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Sindrei [870]
3 years ago
9

Why is the gravitational potential energy of an object 1 meter above the moons surface less than its potential energy 1meter abo

ve earths surface?
a. the objects mass is less on the moon
b. the objects weight is more on the moon
c. the objects acceleration due to gravity is less on the moon
d. both a and c
Physics
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

c

Explanation:

took test

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How will minerals orient when a rock is put under normal stress?
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perpendicular to the direction of the greatest stress

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When you trace the outline of your palm how do you find its area​
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you count the squares or messure it

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3 years ago
Read 2 more answers
A car travels for 5.5 h at an average speed<br> of 75 km/h. How far did it travel?<br> T TE
ANEK [815]

Answer:

\boxed {\boxed {\sf 412.5 \ kilometers }}

Explanation:

Distance is the product of speed and time.

d=s*t

The speed of the car is 75 kilometers per hour. It traveled for 5.5 hours.

s= 75 \ km/hr \\t= 5.5 \ hr

Substitute the values into the formula.

d= 75 \ km/hr * 5.5 \ hr

Multiply. Note that the hours will cancel each other out.

d= 75 km * 5.5 \\d= 412.5 \ km

The car travelled <u>412.5 kilometers.</u>

5 0
3 years ago
C2H4O2 is what ??????
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8 0
3 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
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