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Lera25 [3.4K]
3 years ago
6

PLEASE HELP. NOTICE HOW THEY PUT FUTURE

Physics
1 answer:
MrMuchimi3 years ago
6 0

I think its <u>B</u> because it looks like it might indicate a future rain storm.

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A 5.0-kg box has an acceleration of 2.0 m/s2 when it is pulled by a horizontal force across a surface with μK = 0.50. Find the w
Maslowich

Answer:a) 34.5 N; b) 24.5 N; c) 10 N; d) 1J

Explanation: In order to solve this problem we have to used the second Newton law given by:

∑F= m*a

F-f=m*a where f is the friction force (uk*Normal), from this we have

F= m*a+f=5 Kg*2 m/s^2+0.5*5Kg*9.8 m/s^2= 34.5 N

then f=uk*N=0.5*5Kg*9.8 m/s^2= 24.5N

the net Force = (34.5-24.5)N= 10 N

Finally the work done by the net force is equal to kinetic energy change so

W=∫Force net*dr= 10 N* 0.1 m= 1J

7 0
3 years ago
Read 2 more answers
Help me someone ?...
Burka [1]
Gradpoint? The answer is B
8 0
3 years ago
In which one of the following circumstances could mechanical energy not possibly be conserved, even if friction and air resistan
qwelly [4]

Answer:

A car moves up a hill at a constant velocity

Explanation:

Since the velocity is constant, the speed is also constant and so is the kinetic energy. However, total mechanical energy is sum of gravitational potential energy and kinetic energy, and the car is moving up the hill so its potential energy rises.

Thus, in the circumstances described the mechanical energy cannot be conserved.

The correct answer is A car moving up the hill with constant velocity.

5 0
2 years ago
Find the total displacement of a mouse that travels 1.0 m north and then 0.8 m south.
andreev551 [17]
North: 1m
South: 0,8m

Direction:

1m>0,8m 

so mouse moved north

Distance:

1m-0,8m=0,2m

so mouse traveled 0,2m


Answer: The mouse moved 0,2m north.



"Non nobis Domine, non nobis, sed Nomini tuo da gloriam."



Regards M.Y.





3 0
2 years ago
A and B start walking in the same direction at the same time around a circular park of diameter 4200 m. If A walks 10 m more tha
Liono4ka [1.6K]

Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π      where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π  time for A to again catch B

N = 1.39 * 24700 Answer:

Va = 5000 m / 3600 s = 1.39 m/s

(Va - Vb) 60 = 10

Vb = Va - .167 = 1.22 m/s

(Va - Vb) T = 4200 Π      where T is time for A to complete 1 more lap

.17 T = 4200 Π

T = 24700 Π  time for A to again catch B

N = 1.39 * 24700 Π / (4200 Π) = 8.2   laps

A will make 8 but not 9 rounds before catching B

6 0
1 year ago
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