Explain why L0 is not measured to point X on the spring?

me (5) A taxi is travelling at 15. m/s. Its driver accelerates with acceleration 3 m/s for 4 s. What is its new velocity? A car

photoshop1234 [79]

Answer:

5)

Solution

initial velocity = 15 m/s

acceleration = 3m/s^2

time = 4 s

final velocity = 15+3×4

=15+12

=27m/s

A car accelerates from 20m/s to 30m/s in 10 second

T9 be honest I think some part of the second question is missing

4 0

3 years ago

The average human walks at a speed of 5km per hour if your PE teacher asks you to walk for 30 minutes in gym class how far would

Leokris [45]

Answer:

2.5km

Explanation:

Assuming you also walk with the average human walking speed of 5 km/h. We know that speed=distance/time and now making distance the subject, distance is a product of speed and time. The time is 30 minutes, expressed in hours will be 30/60=0.5 hours. Substituting 5km/h for speed and 0.5 hours for time then the diatance covered will be

Distance=5*0.5=2.5 km

6 0

3 years ago

How far must you be from a 110 W speaker to have an intensity of 0.0439 W/m^2? (Treat the speaker as a point source.)

Lina20 [59]

Answer:

$d=14.12m$

Explanation:

Assume a point sound source that emits a sound power P (in W) evenly in all directions of space. Let us also assume that the medium does not absorb this sound power when it passes through it. At a distance d from the source this power will have been evenly distributed over the surface of a sphere of radius d. Therefore, the acoustic intensity I at distance d will be worth:

$I=\frac{P}{4\pi d^{2}}$

This is the expression of the so-called law of the square of distance: "the intensity is inversely proportional to the square of the distance to the source (considered punctual)".

So

$I=0.0439 \frac{W}{ m^{2}}$

$P=110W$

$d=\sqrt { \frac{P}{4 \pi\\I}$

$d=\sqrt{ \frac{110}{ 4\pi (0,0439)} }$

$d=\sqrt{{199.39}$

$d=14.12m$

3 0

3 years ago

What is Newton's second law of motion

Alexandra [31]

Newton’s second law is that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

8 0

3 years ago

12*3a car is stopped at a traffic light. it then travels along a straight road so that its distance from the light is given by w

Digiron [165]

Missing parts in the text of the exercise:
- The distance from the traffic light is given by $x(t) = bt^2 -ct^3$ , where $b=2.4 m/s^2$ and $c=0.13 m/s^3$

Solution:

part a) calculate the average velocity of the car for the time interval t=0 to t=8.0 s
- The average velocity is given by the ratio between the distance covered in the time interval:
 $v_{ave} = \frac{x(8.0 s)-x(0 s)}{8.0 s-0 s}$
x(0 s), the distance covered after t=0 s, is zero, while the distance after t=8.0 s is
$x(8.0 s)=b(8 s)^2-c(8 s)^3 = (2.4 m/s^2)(8 s)^2 -(0.13 m/s^3)(8 s)^3=$
$=87.04 m$
Therefore, the average velocity is
$v_{ave}= \frac{x(8.0 s)-x(0)}{8.0s-0}= \frac{87.04 m}{8.0s}= 10.88 m/s$

part b) calculate the instantaneous velocity of the car at t=0, t=4.0 s and t=8.0 s
- The instantaneous velocity can be found by performing the derivation of x(t):
 $v(t) = \frac{dx(t)}{dt}=2bt-3ct^2$
So now we just have to substitute t=0, t=4 s and t=8 s:
- t=0: v(0)=0
- t=4 s: $v(4.0 s)=2(2.4 m/s^2)(4.0 s)-3(0.13 m/s^2)(4.0s)^2=12.96 m/s$
- t=8 s: $v(8.0 s)=2(2.4 m/s^2)(8.0 s)-3(0.13 m/s^2)(8.0s)^2=13.44 m/s$

part c) how long after starting from rest is the car again at rest?
- To solve this part we must find the value of t for which v(t)=0, so:
 $2bt-3ct^2=0$
$t(2b-3ct)=0$
The first solution is t=0 s, which corresponds to the beginning of the motion, so we are not interested in this value. The second solution is
 $t= \frac{2b}{3c}= \frac{2\cdot 2.4 m/s^2}{3 \cdot 0.13 m/s^3}=12.31 s$
and this is the time at which the car is at rest again.

8 0

4 years ago