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3 years ago

7

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

1 answer:

balandron [24]3 years ago

6 0

Some examples you could look up on-line so I don't have to:

Speed of sound in sea water . . . 1,531 m/s

Speed of sound in iron . . . 5,130 m/s

Speed of sound in air . . . 340 m/s

This gives us

fastest . . . medium . . . slowest

<em>solid . . . . . liquid . . . . . gas</em>

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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

Can someone help me with science

tamaranim1 [39]

The continuious physical force per unit area that one region of a gas, liquid, or solid exerts on another.

6 0

3 years ago

8. A vacuum tube can be used to

Gnoma [55]

answers to electronic devices quiz

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100% just took quiz

8 0

3 years ago

1. Two blocks travel along a level frictionless surface. Block A is initially moving to the right at 5.0 m/s, while block B is i

ad-work [718]

Answer: 2.67 m/s

Explanation:

Given

Mass of block A is $m_a=2\ kg$

mass of block B is $m_b=3\ kg$

The initial velocity of block A $u_a=5\ m/s$

the initial velocity of block B is $u_b=0$

After collision velocity of block A is $v_a=1\ m/s$

Conserving momentum

$m_au_a+m_bu_b=m_av_a+m_bv_b\\\\2\times 5+3\times0=2\times 1+3\times v_b\\\\v_b=\dfrac{8}{3}=2.67\ m/s$

The momentum of block A after the collision is $P_a=2\times 1=2\ kg.m/s$

Therefore, there is no change in sign.

6 0

3 years ago

If you walk 1.2 km north and then 1.6 km east, what are the magnitude and direction of your resultant displacement?

vodomira [7]

<u>Answer:</u>

Option A is the correct answer.

<u>Explanation:</u>

Let the east point towards positive X-axis and north point towards positive Y-axis.

First walking 1.2 km north, displacement = 1.2 j km

Secondly 1.6 km east, displacement = 1.6 i km

Total displacement = (1.6 i + 1.2 j) km

Magnitude = $\sqrt{1.2^2+1.6^2} = 2 km$

Angle of resultant with positive X - axis = $tan^{-1}(1.2/1.6)=36.87^0$ = 36.87⁰ east of north.

5 0

3 years ago