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wel
3 years ago
5

Distance is the length of a path followed by a particle. The displacement of a particle is defined as its change in position in

some time interval.
a. True
b. False
Physics
1 answer:
worty [1.4K]3 years ago
7 0

Answer:

a. True

Explanation:

Distance is described with only magnitude. It is defined as the total path covered by an object, in other words it  is the length of a path followed by a particle.

Displacement is described with both magnitude and direction. It is distance traveled in a specified direction or  change in position in some time interval.

Therefore, the correct option is " a. True"

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PLEASE HELP URGENT 10 points
melamori03 [73]

Answer:

Both have the same amount. C.

Explanation:

3 0
3 years ago
A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.600 rev
natulia [17]

Answer:

The final velocity \omega_f = 0.4235 \frac{rev}{s}

Explanation:

Given data

Mass of merry go round M_m = 120 kg

Radius = 1.8 m

Initial angular velocity \omega_i = 0.6 \frac{rev}{sec}

Mass of boy M_{boy} = 25 kg

We know that the final velocity is given by

\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }

Put all the values in above formula we get

\omega_f = \frac{\frac{1}{2}(120)  0.6}{25 + \frac{1}{2} (120) }

\omega_f = 0.4235 \frac{rev}{s}

This is the final velocity.

3 0
2 years ago
When an object is dropped from the top of a tall cliff, your teacher tells you to use -10 m/s2 for the approximate acceleration
joja [24]
The displacement should be a negative value.  since your zero point is where you dropped it.  The answer is C
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3 years ago
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(BRAINLIEST W/ WORK SHOWN!!!) Help with Physics question?
VashaNatasha [74]

Answer:

\boxed {3.43 x 10^{3}}

Explanation:

We know that speed is defined as distance moved per unit time hence expressed as v=\frac {d}{t} where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then

d=vt

Substituting 343 m/s for d and 10 s for t then

d= 343\times10= 3430= 3.43 x 10^{3}

Therefore, the distance between speaker and deter is \boxed {3.43 x 10^{3}}

6 0
3 years ago
If an object with a density 0.25 g/mL has a mass of 10 grams, and is added to a graduated cylinder filled with 50 mL of water, h
kramer

Answer : The correct option is, (d) 90 mL

Explanation :

First we have to calculate the volume of an object.

As we know that,

Density=\frac{Mass}{Volume}

Given:

Density of an object = 0.25 g/mL

Mass of an object = 10 g

Now put all the given values in the above formula, we get:

0.25g/mL=\frac{10g}{Volume}

Volume=40mL

Thus, the volume of an object is 40 mL.

Now we have to calculate the height of the water in the graduated cylinder rise.

As we are given that:

The volume of water in graduated cylinder = 50 mL

The volume of an object = 40 mL

The height of the water in the graduated cylinder rise = Volume of water in graduated cylinder + Volume of an object

The height of the water in the graduated cylinder rise = 50 mL + 40 mL

The height of the water in the graduated cylinder rise = 90 mL

Therefore, the height of the water in the graduated cylinder rise will be, 90 mL

8 0
3 years ago
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