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valentinak56 [21]
3 years ago
6

A light string is wrapped around the edge of the smaller disk, and a 1.50 kg block is suspended from the free end of the string.

If the block is released from rest at a distance of 2.80 m above the floor, what is its speed just before it strikes the floor
Physics
1 answer:
LenaWriter [7]3 years ago
7 0

Answer: 7.41 m/s

Explanation: By using the law of of energy, kinetic energy of the brick as it falls equals the potential energy before falling.

Kinetic energy = mv²/2, potential energy = mgh

mv²/2 = mgh

v²/2 = gh

v² = 2gh

v = √2gh

Where g = 9.8 m/s², h = 2.80m

v = √2×9.8×2.8 = 7.41 m/s

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A person wants to fire a water balloon cannon such that it hits a target 100m100m away. if the cannon can only be launched at 45
vladimir2022 [97]
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3 0
3 years ago
The work function for a metal surface is 4.98 eV. What is the largest wavelength of light in nm that will produce photoelectrons
bulgar [2K]

Answer:\lambda =248.99 nm

Explanation:

Given

Work function\left ( \phi \right )=4.98\approx 1.602\times 10^{-19}\times 4.98

h=6.626\times 10^{-34} J

c=2.998\times 10^8

\phi =\frac{hc}{\lambda }

\lambda =\frac{hc}{\phi }

\lambda =\frac{6.626\times 10^{-34}\times 2.998\times 10^8}{4.98\times 1.602\times 10^{-19}}

\lambda =248.99 nm

3 0
3 years ago
Why is a flask is wider on the bottom than on the top? allows for a more precise measurement allows for better thermal equilibri
Genrish500 [490]

Answer:

allows for better thermal equilibrium

Explanation:

Due to the cone shape, most of the liquid will be closer to the bottom than the top.  The large surface area of the bottom allows for faster heating.

8 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
3 years ago
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