it covers a distance of

$x(15)=2.5\times 15^{2}-\farc{15^{3}}{18}=375m$

Thus the remaining 125 meters will be covered with a constant speed of

$v(15)=5\times 15-\frac{15^{2}}{6}=37.5m/s$

in time equalling $t_{2}=\frac{125}{37.5}=3.33seconds$

Thus the total time it requires equals 15+3.33 seconds

t=18.33 seconds

3 0

2 years ago

What is the ratio between driver gear A with 60 teeth and driven gear B with 180 teeth?

Anna71 [15]

The ratio between a and b is 1/3

3 0

3 years ago

What causes a boat to float? My questions are super easy. So you can follow me if you like easy question.

marin [14]

Boats float because the gravity is acting down on it and the buoyant force is acting up on the ship.

6 0

3 years ago

Read 2 more answers