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bekas [8.4K]
2 years ago
9

What is the acceleration of a 32-kg object if a 6.75-N force is applied to it?

Physics
1 answer:
professor190 [17]2 years ago
5 0
F=MA
*newton 2nd law*
F=6.75N
M=32kg

A= F/M
A=6.75/32
equal
a.0.21m/s2
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A person stands 6.00 m from a speaker, and 8.00 m from an identical speaker. What is the wavelength of the first (n=1) interfere
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Answer:

wavelength = 4 m

Explanation:

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for a shift in phase t1 = t - 6/c,

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substituting t1 and t2

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solving using trigonometry identities in radians.

we have,

π - 2πn = w(t - 8/c) - w(t - 6/c)

putting w = 2πf

π - 2πn = 2πf(t - 8/c) - 2πf(t - 6/c)

dividing both sides by π

1 - 2n = 2ft - 16(f/c) - 2ft + 12(f/c)

simplifying we have,

1 - 2n = -4(f/c)

solving for f we have,

f = c/4(2n - 1)

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A 3.06kg stone is dropped from a height of 10.0m and strikes the ground with a velocity of 7.00m/s. What average force of air fr
xxTIMURxx [149]
<span>22.5 newtons. First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as: E = 0.5mv^2 where E = Energy m = mass v = velocity Substituting known values and solving gives: E = 0.5 3.06 kg (7 m/s)^2 E = 1.53 kg 49 m^2/s^2 E = 74.97 kg*m^2/s^2 Now ignoring air resistance, how much energy should the rock have had? We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So 3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2 So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is 299.88 J - 74.97 J = 224.91 J So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division. 224.91 J / 10 m = 224.91 kg*m^2/s^2 / 10 m = 22.491 kg*m/s^2 = 22.491 N Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
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