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3 years ago

If you travel at 3 m/s for 12 seconds, how far did you travel?

Physics

2 answers:

Rom4ik [11]3 years ago

8 0

Answer:

$\boxed {\tt 36 \ meters }$

Explanation:

We want to find how far was traveled, or the <u>distance.</u>

The formula for distance is:

$d=s*t$

where $s$ is the speed and $t$ is the time.

The speed is 3 meters per second and the time is 12 seconds.

$s= 3 \ m/s \\t= 12 \ s$

Substitute the values into the formula.

$d= 3 \ m/s * 12 \ s$

Multiply. Note that when multiply, the seconds, or $s$ will cancel.

$d= 3 \ m * 12$

$d= 36 \ m$

$d= 36 \ meters$

The distance traveled was 36 meters.

ella [17]3 years ago

4 0

Answer:

V=3 m/s

t=12 seconds

S=?

S=V×t

S=3×12

S=36meters

So distance you travel is 36meters.

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3 years ago

A charge of +3.00 μC is located at the origin, and a second charge of -2.00 μC is located on the x-y plane at the point (50.0 cm

romanna [79]

Answer:

0.0567 N

Explanation:

q1 = 3 micro coulomb

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AB = 60 cm

By using Pythagoras theorem in triangle OAB

$OA^{2}=OB^{2}+AB^{2}$

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$F=\frac {Kq_{1}q_{2}}{OA^{2}}= \frac{9 \times 10^{9}}\times 3\times 10^{-6} \times 2 \times 10^{-6}{0.781^{2}}$

F = 0.0885 N

The horizontal component of force is

= F CosФ = $F\times \frac{OB}{OA}$

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3 years ago

a steam engine work on its vicinity 285 k heat is released with the help of 225 c energy absorbed to the system what is the effi

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The efficiency of the steam engine is 78.9% because the rest of the work input is used to overcome friction.

<h3>What is efficiency of machines?</h3>

Efficiency of a machine expresses the useful work done by a a machine as a percentage.

Efficiency of a machine = work output/work input × 100 %%

The efficiency of machines are always less than 100% percent due to energy losses due to friction and heat.

For the steam engine:

Work output = 225 J

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Efficiency = 225/285 × 100% = 78.9 %

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Learn more about efficiency of machines at: brainly.com/question/7536036

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1 year ago

The unit light-year is a measure of

stiv31 [10]

A light-year is a unit of distance.

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3 years ago

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A hungry 169169 kg lion running northward at 77.377.3 km/hr attacks and holds onto a 31.731.7 kg Thomson's gazelle running eastw

navik [9.2K]

Answer: 75,242.9 m/s

Explanation:

from the question we are given the following parameters

mass of Lion (ML) = 169,169 kg

velocity of lion (VL) = 777,377.7 m/s

mass of Gazelle (Mg) = 31,731.7 kg

velocity of Gazelle (Vg) = 63,863.8 kg

mass of Lion and Gazelle (M) = 200,900.7 kg

velocity of Lion and Gazelle (V) = ?

The first figure below shows the motion of the Lion and Gazelle with their direction.

The second diagram shows the motion of the Lion and Gazelle with their directions rearranged to form a right angle triangle.

from the triangle formed we can get the velocity of the Lion and Gazelle immediately after collision using their momentum and Phytaghoras theorem

momentum = mass x velocity

momentum of the Lion = 169,169 x 77,377.3 = 13,089,840,463.7 kgm/s

momentum of the Gazelle = 31,731.7 x 63,863.8 = 2,026,506,942.46 kgm/s

momentum of the Lion and Gazelle = 200,900.7 x V

now applying Phytaghoras theorem we have

13,089,840,463.7 + 2,026,506,942.46 = 200,900.7 x V

15,116,347,406.16 = 200,900.7 x V

V = 75,242.9 m/s

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2 years ago

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