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Hatshy [7]
3 years ago
15

A dog is 60m away while moving at constant velocity of 10m/s towards you. Where is the dog after 4 seconds?

Physics
2 answers:
Kitty [74]3 years ago
6 0
The dog is 20m away.
if the dog moves 10m/s, multiply 10 by 4, subtract 40 from 60. 20
evablogger [386]3 years ago
4 0
20m away

the dog was 60m away from. you subtract 40m since it is 10m/s x 4 seconds
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What is the acceleration of Karla's I Phone being thrown from Mr. Higley's classroom at 0m/s if it hits the wall 1.2 seconds lat
blagie [28]

The acceleration of Karla's I IPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The pace at which a body's velocity varies is represented by acceleration, which is a vector quantity.

The given data in the problem is given by ;

u is the initial speed =  0 m/sec

v is the final speed= 35  m/sec

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a is the acceleration=? m/sec²

The formula for acceleration is;

\rm a=\frac{v-u}{t} \\\\ a= \frac{35-0}{1.2} \\\\ a= 29.16 \ m./s^2

Hence, the acceleration of Karla's iPhone being thrown from Mr. Higley's classroom at 0 m/s will be 29.16 m/s²

To learn more about acceleration, refer to the link;

brainly.com/question/2437624#SPJ2

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5 0
1 year ago
A 90 kg painter is standing on a horizontal wooden scaffolding of length 10 m, which is supported on each end by a rope. The pai
RSB [31]

Explanation:

For equilibrium, \sum M = 0.

So,   8 m \times mg - (10 m) T_{1} = 0

             T_{1} = \frac{8 \times mg}{10}

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                        = 705.6 N

Also, for equilibrium \sum F_{y} = 0

              T_{1} + T_{2} - mg = 0

or,         T_{2} = mg - T_{1}

                        = 90 \times 9.8 - 705.6

                        = 176.4 N

Thus, we can conclude that the tension in the first rope is 176.4 N.

8 0
3 years ago
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