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3 years ago

Do objects need to be touching to transfer change

Physics

2 answers:

ziro4ka [17]3 years ago

7 0

Some do but not all. I agree u should look up on goggle

pogonyaev3 years ago

3 0

I believe so. Hope that helps, look it up on google to be sure

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A ball is thrown horizontally from a platform so that the initial height of the ball is 6.0 m above the level ground below. The

solniwko [45]

We can first obtain time of flight from vertical fall
Initial velocity U=0, d = 6 m, a = 9.8 m/s²
d = ut + 1/2 at²
6.0 = 0 + (1/2 × 9.80 t²)
t = √(12/9.8)
= 1.106 sec
Horizontal velocity = Vh = Dh/t
= 24.0 /1.106 s
= 21.69 m/s
The ball was thrown at a speed of 21.69 m/s

5 0

4 years ago

Which travels at a greater speed, visible or infrared light?

Ksju [112]

Infrared light because it is barely able to be seen

7 0

3 years ago

Will binoculars work properly if their prisms (assume n = 1.50) are immersed in water (nwater= 1.33)? assume that binoculars wor

Amanda [17]

The binoculars will not work properly.

Please refer to the diagram attached.

Binoculars work on the principle of total internal reflection. The incident ray from air enters the prism through on face normally and it strikes the opposite face at an angle of 45°. The ray undergoes Total internal reflection if its critical angle is less than the angle of incidence,which , in this case is 45°.

The critical angle $i_{c}$ is related to the refractive index of the material of glass μ as shown below.

$\mu =\frac{1}{sini_c}$

For a refractive index equal to 1.5, the critical angle is found as shown below.

$sini_c=\frac{1}{\mu} \\ =\frac{1}{1.5} \\ =0.666$

Take the inverse of the value 0.666 to determine the critical angle.

$sin^{-1} (0.666)=41.8^o$

The critical angle in air is less than the angle of incidence, and hence total internal reflection occurs in air.

When the binoculars are immersed in water, the ray passes into the glass through water. therefore, the refractive index of the prism when immersed in water is given by,

$\mu_g_w=\frac{\mu_g}{\mu_w}$

Therefore the critical angle c in this case is given by,

$sinc=\frac{\mu_w}{\mu_g}$

Substitute 1.33 for $\mu_w$ and 1.5 for $\mu_g$ .

$sinc=\frac{\mu_w}{\mu_g} \\ =\frac{1.33}{1.5} \\ =0.8866$

Take the inverse of the value 0.8866.

$c=sin^{-1} (0.8866)\\ =62^o$

Since the critical angle of the prism when immersed in water, is 62°, which is greater than the angle of incidence of 45° required for viewing the object, the binoculars which are set for Total reflection in air, will not function when immersed in water.

4 0

4 years ago

Which part of the carbon cycle is leading to coral bleaching? photosynthesis in plants respiration in plants absorption of carbo

ivann1987 [24]

I would say that the answer is absorption of carbon dioxide into the oceans.

7 0

4 years ago

Read 2 more answers

what is the distance moved by a particle moving along x axis if it starts from x=-7m goes upto x= -12m turns and stops at x=+34m

Marizza181 [45]

Answer: Trajectory=51m

Displacement=41m

Explanation:

Let's begin by stating clear that movement is the change of position of a body at a certain time. So, during this movement, the body will have a trajectory and a displacement, being both different:

The trajectory is the path followed by the body (is a scalar magnitude).

The displacement is the distance in a straight line between the initial and final position (is a vector magnitude).

According to this, in the description of the object placed at x= -7m on a number line and moving some 12m to the left and then to the right, stopping at x=34m; we are talking about the path followed by the object, hence its trajectory. So, 51 m is its trajectory.

But, if we talk about displacement, we have to draw a straight line between the initial position of the object (x=-7m) to its final position (x=+34m).

Now, being this an unidimensional problem, the displacement vector for this object is 41m.

7 0

3 years ago