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monitta
3 years ago
8

Hello, what does the first condition mean?

Physics
2 answers:
Effectus [21]3 years ago
6 0

Answer:

see below

Explanation:

the first condition means that the light ray inside the optically denser medium hits the surface of the optically less dense medium.

this in simple words means, light must be travelling from am optically denser medium to an optically less dense medium

hope this helps :)

Naya [18.7K]3 years ago
4 0

Answer:

An object at rest is described by Newton's First Law of Motion. An object in static equilibrium has zero net force acting upon it. The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes.

Explanation:

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3 0
3 years ago
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What is the mass of a 3,500-N rock?
Viktor [21]
350kg because to get Newton’s it’s mass x Gravity, earths gravity is x10 so 3500 divided by 10 is 350
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3 years ago
What is the velocity of a 0.8kg ball that has a momentum of 3 kg*m/s?
Marysya12 [62]

The velocity is 3.75

5 0
3 years ago
A 1500kg car traveling at 25m/s skids to a stop. The force of friction between the tires and the road is 10500N. How far does th
ale4655 [162]

44.64m

Explanation:

Given parameters:

Mass of the car = 1500kg

Initial velocity = 25m/s

Frictional force = 10500N

Unknown:

Distance moved by the car after brake is applied = ?

Solution:

The frictional force is a force that  opposes motion of a body.

To solve this problem, we need to find the acceleration of the car. After this, we apply the appropriate motion equation to solve the problem.

     -Frictional force = m x a

the negative sign is because the frictional force is in the opposite direction

m is the mass of the car

 a is the acceleration of the car

    a = \frac{frictional force}{mass} = \frac{10500}{1500} = -7m/s²

Now using;

   V² = U² + 2as

   V is the final velocity

   U is the initial velocity

   a is the acceleration

   s is the distance moved

  0² = 25² + 2 x 7 x s

  0 = 625 - 14s

  -625 = -14s

      s = 44.64m

   

learn more:

Velocity problems brainly.com/question/10932946

#learnwithBrainly

7 0
3 years ago
A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for
Monica [59]

Question

A banked highway is designed for traffic moving at v 8 km/h. The radius of the curve = 330 m. 50% Part (a) Write an equation for the tangent of the highway's angle of banking. Give your equation in terms of the radius of curvature r, the intended speed of the turn v, and the acceleration due to gravity g

Part (b) what is the angle of banking of the highway? Give your answer in degrees

Answer:

a. Equation of Tangent

tan(θ) = v²/rg

b. Angle of the banking highway

θ = 0.087°

Explanation:

Given

Radius of the curve = r = 330m

Acceleration of gravity = g = 9.8m/s²

Velocity = v = 8km/h = 8 * 1000/3600

v = 2.22 m/s

a . Write an equation for the tangent of the highway's angle of banking

The Angle is calculated by

tan(θ) = v²/rg

θ = tan-1(v²/rg)

b.

Part (b) what is the angle of banking of the highway? Give your answer in degrees

θ = tan-1(v²/rg)

Substituting the values of v,g and r

θ = tan-1(2.22²/(330 * 9.8)

θ = tan-1(0.001523933209647)

θ = 0.087314873580116°

θ = 0.087°

3 0
3 years ago
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