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andrew-mc [135]
3 years ago
6

Suponga que usted pone en contacto 2 cuerpos, uno con carga +6 y otro con carga -8. ¿Con qué carga queda cada uno al final?

Physics
1 answer:
soldi70 [24.7K]3 years ago
4 0
What is the question?
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The Sun being the largest body has the highest gravitational pull on all planets however planets do not collapse into the sun th
mojhsa [17]

Answer:

A:gravitational of the earth

7 0
2 years ago
How are credit unions different from banks?
aleksandrvk [35]
Answer: B ≈ Credit unions are owned by stockholders rather than partners
7 0
3 years ago
A force pointing in the xx-direction is given by F=ax3/2F=ax3/2, where aa is a constant. The force does 2.01 kJkJ of work on an
damaskus [11]

Answer:

Explanation:

Given that

F=ax^3/2. a is a constant

The force does a work of

W=2.01KJ from x=0 to x=15.2m

We need to find a

Work is give as,

W=∫F.ds

But this is in x direction only then,

W=∫Fdx. from x=0 to x=15.2m

W=∫ax^3/2dx from x=0 to x=15.2m

W=ax^(3/2+1)/(3/2+1).

W=ax^(5/2)/5/2

W=ax^(2/5)/2.5 from x=0 to x=15.2m

Cross multiply

2.5W=ax^2.5. from x=0 to x=15.2m

2.5W= a (15.2^2.5-0)

W=2.01KJ=2010J

2.5×2010=a×900.76

Therefore,

a=5.56

7 0
2 years ago
A skater increases her speed from 5.0 m/s to 10 m/s in 5 seconds. What is the acceleration of the skater?
forsale [732]

Answer:

1 m/s^2

Explanation:

The acceleration of an object is the rate of change of velocity of the object; it is given by:

a=\frac{v-u}{t}

where

u is the initial velocity of the body

v is its final velocity

t is the time elapsed

In this problem, we have:

u = 5.0 m/s is the initial velocity of the skater

v = 10.0 m/s is the final velocity

t = 5 s is the time elapsed

Therefore, the acceleration of the skater is:

a=\frac{10-5}{5}=1 m/s^2

8 0
3 years ago
A man on a road trip drives a car at different constant speeds over several legs of the trip. He drives for 15.0 min at 50.0 km/
JulsSmile [24]

Answer

given

Case 1-Speed of car 50 Km/h for 15 min

Case 2-Speed of car 85 Km/h for 19 min

Case 3 - Speed of car 60 km/h for 55 min

lunch break time = 25 min

a) total distance travel

   we know,

    distance = speed x time

 D = 50 \times \dfrac{15}{60} + 85 \times \dfrac{19}{60} + 60\times \dfrac{55}{60}

  D = 94.42 Km

b) average\ speed = \dfrac{total\ distance}{total\ time}

   average\ speed = \dfrac{94.42}{\dfrac{15}{60}+\dfrac{19}{60}+\dfrac{55}{60}+\dfrac{25}{60}}

average\ speed = 49.69\ km/h

4 0
3 years ago
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