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umka21 [38]
3 years ago
12

A person walks first at a constant speed of 5.10 m/s along a straight line from point circled a to point circled b and then back

along the line from circled b to circled a at a constant speed of 2.75 m/s. (a) what is her average speed over the entire trip? 3.57 correct: your answer is correct. m/s
Physics
1 answer:
Alika [10]3 years ago
4 0
Average speed = total distance / time

1) Distance from A to B

Distance = velocity * time  = V1 * t1 = 5.10 t1

2) Distance from B to A

Distance = velocity * time  = V2*t 2 = 2.75 t2

Distance from A to B = distance from B to A => 5.10t1 = 2.75t2

=> t2 = 5.10t1 / 2.75

3) Average speed = total distance / total time =

total distance = 2 * 5.10 t1 = 10.20 t1

Average speed = [10.20 t1] / [t1 + 5.10 t1 / 2.75]

As you see t1 is factor for the three terms, so you can simplify it

=> Average speed = [10.20 ] / [1 + 5.10/2.75] = 3.57 m/s

Answer: 3.57 m/s

 


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A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​
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5 m/s

Explanation:

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A vehicle is moving with 20m/s towards the east and another is moving 15m/s towards the west​.

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v=v_1+v_2\\\\=20+(-15)\\\\=5\ m/s

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2 years ago
Which statement is correct? (2 points)
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7 0
3 years ago
An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os
aniked [119]

Answer:

(a)  T = 0.412s

(b)  f = 2.42Hz

(c)  w = 15.25 rad/s

(d)  k = 86.75N/m

(e)  vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz

The frequency is 2.42 Hz

(c) The angular frequency is:

\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}

The spring constant is 86.75N/m

(e) The maximum speed is:

v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

F=kA=(86.75N/m)(0.2m)=17.35N

The maximum force that the spring exerts on the block is 17.35N

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3 years ago
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