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3 years ago

How can you measure the distance an object has moved?

1 answer:

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.

You might be interested in

A 7.0-kg rock is subject to a variable force given by the equation f(x)=6.0n−(2.0n/m)x+(6.0n/m2)x2 if the rock initially is at r

stealth61 [152]

For Newton's second law, the force is equal to the product between the mass and the acceleration of the rocket:
$F=ma$
From which we can rewrite the acceleration as
$a(x)= \frac{F}{m} = \frac{1}{m} (6-2x+6x^2)$
where m=7.0 kg.

The velocity of the rocket is the derivative of the acceleration:
$v(x) = \frac{1}{m} (-2+12 x)$
and if we substitute x=9.0 m, we find the rocket velocity after 9.0 m:
$v(9)= \frac{1}{7}(-2+12\cdot 9)=15.1 m/s$

4 0

3 years ago

EMERGENCY! PLS HELP

Karolina [17]

Answer:

0.75 grams

Explanation:

Today we have 6 grams

In 30 years, 3 grams remain

In 60 years 1.5 grams remain

in 90 years 0.75 grams remain

mathematically It would look like this

m = 6(0.5⁽ⁿ/³⁰⁾) where n is number of years

5 0

2 years ago

A mass of 2000 kg. is raised 5.0 m in 10 seconds. What is the potential energy of the mass at this height?

kolbaska11 [484]

Answer:

98,000J

Explanation:

Given parameters :

Mass = 2000kg

Height = 5m

Time = 10s

P. E = mgh

P. E = 2000x 5x 9.8

P.E = 98,000J

5 0

2 years ago

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

7 0

3 years ago

When the bug is stationary and creating waves, how does the frequency of the wave some distance away from the bug compare with t

Brrunno [24]

Answer:

The frequency is the same

Explanation:

When a wave is created by a source which is vibrating at a certain frequency, the frequency of the wave itself is equal to the frequency of the source.

This occurs with every kind of wave. For instance, if we consider the radio waves produced by an antenna, the frequency of the radio waves is equal to the frequency of the antenna.

In this case, the waves are created by the vibrating bug. The bug is vibrating with a certain frequency $f$ : as a consequence, the frequency $f'$ of the waves produced by the bug will be equal to the frequency of vibration of the bug:

$f'=f$ .

3 0

2 years ago