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SVEN [57.7K]
3 years ago
11

How can you measure the distance an object has moved?

Physics
1 answer:
Naily [24]3 years ago
8 0

You must observe the object twice.

-- Look at it the first time, and make a mark where it is.

-- After some time has passed, look at the object again, and
make another mark at the place where it is.

-- At your convenience, take out your ruler, and measure the
distance between the two marks.

What you'll have is the object's "displacement" during that period
of time ... the distance between the start-point and end-point. 
Technically, you won't know the actual distance it has traveled
during that time, because you don't know the route it took.


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A force of 250 N is applied to a 1 kg softball when struck with a bat. what is the acceleration
My name is Ann [436]
A=f/m
a=250N/1kg
a=250m/s^2
8 0
4 years ago
Find the minor measurement of the vernier scale by taking 49, 1mm divisions of the main scale and dividing it into 50 vernier di
olga nikolaevna [1]
<h2>♨ANSWER♥</h2>

length of V-50 = 49mm

length of V-1 = 49/50mm

= 0.98mm

so,

minor measurement = (M-1) - (V-1)

= 1mm -0.98mm

= 0.02mm

☆ Therefore,

The minor measurement of the vernier scale is 0.02mm.

<u>☆</u><u>.</u><u>.</u><u>.</u><u>hope this helps</u><u>.</u><u>.</u><u>.</u><u>☆</u>

_♡_<em>mashi</em>_♡_

6 0
2 years ago
A sound wave has a speed of 345 m/s and a wavelength of 4.15 meters. What is the frequency of this wave?
Debora [2.8K]

Answer:

83 hertz

OPTION A

Explanation:

wavelength =  \frac{speed}{frequency}

Given

wavelength=4.15m

speed=345 m/s

frequency =  \frac{speed}{wavelength} =  \frac{345}{4.15}

frequency = 83 \: hertz

I hope it helped you

5 0
3 years ago
A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring
sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0
3 years ago
Which pair of graphs represent the same motion?
GaryK [48]
Show me the graphs and i would be glad to help u
7 0
4 years ago
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