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3 years ago

What is the restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b?

2 answers:

3 0

Here is the correct answer of the given question above.
The restoring force of a spring with a spring constant of 4a and a stretched displacement of 3b would be -12a/b.
It is solved using the formula F= -kx. Where x is the displacement and k is the spring constant. F = -(4a)(3b) and you get -12a/b. Hope this answer helps.

Tatiana [17]3 years ago

3 0

F=-kx

F= - (4a) (3b)

F = -12ab

-kx

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The earth has a mass ME = 5.98 · 1024 kg and the moon has a mass MM = 7.36 · 1022 kg. The distance from the center of the earth

ivann1987 [24]

Answer:

2 x 10^20 N

Explanation:

Me = 5.98 x 10^24 kg

Mm = 7.36 x 10^22 kg

r = 3.82 x 10^5 km = 3.82 x 10^8 m

The gravitational force between earth and moon is

F = G Me x Mm / r^2

F = (6.67 x 10^-11 x 5.98 x 10^24 x 7.36 x 10^22) / (3.82 x 10^8 x 3.82 x 10^8)

F = 2 x 10^20 N

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4 years ago

You have been working on a new, strong, lightweight ceramic which will be used to replace steel bearing balls. One cubic meter o

marin [14]

Answer:

r=2.6 cm

Explanation:

Hi!

Lets call steel material 1, and the new alloy material 2. You know their densities:

$\rho_1=8.08*10^3\frac{kq}{m^3}\\\rho_2=3.14*10^3\frac{kq}{m^3}$

The volume of a sphere with radius r is given by:

$V=\frac{4}{3}\pi r^3$

Then the masses of the bearings are:

$m_1=\rho_1V_1=\frac{4}{3}\pi r_1^3 \\m_1=\rho_2V_2=\frac{4}{3}\pi r_2^3$

For the masses to be the same:

$\rho_1 V_1 = \rho_2 V_2\\\frac{V_2}{V_1} =\frac{\rho_1}{\rho_2}\\(\frac{r_2}{r_1})^3 = \frac{8.08}{3.14}=2.57\\r_2=\sqrt[3]{2.57}\;r1 =1.37r_1=1.37*1.9cm=2.6cm$

5 0

4 years ago

A sample of O2 occupies 75 L at 1 atm. If the volume of the

Westkost [7]

Answer:

1/2 atm

Explanation:

Givens

The temperature remains constant so the formula is

P = 1 atm

V = 75 L

P1 = ?

V1 = 75 * 2 = 150 L

Formula

P*V = P1* V1

Solution

1 * 75 = P1 * 150 Divide by 150

75/150 = P1

P1 = 1/2 atmospheres

3 0

3 years ago

A hunter shoots a stone from his catapult with an initial velocity of 40 m per seconds add elevation of 60 degree with the aim o

Paladinen [302]

Answer:

approximately 7 seconds

Explanation:

The given scenario is just an example of Projectile Motion.

In this question, we are given:

- Initial Velocity (u) = 40 m/s

- Angle of Elevation (θ) = 60 degrees

- Distance of Bird (s) = 145 meters [Not used for the time of flight]

Time of Flight:

$T = \frac{2uSin(\theta)}{g}$

we will use g = 9.8 m/s/s since we are not told otherwise.
$T = \frac{2(40)Sin(60 degrees)}{9.8}$

$T = \frac{40\sqrt{3}}{9.8}$

T = 4.081(√3) seconds
rounded to 7 seconds

7 0

3 years ago

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s 2. How fast were they going afterward? Choose the right equation for th

Feliz [49]

Answer:

2.72 m/s

Explanation:

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s².

It means,

Distance, s = 3 m

Initial velocity, u = 0.5 m/s

Acceleration, a = 1.2 m/s²

We need to find the final velocity of the person. Using equation of motion to find it as follows :

$v^2-u^2=2as\\\\v^2=2as+u^2\\\\v^2=2\times 1.2\times 3+(0.5)^2\\\\v=2.72\ m/s$

So, the final velocity of the person is 2.72 m/s.

7 0

3 years ago