To discover the laws of reflection it is necessary to use a ...<br> 1 point

An engineer wishes to design a curved exit ramp for a toll road in such a way that a car will not have to rely on friction to ro

Dafna11 [192]

To solve this problem we will make a graph that allows us to understand the components acting on the body. In this way we will have the centripetal Force and the Force by gravity generating a total component. If we take both forces and get the trigonometric ratio of the tangent we would have the angle is,

$T_x = nsinA = \frac{mv^2}{r}$

$T_y = ncosA = mg$

Dividing both.

$tan A = \frac{v^2}{rg}$

$tan A = \frac{11.7^2}{50*9.8}$

$A = tan^{-1} (0.279367)$

$A = 15.608\°$

Therefore the angle that should the curve be banked is 15.608°

7 0

4 years ago

How can you prove the mechanical nature of sound by a simple experiment

Kruka [31]

Answer:

That is, mechanical waves cannot travel through a vacuum. This feature of mechanical waves is often demonstrated in a Physics class. A ringing bell is placed in a jar and air inside the jar is evacuated. Once air is removed from the jar, the sound of the ringing bell can no longer be heard.

6 0

3 years ago

(b) The distance of mass from mass A if there is no gravitational force acted on C

shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

$F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}$

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ = 6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ = 6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

$F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}$

By plugging in the values and removing like terms, we get;

$\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}} = \dfrac{2 \times 0.05}{x^2}$

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0

3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

Olin [163]

The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).

There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction: $\mu m g$ , where $\mu=0.25$ is the coefficient of friction and $m=30.0 kg$ is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
$F=\mu m g=(0.25)(30.0 kg)(9.81 m/s^2)=73.8 N$
And so, this is the force that the worker must apply to the crate.

5 0

3 years ago

Sound waves are a- Transverse b- Longitudinal c- Transverse and longitudinal

Katen [24]

Answer:

Sound waves are longitudinal in nature.

Explanation:

There are many types of waves like transverse, longitudinal, electromagnetic wave etc.

Sound waves are longitudinal in nature. In longitudinal type of wave, the medium particles moves parallel to the propagation of the wave. This type of waves move in the form of compression and rarefaction.

In compression, the particle density at a point is very less while in rarefaction, the particle density at a point is very high.

So, the correct option is (b) "longitudinal wave".

3 0

3 years ago

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To discover the laws of reflection it is necessary to use a ... 1 point