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3 years ago

9

Stone needs 1350 J of energy to crack down. A bullet of 120 gm is fired on

1 answer:

Agata [3.3K]3 years ago

7 0

There is a lot of help on the internet 350

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How many significant figures are in 107 moles of sodium

Blizzard [7]

Answer:

3 significant figures

Explanation:

here zero is counted as one of the significant figures since it lies next to an integer.

hence the number has 3 sfg

8 0

3 years ago

If the frequency of q of a trait is 0.4, what is the frequency of p?

lianna [129]

In genetic traits, p and q represent the relative probabilities of the two alleles manifesting. If these two are the only options (ex. a dominant one and a recessive one), then the probabilities of both must sum up to 1. In this case, since we are given that q = 0.4, then p + q = 1, p + 0.4 = 1, and p = 0.6.

4 0

4 years ago

Read 2 more answers

What two measuring units can i use to measure the distance

riadik2000 [5.3K]

Feet and inches or millimeters or centimeters or meters or miles or kilometers

8 0

3 years ago

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

Nastasia [14]

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

8 0

3 years ago

The gravitational force,F, on a rocket at a distance,r, from the center of the earth isgiven byF=kr2wherek= 1013N·km2. (Newton·k

Brrunno [24]

Answer:

The gravitational force changing velocity is

$\frac{dF}{dt}=-8\frac{N}{s}$

Explanation:

The expression for the gravitational force is

$F=\frac{k}{r^{2}}\\\\k=10x10^{13} N*km^{2}\\\\r=10x10^{4} km\\\\V=0.4 \frac{km}{s}$

Differentiate the above equation

$\frac{dF}{dt}=\frac{k}{r^{2}}\\\frac{dF}{dt}=k*r^{-2}\\\frac{dF}{dt}=-2*k*r^{-3} \frac{dr}{dt}\\\frac{dF}{dt}=\frac{-2k}{r^{3}}\frac{dr}{dt}$

The velocity is the distance in at time so

$V=\frac{dr}{dt}=0.4 \frac{km}{s}$

$\frac{dF}{dt}=\frac{-2*k}{r^{3}}*0.4\\\frac{dF}{dt}=\frac{-8*10x^{13}N*km^{2} }{(10x10^{4}) ^{3}} \\\frac{dF}{dt}=\frac{-8x10^{12} }{1x10^{12}}$

$\frac{dF}{dt}=-8\frac{N}{s}$

8 0

3 years ago