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avanturin [10]
3 years ago
7

complete the analogy: 1. speed: velocity;______:______ a. acceleration:velocity b. distance:displacement c. speed:acceleration d

. time:scalar 2. positive acceleration:_____: zero acceleration:constant velocity a. constant speed b. decreasing velocity c. increasing velocity d. zero velocity
Physics
2 answers:
Paraphin [41]3 years ago
7 0

Velocity is the vector,
     made up of the scalar speed + direction of speed

b).  Displacement is the vector,
           made up of the scalar distance + direction of distance
maks197457 [2]3 years ago
6 0
B.
distance is used to calculate speed

displacement is used to calculate velocity

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wolverine [178]

4.53 billion years

1.653 trillion days

39.68 trillion hours

2.381 × 10^15 minutes

1.429 × 10^17 seconds

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3. An object of mass 90 kg travels down a slide.
Gwar [14]

Answer:

3) Ep = 13243.5[J]

4) v = 17.15 [m/s]

Explanation:

3) In order to solve this problem, we must use the principle of energy conservation. That is, the energy will be transformed from potential energy to kinetic energy. We can calculate the potential energy with the mass and height data, as shown below.

m = mass = 90 [kg]

h = elevation = 15 [m]

Potential energy is defined as the product of mass by gravity by height.

E_{p}=m*g*h\\E_{p}=90*9.81*15\\E_{p}=13243.5[J]

This energy will be transformed into kinetic energy.

Ek = 13243.5 [J]

4) The velocity can be determined by defining the kinetic energy, as shown below.

E_{k}=\frac{1}{2} *m*v^{2}  \\v = \sqrt{\frac{2*E_{k} }{m} }\\ v= \sqrt{\frac{2*13243.5 }{90} }\\v=17.15[m/s]

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3 years ago
a 46 kilogram student climbs 11 meter up a rope at a constant speed if the student power output is 230 watts how long in seconds
Bezzdna [24]
230×46=10580÷11=961 second
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True or false atoms may be held together by an exchange of electrons​
diamong [38]
The answer is true...............
8 0
3 years ago
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Temperature and pressure of a region upstream of a shockwave are 295 K and 1.01* 109 N/m². Just downstream the shockwave, the te
seraphim [82]

Answer:

change in internal energy 3.62*10^5 J kg^{-1}

change in enthalapy  5.07*10^5 J kg^{-1}

change in entropy 382.79 J kg^{-1} K^{-1}

Explanation:

adiabatic constant \gamma =1.4

specific heat is given as =\frac{\gamma R}{\gamma -1}

gas constant =287 J⋅kg−1⋅K−1

Cp = \frac{1.4*287}{1.4-1} = 1004.5 Jkg^{-1} k^{-1}

specific heat at constant volume

Cv = \frac{R}{\gamma -1} = \frac{287}{1.4-1} = 717.5 Jkg^{-1} k^{-1}

change in internal energy = Cv(T_2 -T_1)

                            \Delta U = 717.5 (800-295)  = 3.62*10^5 J kg^{-1}

change in enthalapy \Delta H = Cp(T_2 -T_1)

                                 \Delta H = 1004.5*(800-295) = 5.07*10^5 J kg^{-1}

change in entropy

\Delta S =Cp ln(\frac{T_2}{T_1}) -R*ln(\frac{P_2}{P_1})

\Delta S =1004.5 ln(\frac{800}{295}) -287*ln(\frac{8.74*10^5}{1.01*10^5})

\Delta S = 382.79 J kg^{-1} K^{-1}

7 0
3 years ago
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