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blagie [28]
3 years ago
7

You have two photos of a person walking. One shows the person at the corner of Third and Main streets, the other shows the perso

n at the corner of Tenth and Main streets. There are lampposts at every corner in this town, and the first picture shows it to be 10:32:00 exactly. The second picture shows it to be 10:49:30. You know three facts: (1) All of the clocks are synchronized; (2) there are exactly 12 equal-sized blocks per kilometer in this town; and (3) the streets that cross Main in this area are numbered consecutively, with no interruptions. What is the person’s average speed in kilometers per hour?
Physics
1 answer:
Lorico [155]3 years ago
3 0

Answer:

average speed

= [(10-3)/12 km] / [(49.5-32.0)/60 hour]

= 5*7 / 17.5

= 2 km/h .

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You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses ont
choli [55]

Answer:

490.5\ \text{N/m}

Explanation:

m = Mass attached to spring = 14 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

x = Displacement of spring = 78-50=28\ \text{cm}

k = Spring constant

The force balance of the system is given by

kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}

The spring constant for that spring is 490.5\ \text{N/m}.

6 0
3 years ago
the process of making alloys involves _____ pure metals to remove impurities. Then the pure metals are ____ with other component
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I think this is the answer for the first line(Cooling ,Heating or mixing ) and for the second line is(broken down,cooled,mixed)
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3 years ago
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Kinetic energy is stored in a stretched rubber
IRISSAK [1]

Answer:

potential until released

4 0
3 years ago
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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
krek1111 [17]

Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

4 0
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stellarik [79]
A would be the answer 
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