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jeyben [28]
3 years ago
14

#9 help plssssssssss

Physics
1 answer:
Klio2033 [76]3 years ago
8 0
Because the output is usually the divided side of the machine

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Where is the largest National Park?
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It is the Wrangell-St.Elias in Alaska.
It is 8,323,148 acres
It was established in December of 1980
7 0
4 years ago
Friction between solids can be most accurately defined as the force that
ki77a [65]

Answer:

True

Explanation:

Friction between solids can be most accurately defined as the force that

opposes the sliding motion of two surfaces that are touching each other.

6 0
3 years ago
Write the dimensional formula of<br>presor and gravitation constant<br>​
deff fn [24]
Or, G = [M1 L1 T-2] × [L]2 × [M]-2 = [M-1 L3 T-2]. Therefore, the gravitational constant is dimensionally represented as M-1 L3 T-2.
8 0
3 years ago
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.

(c) the length of the string is 2.74 m

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

(c) From any of the equations, solve for Length of the string (L);

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.

7 0
3 years ago
How can a large force result in a relatively small power?
Sav [38]

Answer:

hey mate here is your answer

So if an object has a very small velocity (not moving very far over time, even though a large force may be applied to it, the Power will remain small. ... Stepping on the gas, or "speeding up" the car, is applying a force which will increase velocity and increase power.

please mark me as a brainliest

8 0
3 years ago
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