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Brut [27]
3 years ago
14

Give two examples of contact and non-contact forces and explain why they are contact and non-contact forces respectively.​

Physics
1 answer:
Sever21 [200]3 years ago
3 0

The Two examples of contact forces are:

  • frictional force
  • Contact force.

The two examples of non contact forces are:

  • Gravitational force
  • magnetic force.

Contact forces happens due to the contact between two objects

Non Contact forces happens because there is no contact between two objects. There is no attraction.

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Why do you think it is a driving force of weather around the world?
Finger [1]

Answer:

The sun.

Explanation:

The sun provides energy for living organisms, and it drives our planet’s weather and climate patterns.

Remember, Earth is spherical and the energy from the sun does not reach all areas with equal intensity. Areas exposed to the sun are directly on the sun’s rays (i.e. those nearest to the equator) and hence, receive greater solar input. In contrast, those in higher latitudes receive sunlight that is spread over a larger area and that has taken a longer path through the atmosphere. As a result, these higher latitudes receive less solar energy.

Also, ocean circulation and precipitation are all factors of weather

4 0
3 years ago
When sound travels through air, the particles _____. a. vibrate along the direction the wave travels b. vibrate but not in any f
Mandarinka [93]
Vibrate along the direction the wave travels
8 0
4 years ago
a ball is thrown inclined in to the air with a initial velocity of u. if it reaches the maximum height in 6 seconds , find the r
olga55 [171]

Answer:

11:1

Explanation:

At constant acceleration, an object's position is:

y = y₀ + v₀ t + ½ at²

Given y₀ = 0, v₀ = u, and a = -g:

y = u t − ½g t²

After 6 seconds, the ball reaches the maximum height (v = 0).

v = at + v₀

0 = (-g)(6) + u

u = 6g

Substituting:

y = 6g t − ½g t²

The displacement between t=0 and t=1 is:

Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]

Δy = 6g − ½g

Δy = 5½g

The displacement between t=6 and t=7 is:

Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]

Δy = (42g − 24½g) − (36g − 18g)

Δy = 17½g − 18g

Δy = -½g

So the ratio of the distances traveled is:

(5½g) / (½g)

11 / 1

The ratio is 11:1.

8 0
4 years ago
A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa
gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

F-mg=ma

Solving for the force F, we obtain that:

F=ma+mg=m(a+g)

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0
4 years ago
If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?
HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
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