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Sergeu [11.5K]
3 years ago
10

base your answer to this question on the information below and on your knowledge of physics. A toy launcher that is used to laun

ch 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. 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.
Physics
1 answer:
Nadya [2.5K]3 years ago
3 0

Answer: v = 2.24 m/s

Explanation: The <u>Law</u> <u>of</u> <u>Conservation</u> <u>of</u> <u>Energy</u> states that total energy is constant in any process and, it cannot be created nor destroyed, only transformed.

So, in the toy launcher, the energy of the compressed spring, called <u>Elastic</u> <u>Potential</u> <u>Energy (PE)</u>, transforms into the movement of the plastic sphere, called <u>Kinetic</u> <u>Energy (KE)</u>. Since total energy must be constant:

KE_{i}+PE_{i}=KE_{f}+PE_{f}

where the terms with subscript i are related to the initial of the process and the terms with subscript f relates to the final process.

The equation is calculated as:

\frac{1}{2}kx^{2}+0=0+\frac{1}{2}mv^{2}

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

\frac{1}{2}50(0.1)^{2}=\frac{1}{2}(0.1)v^{2}

v^{2}=\frac{50(0.1)^{2}}{0.1}

v=\sqrt{50(0.1)}

v=\sqrt{5}

v = 2.24

The maximum speed the plastic sphere will be launched is 2.24 m/s.

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How do you find the valume for a cylinder
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here is the answer Calculate the area of the base (which is a circle) by using the equation πr² where r is the radius of the circle. Then, multiply the area of the base by the height of the cylinder to find the volume.

3 0
3 years ago
Which is faster a velocity of + 3 m /s or -4 m / s
irina1246 [14]

Answer: Velocity of 3m is the faster

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7 0
3 years ago
If two variables have a non-related relationship and one of the variables is changed, how will the other variable change?
PolarNik [594]

If one of the variables is changed, that tells nothing about what happens to the other one, or IF anything happens, or when, or how long it lasts. Because they are UN-RELATED. You just said so yourself.

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6 0
3 years ago
A 4 kg block is launched up a 30° ramp with an initial speed of 5 m/s. The coefficient of kinetic friction between the block and
zlopas [31]

Answer:

The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

Explanation:

Given;

mass of block, m =  4 kg

coefficient of kinetic friction, μk = 0.25

angle of inclination, θ = 30°

initial speed of the block, u = 5 m/s

From Newton's second law of motion;

F = ma

a = F/m

Net horizontal force;

∑F = mgsinθ + μkmgcosθ

a = \frac{F_{NET}}{m} = \frac{mgsin \theta + \mu_kmgcos \theta}{m} \\\\a = gsin \theta + \mu_kgcos \theta\\\\a = 9.8sin (30) + 0.25*9.8cos(30)\\\\a = 4.9 + 2.1217\\\\a = 7.022 \ m/s^2

At the  top of the ramp, energy is conserved;

Kinetic energy = potential energy

¹/₂mv² = mgh

¹/₂ v² = gh

¹/₂ x 5² = 9.8h

12.5 = 9.8h

h = 12.5/9.8

h = 1.28 m

Height of the ramp is 1.28 m

Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;

v² = u² + 2ah

v²  = 5² + 2 x 7.022 x 1.28

v²  = 25 + 17.976

v²  = 42.976

v = √42.976

v = 6.56 m/s

Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.

4 0
4 years ago
Q|C A string with linear density 0.500 g/m is held under tension 20.0 N. As a transverse sinusoidal wave propagates on the strin
julia-pushkina [17]

A string with linear density 0.500 g/m.

Tension 20.0 N.

The maximum speed  v_{y, max}

The energy contained in a section of string 3.00 m long as a function of v_{y, max}.

We are given following data for string with linear density held under tension :

μ = 0.5 \frac{g}{m}

  = 0.5 x 10⁻³ \frac{kg}{m}

T = 20 N

If string is L = 3m long, total energy as a function of v_{y, max} is given by:

E = 1/2 x μ x L x ω² x A²

  = 1/2 x μ x L x v^{2} _{y, max}

  = 7.5 x 10⁻⁴ v^{2} _{y, max}

So, The total energy as a function of  v^{2} _{y, max} = 7.5 x 10⁻⁴ v^{2} _{y, max}

Learn more about linear density problem here:

brainly.com/question/17190616

#SPJ4

3 0
2 years ago
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