All categories

3 years ago

Select the correct answer.

1 answer:

6 0

The correct answer is C:Waves transfer energy, but not matter. A wave does not move matter in the direction of its propagation. It only transfers energy just like the ocean wave traveling many miles away with the water just moving up and down.

You might be interested in

At 20 C, a steel rod of length 40.000 cm and a brass rod

balandron [24]

Answer:

a. stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

b. new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

Explanation:

Step 1: identify the given parameters and standard parameters

⇒Steel rod: length of rod = 40.000 cm

Young modulus(Υ) = 20 X 10¹⁰ pa

coefficient of linear expansion(α) = 1.2 X 10⁻⁵ K⁻¹

stress in the rod =?

⇒Brass rod: length of rod = 30.000 cm

Young modulus(Υ) = 9 X 10¹⁰ pa

coefficient of linear expansion(α) = 2.0 X 10⁻⁵ K⁻¹

stress in the rod =?

--------------------------------------------------------------------------------------------

Step 2: calculate the stress in each rod

⇒Steel rod: stress in the rod = -Y*α*ΔT

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(60-20)K

= (-20 X 10¹⁰ pa) (1.2 X 10⁻⁵ K⁻¹)(40)K

= -9.6 X 10⁷ pa

--------------------------------------------------------------------------------------------

⇒Brass rod: stress in the rod = -Y*α*ΔT

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(60-20)K

= (-9 X 10¹⁰ pa) (2.0 X 10⁻⁵ K⁻¹)(40)K

= -7.2 X 10⁷ pa

--------------------------------------------------------------------------------------------

∴ stress in steel rod is -9.6 X 10⁷pa while stress in brass rod is -7.2 X 10⁷pa

--------------------------------------------------------------------------------------------

Step 3: calculate the new length of each rod

⇒Steel rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (1.2 X 10⁻⁵ K⁻¹)(40cm)(40)K

ΔL = 1920 X 10⁻⁵ cm = 0.0192cm

New length = ΔL + L₀ = 0.0192cm + 40cm

New length = 40.0192cm

--------------------------------------------------------------------------------------------

⇒Brass rod: New length = ΔL + L₀

ΔL = α*L₀*ΔT

ΔL = (2.0 X 10⁻⁵ K⁻¹)(30cm)(40)K

ΔL = 2400 X 10⁻⁵ cm = 0.024cm

New length = ΔL + L₀ = 0.024cm + 30cm

New length = 30.024cm

--------------------------------------------------------------------------------------------

Therefore, new junction from steel rod is 40.0192cm while new junction from brass rod is 30.024cm

8 0

3 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$