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2 years ago

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A 0.28-kg stone you throw rises 34.3 m in the air. The magnitude of the impulse the stone received from your hand while being th

rown is

1 answer:

goldfiish [28.3K]2 years ago

5 0

Answer:

7.3 kg m/s

Explanation:

First of all, let's calculate the gravitational potential energy of the stone as it reaches its highest point:

$U=mgh=(0.28 kg)(9.8 m/s^2)(34.3 m)=94.1 J$

For the law of conservation of energy, this is equal to the initial kinetic energy of the stone at ground level (where the potential energy is zero), just after the stone leaves your hand:

$K=\frac{1}{2}mv^2=94.1 J$

From this equation we can find the velocity of the stone as it leaves your hand:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(94.1 J)}{0.28 kg}}=25.9 m/s$

The initial velocity of the stone (before leaving your hand) is zero:

$u=0$

The impulse received by the stone is equal to its change in momentum, so:

$I=\Delta p=m\Delta v=m(v-u)=(0.28 kg)(25.9 m/s-0)=7.3 kg m/s$

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2 years ago

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A toy car of mass 0.15kg accelerates from a speed of 10 cm/s to a speed of 15 cm/s. What is the impulse acting on the car?

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v=v2-v1=15-10=5 cm/s

p=mv=0.15•5=0.75 kg•cm/s

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A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat

Zinaida [17]

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

$v=u+at$ ----------1

$s=ut+\dfrac{1}{2}at^2$ ------2

From equation 1 and 2

$s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )$

So we can say that

$s= \left(\dfrac{v+u}{2}\right)t$

Given that the velocity of car at final condition will be zero (v=0)

$s= \left(\dfrac{0+u}{2}\right)t$

$s= \left(\dfrac{u}{2}\right)t$

From the above equation we can find the initial velocity of car without finding the acceleration

$20= \dfrac{u}{2}\times 4$

u = 10 m/s

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2 years ago

This is really urgent

hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

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7 0

2 years ago

You are driving to the grocery store at 20 m/s. You are 110m from an intersection when the traffic light turns red. Assume that

oksano4ka [1.4K]

As we know that reaction time will be

$t = 0.50 s$

so the distance moved by car in reaction time

$d = vt$

$d = 20 \times 0.50$

$d = 10 m$

now the distance remain after that from intersection point is given by

$d = 110 - 10 = 100 m$

So our distance from the intersection will be 100 m when we apply brakes

now this distance should be covered till the car will stop

so here we will have

$v_f = 0$

$v_i = 20 m/s$

now from kinematics equation we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 20^2 = 2(a)100$

$a = \frac{-400}{200} = -2 m/s^2$

so the acceleration required by brakes is -2 m/s/s

Now total time taken to stop the car after applying brakes will be given as

$v_f - v_i = at$

$0 - 20 = -2 (t)$

$t = 10 s$

total time to stop the car is given as

$T = 10 s + 0.5 s = 10.5 s$

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2 years ago