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Elodia [21]
3 years ago
12

oscillating spring mass systems can be used to experimentally determine an unknown mass without using a mass balance. a student

observes that a particular spring-mass system has a frequency of oscillation of 10 Hz. the spring constant of the spring is 250 N/m. what is the mass?​
Physics
1 answer:
pochemuha3 years ago
6 0

Answer:

m = 63 grams

Explanation:

ω = 10 cycles/s(2π radians/cycle) = 20π rad/s

ω = √(k/m)

m = k/ω² = 250/(20π)² = 0.06332... kg

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what is the dot product and cross product of of two vectors if the angle is between them is 90 degree?​
DaniilM [7]

\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}

  • The angle between the two vectors is 90° .

\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}

  1. The dot product of two vectors .
  2. The cross product of two vectors .

\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}

⚡ Let \rm{\vec{a}} and \rm{\vec{b}} are the two vectors .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}

  • cos 90° = <u>0</u>

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}

\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:0\:}

\rm{\red{\therefore}} [1] The dot product of two vectors is “ <u>0</u> ” .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}

Where,

  • θ = 90°

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\sin{90^{\degree}}\:}

  • sin 90° = <u>1</u>

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}

\rm{\red{\therefore}} [2] The cross product of two vectors is “ <u>ab</u> ” .

3 0
3 years ago
Read 2 more answers
While walking past a construction site, a person notices a pipe sticking out of a second floor window with water pouring out. As
tia_tia [17]

Answer:

Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.

P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0

Explanation:

P1 = P2 = atmospheric pressure so, P1 - P2 = 0

h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²

From the continuity equation for fluids

A1v1 = A2v2

v2 = A1v1/A2

Substituting into the equation above

(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho

Making A2² the subject of the formula,

A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}

The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.

Thank you for reading.

4 0
3 years ago
20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solu
Anastasy [175]

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

7 0
2 years ago
look at the circuit in the figure. find the current, voltage, and power in each resistor. please list answer
sashaice [31]

Answer:

fhcjctfkbraf gdovtckcrhha6g

6 0
2 years ago
A force of 14 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i
enyata [817]

Answer:

W = 19.8 J

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

So we have

F = 14 lbf

F = 6.35 \times 9.8 N

F = 62.3 N

stretch in the spring is given as

x = 4 in = 0.1016 m

now we will have

F = kx

62.3 = k(0.1016)

k = 613.125 N/m

Now we need to find the work to stretch it by x = 10 in = 0.254 m

so we have

W = \frac{1}{2}kx^2

W = \frac{1}{2}(613.125)(0.254)^2

W = 19.8 J

7 0
3 years ago
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