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drek231 [11]
3 years ago
9

Tiny mass of 67kg and tubby mass of 83 kg are now on the opposite ends of a see saw and are both 1.4m from the balance point whe

re should their sister teeny whose mass is 32 kg sit so the see saw balances
Physics
1 answer:
Fittoniya [83]3 years ago
6 0

Answer:

0.7 m between the balance point and Tiny

Explanation:

It's intuitive to say she sould be on the side the smaller mass is placed. How far from the balance point?

Let's call x that distance. and assume that the contraption is "oriented" so that being on the side with tiny (and teeny) gives you a positive distance from the balance point, and negative on the other side.

At equilibrium, the sum of all momentums (momenta?) will be zero:

67\times 1.4 +32\times x + 83 \times (-1.4) = 0\\32x -16(1.4) = 0 \rightarrow 2x=1.4 x= 0.7

In layman's term, she has to be halfway between the balance point and Tiny.

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Heights of men on a baseball team have a​ bell-shaped distribution with a mean of 166 cm 166 cm and a standard deviation of 5 cm
kaheart [24]

Answer:

95 %

99.7 %

Explanation:

\mu = 166 cm = Mean

\sigma = 5 cm =  Standard deviation

a) 156 cm and 176 cm

166-5\times 2=156

166+5\times 2=176

From the empirical rule 95% of all values are within 2 standard deviation of the mean, so about 95% of men are between 156 cm and 176 cm.

b) 151 cm and 181 cm

166-5\times 3 =151

166+5\times 3=181

The empirical rule tells us that about 99.7% of all values are within 3 standard deviations of the mean, so about 99.7% of men are between 151 cm and 181 cm.

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3 years ago
What is one defining feature of a prokaryotic cell
trapecia [35]
A. nucleus
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5 0
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Which feature of the ocean floor lies just off the beach and is a part of continental crust
4vir4ik [10]
The coastline/shoreline
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8 0
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un astronauta cuya masa es 80kg permanece inmovil en el espacio exterior, al activar la unidad propulsora que lleva en la espald
beks73 [17]

Answer:

Explanation:

80(0) = 2(15) + (80-2)v

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5 0
3 years ago
How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances
Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = x+y=149.6\times 10^6\ m

G = Gravitational constant

M_s = Mass of Sun = 1.989\times 10^{30}

M_e = Mass of Earth = 5.972\times 10^{24}\ kg

According to the question

\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y

x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m

The probe should be 258774.9441 m from Earth

7 0
3 years ago
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