What would happen if there were no action-reaction forces in effect and you tried walking down a sidewalk

The slope of the line on any speed vs time graph is equal to the objects ?

vivado [14]

i think its a bar graph

i hope this works

3 0

3 years ago

g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7

bogdanovich [222]

Answer:

When they are approaching each other

$f_a = 2228.7 \ Hz$

When they are passing each other

$f_a = 2100Hz$

When they are retreating from each other

$f_a = 1980.7 Hz$

Explanation:

From the question we are told that

The velocity of car one is $v_1 = 13.0 m/s$

The velocity of car two is $v_2 = 7.22 m/s$

The frequency of sound from car one is $f_e = 2.10 kHz$

Generally the speed of sound at normal temperature is $v = 343 m/s$

Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as

$f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ]$

Where $v_s$ is the velocity of the source of sound

$v_o$ is the velocity of the observer of the sound

$f_o$ is the actual frequence

$f_a$ is the apparent frequency

Considering the case when they are approaching each other

$f_a = f_o [\frac{v + v_o}{v - v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ]$

$f_a = 2228.7 \ Hz$

Considering the case when they are passing each other

At that instant

$v_o = v_s = 0m/s$

$f_o = f_e$

$f_a = f_o [\frac{v }{v } ]$

$f_a = f_o$

Substituting value

$f_a = 2100Hz$

Considering the case when they are retreating from each other

$f_a = f_o [\frac{v - v_o}{v + v_s} ]$

$v_o = v_2$

$v_s = v_1$

$f_o = f_e$

Substituting value

$f_a = 2100 [\frac{343 - 7.22}{343 + 13} ]$

$f_a = 1980.7 Hz$

7 0

3 years ago

Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One o

Zarrin [17]

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

qA = ½ q

qB = ½ q

qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

qA = 1/2 (q / 2) = ¼ q

qC = ¼ q

qB = ½ q

At this point all spheres have a charge,

qA = ¼ q

qb = ½ q

qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

Q = ½ q + ¼ q = ¾ q

When splitting the charge

qB = ½ ¾ q = 3/8 q

qC = ½ ¾ q = 3/8 q

qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

Q = qA + QB + QC

Q = ½ q + 3/8 q + ¼ q

Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

8 0

3 years ago

A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Tems11 [23]

Answer:

a) $V_{wf}$ = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

$m_{b}V{b_{i} } + V_{wi} = m_{w} V_{wf} + m_{b}V_{bf}$

where $m_{b},V{b_{i}$ are the mass and the initial velocity of the bullet, $m_{w}$ and $V_{wi}$ are the mass and the initial velocity of the wooden block, and $V_{wf}$ and $V_{bf}$ are the final velocities of the wooden block and the bullet