A negative ion has more question 1 options:

The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the

tigry1 [53]

Answer:

A) The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) The maximum speed of the plastic sphere will be 2.2 m/s

Explanation:

Hi there!

I´ve found the complete problem on the web:

A toy launcher that is used to launch small plastic spheres horizontally contains a spring with a spring constant of 50. newtons per meter. The spring is compressed a distance of 0.10 meter when the launcher is ready to launch a plastic sphere.

A) Determine the elastic potential energy stored in the spring when the launcher is ready to launch a plastic sphere.

B) The spring is released and a 0.10-kilogram plastic sphere is fired from the launcher. Calculate the maximum speed with which the plastic sphere will be launched. [Neglect friction.] [Show all work, including the equation and substitution with units.]

A) The elastic potential energy (EPE) is calculated as follows:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compressing distance

EPE = 1/2 · 50 N/m · (0.10 m)²

EPE = 0.25 J

The elastic potential energy stored in the spring when it is compressed 0.10 m is 0.25 J.

B) Since there is no friction, all the stored potential energy will be converted into kinetic energy when the spring is released. The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the sphere.

v = velocity

The kinetic energy of the sphere will be equal to the initial elastic potential energy:

KE = EPE = 1/2 · m · v²

0.25 J = 1/2 · 0.10 kg · v²

2 · 0.25 J / 0.10 kg = v²

v = 2.2 m/s

The maximum speed of the plastic sphere will be 2.2 m/s

6 0

4 years ago

3 Select the correct answer. What is a substance?

USPshnik [31]

Answer:

physical material from which something is made or which has discrete existence

Explanation:

5 0

3 years ago

An object moves 2.5 m. This is an example of a _______. Question 3 options: direction distance velocity speed

Hitman42 [59]

Distance, since distance represents how far something has travelled, which would be in our case 2.5m.

5 0

3 years ago

Read 2 more answers

A single-turn circular loop of radius 14 cm is to produce a field at its center that will just cancel the earth's magnetic field

Olin [163]

Answer:

49 A

Explanation:

We are given that

Number of turns,N=1

Radius,r=14 cm= $\frac{14}{100}=0.14 m$

1 m=100 cm

Magnetic field,B=0.7 G= $0.7\times 10^{-4} T$

$1G=10^{-4} T$

$\mu_0=4\pi\times 10^{-7} Tm/A$

We know that magnetic field

$B=\frac{\mu_0IN}{2\pi r}$

Substitute the values

$0.7\times 10^{-4}=\frac{4\pi\times 10^{-7}\times 1\times I}{2\pi\times 0.14}$

$I=\frac{0.7\times 10^{-4}\times 2\pi\times 0.14}{4\pi\times 10^{-7}\times 1}$

$I=49A$

6 0

3 years ago

A cylindrically shaped piece of collagen (a substance found in the body in connective tissue) is being stretched by a force that

Marat540 [252]

Answer:

Part(a): The value of the spring constant is $3.11 \times 10^{2}~Kg~s^{-2}$ .

Part(b): The work done by the variable force that stretches the collagen is $1.5 \times 10^{-6}~J$ .

Explanation:

Part(a):

If ' $k$ ' be the force constant and if due the application of a force ' $F$ ' on the collagen ' $\Delta l$ ' be it's increase in length, then from Hook's law

$F = k~\Delta l....................................................................(I)$

Also, Young's modulus of a material is given by

$Y = \dfrac{F/A}{\Delta l/l}...............................................................(II)$

where ' $A$ ' is the area of the material and ' $l$ ' is the length.

Comparing equation ( $I$ ) and ( $II$ ) we can write

$&& Y = \dfrac{l~k}{A}\\&or,& k = \dfrac{Y~A}{l}\\&or,& k = \dfrac{Y~(\pi~r^{2})}{l}$

Here, we have to consider only the circular surface of the collagen as force is applied only perpendicular to this surface.

Substituting the given values in equation ( $III$ ), we have

$k = \dfrac{3.10 \times 10^{6}~N~m^{-2} \times \pi \times (0.00093)^{2}~m^{2}}{.027~m} = 3.11 \times 10^{2}~Kg~s^{-2}$

Part(b):

We know the amount of work done ( $W$ ) on the collagen is stored as a potential energy ( $U$ ) within it. Now, the amount of work done by the variable force that stretches the collagen can be written as

$W = \dfrac{1}{2}k~x^{2} = \dfrac{(\dfrac{F}{k})^{2}k}{2} = \dfrac{F^{2}}{2~k}...................................(IV)$

Substituting all the values, we can write

$W = \dfrac{(3.06 \times 10^{-2})^{2}~N^{2}}{2 \times 3.11 \times 10^{2}~Kg~s^{-2}} = 1.5 \times 10^{-6}~J$

3 0

3 years ago