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3 years ago

4. Identify Problems A student is describing a longitudinal wave in his notebook. He

Physics

1 answer:

andrew11 [14]3 years ago

3 0

Answer:

"The distance between crests is 3 cm."

Explanation:

If he writes down "The distance between crests is 3 cm."

That means he is describing the wavelength of a wave and not longitudinal wave. He ought to write something about " direction "

Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or parallel to, the direction of propagation of the wave. While

Wavelength is the distance between the two successfully Crest or trough

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a ball is rolled uphill a distance of 3 meters before it slows,stops,and begins to roll back. the ball rolls downhill 6 meters b

Leona [35]

The answer to your question is 3 meters

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3 years ago

1. A car with a mass of 1,500 kilograms is moving around a circular curve at a uniform velocity of 16 m/s. The curve has a radiu

svet-max [94.6K]

<u>Answers</u>

1) 3,840 N

2) 80 m

<u>Explanations</u>

Centripetal force is the force that maintenance objects that are moving in a circular motion.

F = mv²/r

= (1500 × 16²)/100

= (1500×256)/100

= 384000/100

= 3,840 N

2) F = (mv²)/r

6000 N = (1200 × 20²)/r

r = (1200 × 400)/6000

= 480,000/6000

= 80 m

5 0

4 years ago

Read 2 more answers

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

nlexa [21]

Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration $a_1 = 2.65~m/s^2$ and for a total time of $t_1=17~s$ . The body is initially at rest, so the distance covered is given by
$S= \frac{1}{2}a_1t_1^2=382.9~m$
Calling $v_f$ and $v_i$ the final and initial velocity, and since the $v_i=0~m/s$ because the body starts from rest, we can use
$a= \frac{v_f-v_i}{t}$
to find the final velocity after this first leg:
$v_{f}=v_i+a_1t_1=45~m/s$
And the average velocity in this first leg is
$v_1= \frac{v_f+v_i}{2}=22.5~m/s$

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: $v_2=45~m/s$ . This is also the average velocity of the second leg.
The total time of this second leg is $t_2=1.60~min = 96~s$ . The distance covered is given by
$S_2=v_2t_2=45~m/s \cdot 96~s=4320~m$

3) Uniformly decelerated motion, with constant deceleration of $a_3=-9.39~m/s^2$ and for a total time of $t_3=4.8~s$ . Here, the initial velocity of the body is the final velocity of the previous leg, i.e. $v_i=45~m/s$ . Therefore, the distance covered in this leg is given by
$S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m$
The final velocity in this leg is given by
$v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s$
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
$v_3 = \frac{1}{2}(v_f+v_i)= \frac{1}{2}(-0.07~m/s+45~m/s)= 22.5~m/s$

4) Finally, the total distance covered in the motion is
$S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m$
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
$v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s$

8 0

3 years ago

A pitcher throws a 0.143-kg baseball toward the batter so that it crosses home plate horizontally and has a speed of 42 m/s just

o-na [289]

Answer:

200N appox

Explanation:

Step one:

Given data

mass of ball= 0.143kg

initial velocity of ball u= 42m/s

final velocity of the ball= 49m/s

time of impact= 0.005s

Step two

From the relation Ft=mΔv

we can find the average force as

F=mΔv/t

substitute

F=0.143*(49-42)/0.005

F=0.143*7/0.005

F=1.001/0.005

F=200.2

F= 200N appox

The average force is 200N

5 0

3 years ago

A 7.2 magnitude earthquake, with an epicenter in Northern Mexico, was felt over 100 miles away in Southern California because wa

ryzh [129]

Answer:

because waves generated

7 0

1 year ago