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3 years ago

How much force does it take to stop a 1,000 kg car traveling at 20 m/s in 10 seconds?

1 answer:

5 0

The initial momentum of the car is

(mass) x (speed) = 20,000 kg-m/s

"Impulse" is (force exerted) x (time the force lasts)
and it's equal to the change in momentum.

(Force) x (10 sec) = 20,000 kg-m/s

Divide each side by (10 sec) and you have

Force = (20,000 kg-m/s) / (10 sec)

= 2,000 kg-m/s² = 2,000 Newtons

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Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.

gregori [183]

First one second one and last one

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4 years ago

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A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ

Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3$

$195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m$

(a) the fundamental frequency is calculated as;

$f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o = 65 \ Hz$

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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3 years ago

1. A pendulum is 0.25 m long. What is the frequency of its oscillations?<br> How do you solve this

lidiya [134]

Answer:

Explanation:

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2 years ago

If you comb your hair and the comb becomes negatively charged, your hair becomes?

slamgirl [31]

A positive charge cause dosent your hair static up when that happens or no?

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3 years ago

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Anna drew a diagram to compare the strong and weak force. Which labels belong in the areas marked X, Y, and Z? X: infinite range

Maslowich

Answer:

For areas marked X, Y, Z, X is attractive only, Y has a very small range, and Z is attractive and repulsive

Explanation:

Solution

Given that:

From the question stated, Anna drew a diagram to compare forces that are strong and weak.

Now,

We are to find which labels are grouped in areas marked as X, Y, Z respectively.

Thus,

For X, Y, Z it is marked as:

X: Always attractive or attractive only

Y: Very small range

Z: Repulsive and attractive

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3 years ago