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3 years ago

I have no clue how to do this at all

Physics

1 answer:

Daniel [21]3 years ago

6 0

Answer:

Below

Explanation:

There are 2 sig figs in 120000. They are the 1 and 2.

Zeros at the end of a whole number are NOT considered significant. And since there are no decimals that come before the trailing zeros they are not significant.

Hope this helps!

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Hitting a ball off a tee would be considered which of the following?

Keith_Richards [23]

I think the answer would be B

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4 years ago

Alex heated some water in a pan until the entire water changed into vapor. The mass of the total vapor formed was 150 g. What wa

ELEN [110]

Answer:

B 150g

Explanation:

b 150g mass of water remains the same when it changes state

4 0

3 years ago

Determine the minimum amount of energy needed to move a satellite from its orbit around the Earth to an infinite distance from t

zmey [24]

Answer: $19.80\times 10^{10} J$

Explanation:

Given

mass of satellite $m=7.45 \times 10^3 kg$

orbital radius $R=7.50\times 10^6 m$

mass of Earth $M=5.97\times 10^{24} kg$

Minimum amount of Energy to move Satellite from its orbit to an infinite distance is sum of Potential Energy + Kinetic Energy of Satellite

$W=U+K.E.$

$U=-G\frac{Mm}{R}$

$U=-6.67\times 10^{-11}\times \frac{5.97\times 10^{24}\times 7.45 \times 10^3}{7.50\times 10^6}$

$U=-\frac{296.658\times 10^{16}}{7.5\times 10^6}$

$U=-39.55\times 10^{10} J$

$K.E.=\frac{1}{2}\times mv^2$

Where $v=orbital\ velocity$

$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.67\times 10^{-11}\times 5.97\times 10^{24}}{7.50\times 10^6}}$

$v^2=5.30\times 10^7 m/s$

$K.E.=\frac{1}{2}\times mv^2$

$K.E.=\frac{1}{2}\times 7.45\times 10^3\times (5.30\times 10^7)=19.74\times 10^{10} J$

$W=-39.55\times 10^{10}+19.74\times 10^{10}$

$W=-19.80\times 10^{10} J$

i.e. $19.80\times 10^{10} J$ Energy is required to provide to move Satellite out of its orbit

8 0

3 years ago

A car battery can deliver up to 75 amps of current at a voltage of 12 volts. How much power is this?

Alex_Xolod [135]

A.900 watts That would be your correct answer

4 0

3 years ago

Read 2 more answers

A 150 N is applied on a 20kg object to the right (horizontally). if the force of friction is 40N

Tema [17]

Answer:

a) net force = 110N to the right.

b) acceleration = 5.5 m/s/s

Explanation:

a) 150N - 40N = 110N

b) F = ma

110N = 20kg x acceleration

acceleration = 110/20 = 5.5

6 0

3 years ago