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3 years ago

What type of telescope is shown in Figure 24-2

Physics

2 answers:

lesantik [10]3 years ago

7 0

Refractor, It's a refractor-esque telescope

djyliett [7]3 years ago

4 0

It's a refractor telescope

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Distance meters Time (seconds) Trial Tria 2 Trials Average 3 053 059 0.52 055 6 106 719 104 Uning the values trom the data table

kumpel [21]

Jsisisusuagahannananana this is worth 22 points g that’s crazy 25

5 0

1 year ago

A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5m/s2 for 4s. After t

Ber [7]

Answer:

+7.0 m/s

Explanation:

Let's take rightward as positive direction.

So in this problem we have:

a = -2.5 m/s^2 acceleration due to the wind (negative because it is leftward)

t = 4 s time interval

v = -3.0 m/s is the final velocity (negative because it is leftward)

We can use the following equation:

v = u + at

Where u is the initial velocity

We want to find u, so if we rearrange the equation we find:

$u = v - at = (-3.0 m/s) - (-2.5 m/s^2)(4 s)=+7.0 m/s$

and the positive sign means the initial direction was rightward.

6 0

3 years ago

An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of

Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

$3\times \frac{\lambda}{2} = 0.39$

$\lambda = 0.26 m$

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

$d = \frac{\lambda}{4}$

$d = \frac{0.26}{4} = 0.065 m = 6.5 cm$

So the distance will be 6.5 cm

3 0

2 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

3 years ago

A student wants to design an experiment to study the transformation of mechanical energy. Which object can be used to investigat

Vsevolod [243]

I guess it’s the 3rd one

7 0

2 years ago

Read 2 more answers