R=1m. Vt=+- 8m/s atot (tan √3)

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton

pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:

$F = K * \Delta{x}$

$2\:N = 4\:N/cm*\Delta{x}$

$4\:N/cm*\Delta{x} = 2\:N$

$\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}$

simplify by 2

$\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}$

$\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark$

Answer:

B.) 1/2 cm

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I Hope this helps, greetings ... Dexteright02! =)

7 0

3 years ago

15 points. give me the method.

AveGali [126]

Answer:

$\boxed{{160 \: m(s)}^{ - 1} }$

Explanation:

$if \:the \: frequencies \: are \to \\ f_{1} = 640Hz \\ and \\f_{2} = 480Hz \: \\ but \: \boxed{v = f \gamma }: f = \frac{v}{ \gamma } \\ if \: \gamma_{1} - \gamma _{2} = 1 = \gamma \\ f_{1} - f_{2} = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 = \boxed{{160 \: m(s)}^{ - 1} }$

5 0

3 years ago

Calculate the volume (in cm³) of 38.5 grams of platinum if the density of platinum is 21.5 g/cm³. Give your answer to 2 decimal

Alexeev081 [22]

Answer:

1.79 cm^{3}

Explanation:

by the definition ,

$density =\frac{mass}{volume}\\ volume=\frac{mass}{density}\\=\frac{38.5 g}{21.5 gcm^{-3} } \\=1.79 cm^{3}$

6 0

4 years ago

if the vessel in the sample problem accelerates fir 1.00 min, what will its speed be after that minute ?

LUCKY_DIMON [66]

Answers:

a) 154.08 m/s=554.68 km/h

b) 108 m/s=388.8 km/h

Explanation:

The complete question is written below:

In 1977 off the coast of Australia, the fastest speed by a vessel on the water was achieved. If this vessel were to undergo an average acceleration of $1.80 m/s^{2}$ , it would go from rest to its top speed in 85.6 s.

a) What was the speed of the vessel?

b) If the vessel in the sample problem accelerates for 1.00 min, what will its speed be after that minute?

Calculate the answers in both meters per second and kilometers per hour

a) The average acceleration $a_{av}$ is expressed as:

$a_{av}=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{\Delta t}$ (1)

Where:

$a_{av}=1.80 m/s^{2}$

$\Delta V$ is the variation of velocity in a given time $\Delta t$ , which is the difference between the final velocity $V$ and the initial velocity $V_{o}=0$ (because it starts from rest).