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elena55 [62]
3 years ago
15

Describe the relationship of attractive forces between molecules

Physics
1 answer:
Ede4ka [16]3 years ago
4 0
<span>Describe the relationship of attractive forces between molecules and the ability of a solvent to dissolve a substance. Solvents can dissolve a substance only if the attraction of the solvent molecules is greater than the attraction between the molecules of the substance.</span>
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5. If one object has a greater speed than a second object. does the first necessarily have a greater acceleration? Explain, usin
Sveta_85 [38]

Answer:

Explanation:

5. not necessarily so that the first object could have left with initial velocity and the second not, so even if the second has a greater acceleration its velocity is less than that of the first

6. The acceleration of the motorcycle is

     SI System Reductions

     Vo = 80 km / h (1000m / 1km) (1h / 3600s) = 22.2 m / s

     Vf = 90 km / h (1000m / 1km) (1h / 3600s) = 25 m / s

     Vf = Vo + at at = Vf-Vo

     am = (Vf-Vo) / t

     am = (25 -22.2) / t = 2.8 / t

      am= 2.8/t

For the bike we have

      Vf = 10 km / h (1000m / 1km) (1h / 3600s) = 2.78 m / s

      Vo = 0

      ab = (Vf -Vo) / t

      ab = (2.78 -0) / t

      ab = 2.8/t

Since time is the same for both of us, if we round to Significant figures the two accelerations are equal

7. If when an object is slowing or slowing down.

     For example, a car goes north and must stop at the traffic light, the acceleration of the brakes goes south

8. Yes, since an object can go to the left and the acceleration to the right, but the object will lose speed over time

9. in the launch of projectiles the acceleration is negative and the speed after half the path is also negative

10. Car B must be moving to car A, because if they leave together B has more acceleration, bone that travels the distance at the same time

11. When we have friction, the velocity of an object increases by an external force, but the friction also increases the acceleration, but since it is positive, the velocity increases until the acceleration is zero and hence the velocity remains constant.

8 0
4 years ago
A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w
nignag [31]

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

  • h is height of the cliff
  • v is velocity
  • t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

  • vyf is the final vertical velocity
  • vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

  • vxf is the final horizontal velocity
  • vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

  • vf is the speed of the projectile when it strikes the ground below

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1  

3 0
2 years ago
Ben helped his parents load boxes and furniture onto the moving truck. He easily carried several boxes out of the house and up a
NikAS [45]
D) because the refrigerator was much heavier than the other objects
5 0
4 years ago
Read 2 more answers
Is O2 considered one atom or 2 atoms?
natulia [17]
O2 is considered 2 atoms because O2 is 2 oxygen atoms.

4 0
3 years ago
Read 2 more answers
Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original v
Gelneren [198K]

Answer:

F' = F/12

Therefore, the electrostatic force is reduced to one-twelve of its original value.

Explanation:

The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:

F = kq₁q₂/r²   ---------- equation 1

where,

F = Electrostatic Force

k = Coulomb's Constant

q₁ = magnitude of first charge

q₂ = magnitude of 2nd charge

r = distance between charges

Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:

F' = kq₁'q₂'/r'²

where,

q₁' = (1/3)q₁

q₂' = q₂

r' = 2r

Therefore,

F' = k(1/3 q₁)(q₂)/(2r)²

F' = (1/12)kq₁q₂/r²

using equation 1:

<u>F' = F/12</u>

<u>Therefore, the electrostatic force is reduced to one-twelve of its original value.</u>

6 0
3 years ago
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