What current is produced with a voltage of 6.0 V and a resistance of 445 ohms?17.4 mA

Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

3 years ago

How much time does it take a trucker to deliver his shipment 480 km away travelling at an average speed of 120 km/hr?

kifflom [539]

Answer: time=4 hours

Explanation:

formula:

speed = $\frac{distance}{time}$

time = $\frac{distance}{speed}$

time= x

speed = 120km/hr

distance = 480 km

x = $\frac{480 km}{120km/hr\\}$

= $\frac{48km*hr}{12km}$

cut the zeros

and remove the signs

then

$\frac{48}{12}$ = 4

therefore, time taken to deliver the shipment by trucker is 4 hours

8 0

2 years ago

What is the acceleration of a softball if it hits the ground with a force 0.50 kg and hits the catchers glove with a force of 25

Rom4ik [11]

Acceleration of the ball is $50 m/s^2$

Explanation:

The acceleration of the ball can be found by using Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:

$F=ma$

where

F is the net force

m is the mass

a is the acceleration

For the ball in this problem, we have

m = 0.50 kg (mass)

F = 25 N (force)

thereofre, the acceleration of the ball is

$a=\frac{F}{m}=\frac{25}{0.50}=50 m/s^2$

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

Calculate the speed of a car that travels 355km in 15 hours , a good answer please

Ludmilka [50]

Answer:

23.67 km / hr

Explanation:

car travels 355 km (d)

duration = 15 hrs (t)

average speed formula = v = d / t

v = 355 km / 15 hr

v = 23.67 km / hr

8 0

3 years ago

Read 2 more answers

The S-strain Pneumococcus bacteria had a smooth surface, because ____________________.

drek231 [11]

The S strain Pneumococcus bacteria had a smooth surface because IT IS SURROUNDED BY A CARBOHYDRATE CAPSULE CALLED THE S STRAIN. The other form, the R strain has a rough surface and no capsule. It is only the S strain that exhibits virulence.

4 0

3 years ago