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Goshia [24]
4 years ago
11

College students living off-campus frequently consume large amounts of ramen noodles and boxed macaroni and cheese. When they fi

nish school and start careers, their consumption of both goods frequently declines. This suggests that ramen noodles and boxed macaroni and cheese are:
Physics
1 answer:
tigry1 [53]4 years ago
4 0

Answer:

Ramen Noodles and boxed Macaroni and Cheese are COMPLEMENTARY Goods

Explanation:

Complementary Goods are goods that are demanded jointly (together) to satisfy a particular want.

Ramen Noodles, Boxed Macaroni, Cheese are demanded together by off campus college students. So, they are complementary goods.

Also, their demand are directly related - demand of one good increases, demand of other good increases & demand of one good decreases , demand of other good decreases ; because they are demanded together.

After school end , consumption of all goods declining together - is also their characteristic pointing towards them being Complementary Goods .

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.2, A car starting from rest has an acceleration of
riadik2000 [5.3K]

Answer: 2.5 m/s and 6.25 m

Explanation:

u = 0

a = 0.5 m/s²

t = 5 s

v = u + at

=  0 + 0.5 × 5

= <u>2.5 m/s</u>

s = ut + 1/2 at²

= 1/2 × 2.5 × 5

=<u> 6.25 m</u>

7 0
4 years ago
Is it more fun on a swing with low or high amplitude? Explain.
marusya05 [52]
A pendulum is an object hung from a fixed point that swings back and forth under the action of gravity. In the example of the playground swing, the swing is supported by chains that are attached to fixed points at the top of the swing set. When the swing is raised and released, it will move freely back and forth due to the force of gravity on it. The swing continues moving back and forth without any extra outside help until friction (between the air and the swing and between the chains and the attachment points) slows it down and eventually stops it.
7 0
3 years ago
An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an
kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

v_{f1} = \sqrt{2*a_{1}*d_{1}  }  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

v_{f1} = \sqrt{2*3.2*3}  }

v_{f1} =\sqrt{2*3.2*3}

v_{f1} = 4.38 m/s

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0
4 years ago
If a motorcycle starts from rest and maintains a constant acceleration of 3 m/s^2, what will its velocity be after 10 s?
zvonat [6]
In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

hope this helps XD
5 0
3 years ago
Read 2 more answers
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mote1985 [20]

Answer:

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Explanation:

5 0
3 years ago
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