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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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How does increased pressure result in the formation of a mineral?
drek231 [11]
The best answer is 
A) <span>The atoms in the mineral get rearranged

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5 0
3 years ago
How you use physics in your<br> everyday life.
stepan [7]

Answer:

Explanation:

Physics gets involved in your daily life right from you wake up in the morning. The buzzing sound of an alarm clock helps you wake up in the morning as per your schedule. The sound is something that you can't see, but hear or experience. Physics studies the origin, propagation, and properties of sound

5 0
3 years ago
A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region
zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with  due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

K_{1} = K_{2} + W_{f}

Where:

K_{1}, K_{2} are the initial and final translational kinetic energies of the tobbogan, measured in joules.

W_{f} - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})

Where:

f - Friction force, measured in newtons.

\Delta s - Distance travelled by the toboggan in the rough region, measured in meters.

m - Mass of the toboggan, measured in kilograms.

v_{1}, v_{2} - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}

If m = 375\,kg, v_{1} = 4.50\,\frac{m}{s}, v_{2} = 1.20\,\frac{m}{s} and \Delta s = 5.40 \,m, then:

f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}

f = 653.125\,N

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%

\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%

\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%

\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%

\%K_{loss} = 92.889\,\%

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%

\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%

If v_{1} = 4.50\,\frac{m}{s} and v_{2} = 1.20\,\frac{m}{s}, then:

\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%

\%v_{loss} = 73.333\,\%

The speed of the toboggan is reduced in 73.333 %.

5 0
3 years ago
The fixed hydraulic cylinder C imparts a constant upward velocity v = 2.2 m/s to the collar B, which slips freely on rod OA. Det
Olenka [21]

Answer:

so angular velocity is 7.13128 sec−1

Explanation:

velocity v = 2.2 m/s

displacement s = 220 mm = 0.220 m

distance d = 510 mm = 0.510 m

to find out

angular velocity

solution

we know that

angular velocity will be velocity ( v)  / (displacement²  +  distance²)   .....1

now put all these value in equation 1 and we get angular velocity i.e.

angular velocity =  velocity ( v)  / (displacement²  +  distance²)

angular velocity = 2.2  / (0.22²  +  0.51²)

angular velocity = 2.2 / 0.3085

angular velocity = 7.13128

so angular velocity is 7.13128 sec−1

6 0
3 years ago
A rectangular coil with 50 turns of conducting wire and a total resistance of 10.0 Ω initially lies in the yz-plane at time t =
castortr0y [4]

Answer:

a) 43.20V

b) 2.71W/s

c) 40.25s

d) 7.77Nm

Explanation:

(a) The emf of a rotating coil with N turns is given by:

emf=NBA\omega sin(\omega t)

N: turns

B: magnitude of the magnetic field

A: area

w: angular velocity

the emf max is given by:

emf_{max}=NBA\omega=(50)(1.80T)(0.200m*0.100m)(24.0rad/s)\\\\emf_{max}=43.20V

(b) the maximum rate of change of the magnetic flux is given by:

\frac{d\Phi_B}{dt}=\frac{d(A\cdot B)}{dt}=\frac{d}{dt}(ABcos\omega t)=AB\omega sin(\omega t)\\\\\frac{d\Phi_B}{dt}_{max}=(\pi(0.200*0.100))(1.80T)(24.0rad/s)=2.71\frac{W}{s}

(c) emf(t=0.050s)=(50)(1.80T)(0.200m*0.100m)(24rad/s)sin(24.0rad/s(0.050s))\\\\emf(t=0.050s)=40.26V

(d) The torque is given by:

\tau=NABIsin\theta\\\\NAB\omega=emf_{max}\\\\\tau=\frac{emf_{max}}{\omega}\frac{emf_{max}}{R}\\\\\tau=\frac{(43.20V)^2}{(24.0rad/s)(10.0\Omega)}=7.77Nm

3 0
3 years ago
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