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bogdanovich [222]
3 years ago
15

Examples of Laws Of Motion

Physics
1 answer:
Crazy boy [7]3 years ago
6 0

Answer:

Explanation:

1. The law that best illustrate this is Newton's first law of motion, which is also called the law of inertia

The law states that every object will continue in the sate of rest or in uniform motion except it is acted upon by an external force. The bike stop moving because it hits a curb.

Inertia: This is the reluctance or the tendency for a body to be in its state of rest or uniform motion.

And Maya is thrown into the air due to it inertia.

2. The Law that best illustrate this is Newton's  Third law of motion.

This law states that to every action there is an equal and opposite reaction.

As the bird uses his wind to push air downward, air pushes upward on the bird allows it to fly. the upward and downward movement of air are both equal but opposite reaction.

3. The Law that illustrate this is Newton's second law of motion.

This law can be summarize as force is directly proportional to mass and acceleration.

Because the mass of the second box is smaller than the first box that was why its accelerate at a greater rate.

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How many lenses with different focal lengths can be obtained by combining two surfaces whose radii of curvature are 4.00 cm and
Dovator [93]

Answer:

The lenses with different focal length are four.

Explanation:

Given that,

Radius of curvature R₁= 4

Radius of curvature R₂ = 8

We know ,

Refractive index of glass = 1.6

When, R₁= 4, R₂ = 8

We need to calculate the focal length of the lens

Using formula of focal length

\dfrac{1}{f}=(n-1)(\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}})

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}+\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{9}{40}

f=4.44\ cm

When , R₁= -4, R₂ = 8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}+\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{3}{40}

f=-13.33\ cm

When , R₁= 4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{4}-\dfrac{1}{8})

\dfrac{1}{f}=\dfrac{3}{40}

f=13.33\ cm

When , R₁= -4, R₂ = -8

Put the value into the formula

\dfrac{1}{f}=(1.6-1)(\dfrac{1}{-4}-\dfrac{1}{8})

\dfrac{1}{f}=-\dfrac{9}{40}

f=-4.44\ cm

Hence, The lenses with different focal length are four.

8 0
3 years ago
What parts of an atom determines which element it is?
uysha [10]

Explanation:

The protons determines the atomic number which helps in identifying an atom of an element

4 0
3 years ago
Read 2 more answers
A car travels 200km in 3.0 hours. Determine the average velocity of the car
Masja [62]

Answer:

<h2>66.67 km/hr</h2>

Explanation:

The average velocity of the car can be found by using the formula

a =  \frac{d}{t }  \\

d is the distance

t is the time taken

From the question we have

a =  \frac{200}{3}  \\  = 66.66666...

We have the final answer as

<h3>66.67 km/hr</h3>

Hope this helps you

4 0
3 years ago
Two simple pendulum of slightly different length , are set off oscillating in step is a time of 20s has elasped , during which t
adell [148]

Answer:

Length of longer pendulum = 99.3 cm

Length of shorter pendulum = 82.2 cm

Explanation:

Since the longer pendulum undergoes 10 oscillations in 20 s, its period T = 20 s/10 = 2 s.

From T = 2π√(l/g), the length of the pendulum. l = T²g/4π²

substituting T = 2s and g = 9.8 m/s² we have

l = T²g/4π²

= (2 s)² × 9.8 m/s² ÷ 4π²

= 39.2 m ÷ 4π²

= 0.993 m

= 99.3 cm

Now, for the shorter pendulum to be in step with the longer pendulum, it must have completed some oscillations more than the longer pendulum. Let x be the number of oscillations more in t = 20 s. Let n₁ = number of oscillations of longer pendulum and n₂ = number of oscillations of longer pendulum.

So, n₂ = n₁ + x. Also n₁ = t/T₁ and n₂ = t/T₂ where T₂ = period of shorter pendulum.

t/T₂ = t/T₁ + x

1/T₂ = 1/T₁ + x  (1)

Also, the T₂ = t/n₂ = t/(n₁ + x)  (2)

From (1) T₂ = T₁/(T₁ + x) (3)

equating (2) and (3) we have

t/(n₁ + x) = T₁/(T₁ + x)

substituting t = 20 s and n₁ = 10 and T₁ = 2s, we have

20 s/(10 + x) = 2/(2 + x)

10/(10 + x) = 1/(2 + x)

(10 + x)/10 = (2 + x)

(10 + x) = 10(2 + x)

10 + x = 20 + 10x

collecting like terms

10x - x = 20 - 10

9x = 10

x = 10/9

x = 1.11

x ≅ 1 oscillation

substituting x into (2)

T₂ = t/n₂ = t/(n₁ + x)

= 20/(10 + 1)

= 20/11

= 1.82 s

Since length l = T²g/4π²

substituting T = 1.82 s and g = 9.8 m/s² we have

l = T²g/4π²

= (1.82 s)² × 9.8 m/s² ÷ 4π²

= 32.46 m ÷ 4π²

= 0.822 m

= 82.2 cm

6 0
3 years ago
a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit
Darina [25.2K]
Here is my step-by-step-work. Let me know if you have any questions! :)

3 0
3 years ago
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