According to Newton’s Second Law of Motion, an object will accelerate if you apply what kind of force?

1 answer:

6 0

If you got this question from a book or homework sheet, then
you should take it back and comp[lain. It's a defective question.

It's fishing for the answer "unbalanced force", but that's a misleading
and misuse of the term. There's no such thing as a "balanced force"
or an "unbalanced force".

A GROUP of two or more forces can be balanced ... if the forces
all cancel each other out and add up to zero ... or unbalanced ...
if they don't. But it's never correct to apply that description to
a single force.

According to Newton's second law of motion, an object will
accelerate if the forces on it are unbalanced.

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The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm

icang [17]

Fair enough, but you'll have to tell us the volume of the bar first.

5 0

3 years ago

A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

Explanation:

For an open pipe, the velocity is given by

$v=\frac {2Lf}{n}$

Making L the subject then

$L=\frac {nV}{2f}$

Where f is the frequency, L is the length, n is harmonic number, v is velocity

Substituting 1 for n, 320 Hz for f and 331 m/s for v then

$L=\frac {1*331}{2*320}=0.5171875\approx 0.52m$

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, $f2=2\times \frac {v}{2L}$ and $f3=3\times \frac {v}{2L}$

$f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz$

(c)

When v=367 m/s then

$f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz$

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3 years ago

A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?

Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

$W=(mg)h$