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PolarNik [594]
3 years ago
13

My Notes Ask Your Teacher 6. Six blocks with different masses, m, each start from rest at the top of smooth, frictionless inclin

es having length d and vertical height h and slide down. Rank the order, from greatest to smallest, of the final kinetic energies of the masses when they reach the bottom of the inclines after having traveled their full lengths. If any of the situations yield the same kinetic energies, give them the same ranking. (Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.) a. m 70 kg; h = 8 m; and d = 40 m b. m 70 kg; h 8 m; and d 20 m c. m 35 kg; h 4 m; and d 40 m d. m 7 kg; h 16 m; and d 20 m e. m-7 kg;h-4 m; and d=20 m f. m105 kg; h-6.0 m; and d 30.0 m
Physics
1 answer:
Tresset [83]3 years ago
8 0

Answer:

Kf > Ka = Kb > Kc > Kd > Ke

Explanation:

We can apply

E₀ = E₁

where

E₀: Mechanical energy at the beginning of the motion (top of the incline)

E₁: Mechanical energy at the end (bottom of the incline)

then

K₀ + U₀ = K₁ + U₁

If v₀ = 0  ⇒  K₀

and  h₁ = 0   ⇒    U₁ = 0

we get

U₀ = K₁    

U₀ = m*g*h₀ = K₁

we apply the same equation in each case

a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J

c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J

d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J

e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J

f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J

finally, we can say that

Kf > Ka = Kb > Kc > Kd > Ke

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The table below shows two types of electromagnetic waves and three random applications of electromagnetic waves.
Musya8 [376]

Answer:

a. Microwaves—3 and infrared waves—1

Explanation:

Microwaves and infrared waves are both part of the electromagnetic spectrum, but they have different frequency and wavelength.

In particular:

- Microwaves are long-wavelength electromagnetic waves, with wavelength between 1 mm and 1 m. Their wavelength is longer than visible light

- Infrared waves are also long-wavelength electromagnetic waves, but their wavelength is shorter than microwaves: between 700 nm and 1 mm. Their wavelength is also longer than visible light.

The two types of waves are also used for different purposes. In particular:

- Infrared waves are emitted by any hot object, and their intensity depends on the temperature of the object. Therefore, they are used in astronomy to show the heat released by astronomical objects (option 1)

- Microwaves are used to study the Cosmic Microwave Background (CMB). This is electromagnetic radiation that permeates the whole universe, and its wavelength depends inversely on the local temperature. Therefore, areas with longer wavelength have lower temperature, and viceversa. Therefore, microwaves are used to measure temperature differences in space (option 3).

7 0
3 years ago
How does deformation by drawing of a semicrystalline polymer affect its tensile strength?
Elenna [48]
This effect is explained by increased chain entanglements at higher molecular weights. Increasing the degree of crystallinity of a semicrystalline polymer leads to an enhancement of the tensile strength. Deformation by drawing increases the tensile strength of a semicrystalline polymer.
4 0
3 years ago
Hockey puck B rests on a smooth ice surface and is struck by a second puck A, which has the same mass. Puck A is initially trave
Marina86 [1]

Answer:

a.v_{b2}=6.8544 \ m/s\\\\b. v_{a2}=12.891 \ m/s\\\\c. \theta _b=62\textdegree

Explanation:

Puck A's initial speed is v_a_1=14.6m/s and move in a direction \theta_b=28.0\textdegree after the collision.

#P_1=P_2 since there's no external force on the system(P=mv).

#The collision equation can be written as;

m_av_a_1+m_bv_b_1=m_av_a_2+m_bv_b_2

The kinetic energies before and after the collision are expressed as:

K_a_1+K_{b1}=K_a_2+K_{b2}, \ K=0.5mv^2

0.5m_av_a_1+0.5m_b(0)=0.5m_av_a_2+0.5m_bv_b_2\\\\14.6^2=v_{a2}^2+v_{b2}^2\\\\v_{b2}^2=213.16-v_{a2}^2

Let +x along A's initial direction and +y along A's final direction makes the angle 62\textdegree

\dot v_{a1}=14.6i\\\\\dot v_{a2}=(v_{a2} \ cos \ 28\textdegree)i+(v_a_2\ sin \ 28\textdegree)j\\\\\dot v_{b2}=v_{b2x}i+v_{b2y}j

#Substitute in v_{b2}^2=213.16-v_{a2}^2:

\dot v_{b2}=(14.6i)-\dot v{a2}\\\\\dot v_{b2}.\dot v_{b2}=v_{b2}^2\\\\\#Right \ side\\\\(14.6i-\dot v{a2}).(14.6i-\dot v{a2})=(14.6i)^2+\dot v_{a2}^2.\dot v_{a2}^2-2(14.6i).\dot v_{a2}\\\\=213.16+v_{a2}^2-2(14.6i).(v_{a2}\ cos 28\  \textdegree i+v_{a2} \ sin \ 28\textdegree j)\\\\v_{b2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\ \  v_{b2}^2=213.16-v_{a2}^2\\\\213.16-v_{a2}^2=213.16+v_{a2}^2-29.2\ cos \ 28\textdegree v_{a2}\\\\2v_a2}^2=29.2\ cos\ 28 \textdegree v_{a2}\\\\

v_{a2}=12.891\ m/s

Hence, the the speed of puck A after the collision is 12.891  m/s

#. The velocity of A after the collision is;

\dot v_{a2}=12.891 \ cos \ 28 \textdegree i+12.891 \ sin \ 28\textdegree j\\\\=11.382i+6.052j

Substitute \dot v_{a2} into \dot v_{b2}=(14.6i)-\dot v{a2}:

\dot v_{b2}=14.6i-(11.382i+6.052j)\\\\=3.218i-6.052j

This is the velocity of puck B after the collision, it's speed is:

v_{b2}=\sqrt{v_{b2x}^2+v_{b2y}^2}\\\\=\sqrt{3.218^2+(-6.052)^2}\\\\v_{b2}=6.8544 \ m/s

The velocity of puck B after the collision is 6.8544 m/s

c. The direction of puck B after the collision is:

\theta _b=tan^{-1}\frac{v_{b2y}}{v_{b2x}}\\\\=tan^{-1} \frac{-6.052}{3.218}\\\\\approx 62\textdegree

Hence, the direction of B's velocity after the collision is 62°

4 0
4 years ago
The 30-kg disk is originally spinning at v = 125 rad>s. If it is placed on the ground, for which the coefficient of kinetic f
irga5000 [103]

Answer:

t = 3.82 s

Ax = 147 N  (←)

Ay = 294 N   (+↑)

Explanation:

Given

m = 30.0 Kg

ωinitial = 125 rad/s

ωfinal = 0 rad/s

μC = 0.5

R = 0.3 m

t = ?

Ax = ?

Ay = ?

For the disk, we can apply

∑ τ = I*α

where I = m*R²/2

then

⇒ R*Ffriction = (m*R²/2)*α

⇒ R*(-μC*N) = R*(-μC*m*g) = (m*R²/2)*α

⇒ α = -2*μC*g / R

⇒ α = -2*(0.5)*(9.8) / 0.3 = -32.666 rad/s²

we can use the equation to get t:

α = Δω / t      ⇒   t = Δω / α = (0 - 125) / (-32.666)

⇒   t = 3.82 s

The horizontal and vertical components of force which the member AB exerts on the pin at A during this time are

∑ Fx = 0  (+→)

Ax - Ffriction = 0

⇒  Ax = Ffriction = μC*m*g = (0.5)*(30)*(9.8) = 147 N

⇒   Ax = 147 N  (←)

∑ Fy = 0   (+↑)

⇒  Ay - m*g = 0

⇒  Ay = m*g

⇒  Ay = 30*9.8 = 294 N

⇒  Ay = 294 N   (+↑)

5 0
3 years ago
Great Sand Dunes National Park in Colorado is famous for its giant sand dunes. Sand dunes are landforms that are found in desert
Sliva [168]

Answer:

Wind deposits sand into a small mound. So the answer is Deposition

7 0
3 years ago
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