Me amor

Which transformation of energy takes place when a slingshot launches a stone? (PE stands for potential energy.) A. elastic PE to

boyakko [2]

B. Elastic potential to kinetic energy

The elastic potential energy in the slingshot will be transferred to the stone as kinetic energy as the stone is launched.

6 0

2 years ago

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to start an avalanche on a mountain slope, an artillery shell is fired with an initial velocity of 290 m/s at 53.0° above the ho

kap26 [50]

So this is easy to calculate when you split the velocity into x and y components. The x component is going to equal cos(53) * 290 and the y component is going to equal sin(53)*290.

The x location therefore is 290*cos(53)*35 = 6108.4m

The y location needs to factor in the downwards acceleration of gravity too, which is 9.81m/s^2. We need the equation dist. = V initial*time + 0.5*acceleration*time^2.

This gives us d=290*sin(53)*35 + (0.5*-9.81*35^2)=2097.5m

So your (x,y) coordinates equals (6108.4, 2097.5)

5 0

3 years ago

Find the change in velocity of a 369 g hockey puck subject to the force shown below.

Elena-2011 [213]

Answer:

The change in velocity is 15.83 [m/s]

Explanation:

Using the Newton's second law we have:

ΣF = m*a

The force in the graph is 185 N, therefore:

$185=0.369*a\\Where\\a=acceleration made it by the force [m/s^2]$

$a=501.35[m/s^2]$

Now using the following kinematic equation:

$V^{2}=Vi^{2} + 2*a*(x-xi) \\where\\V=final velocity [m/s]\\Vi= initial velocity [m/s] = 0 the hockey disk is in rest when receives the hit.\\ x = Final position [m] = 0.4 m\\xi = initial position [m] = 0.15m\\$

Now replacing the values:

$V^{2}=0 + 2*501.35*(0.4-0.15)\\ \\V= 15.83[m/s]$

3 0

2 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

5 0

3 years ago

Indicate the types of forces that are involved between the solute and solvent when forming a homogeneous solution between ch3ch2

Alisiya [41]

Ion-dipole force is involved in the chemical reaction above.

7 0

2 years ago