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zloy xaker [14]
2 years ago
13

FONTE: JBC w our tree to eat por Pirage B097 VO DED expleyo 2. How do you feel about it? (Are you happy? A little bit sad? Conte

nted? Worried? etc.) Why? Suggested Time Allotment: (30 minutes)​
Physics
1 answer:
xxTIMURxx [149]2 years ago
5 0

Answer:

sorry kilqngqn kolqng ng point

Explanation:

sorry tlaga please pa heart nlang please

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An electric grinder uses a grinding wheel
luda_lava [24]
(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s² 

<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>

<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
3 0
2 years ago
Read 2 more answers
if you use the compound pulley, you will need to pull twice the distance but with less force. the force you need is equal to one
algol13

Answer:

F= 25/2 = 12.5N

Explanation:

When you use a compound pulley the force required depends on the mechanical advantage of the compound pulley. This is known as rate of loss of distance or the ratio of the force to the load.

M.A = Effort distance /Load distance. OR M.A = Load/Effort

6 0
3 years ago
While spending the weekend in your cabin, you burn wood in your pot-bellied stove to heat a kettle of water for tea.
Olegator [25]

Answer: 1 = Heat

2=gas

3=it gets hot enough to boil because the metal conducts the heat into the water to heat it up and eventually boil.

Explanation: its common sense

3 0
2 years ago
What is the average speed of a car traveling for 1000 miles in 6 hours
nlexa [21]

You would divide 6 into 1000 so your answer is B. 166

8 0
2 years ago
In a photoelectric effect experiment, electrons emerge from a copper surface with a maximum kinetic energy of 1.10 eV when light
Sladkaya [172]

Answer:A) 220 nm

Explanation:

Given

Maximum Kinetic Energy K.E.=1.10 eV

Work Function W=4.65 eV

from Einstein Equation

h\mu =W+K.E.

h\cdot \frac{c}{\lambda }=W+K.E.

h\cdot \frac{c}{\lambda }=4.65+1.10

6.626\times 10^{-34}\cdot \frac{3\times 10^8}{\lambda }=5.75

1 eV=1.6\times 10^{-19} J

thus 5.75 eV=9.2\times 10^{-19} J

\lambda =\frac{6.626\times 10^{-34}\time 3\times 10^8}{9.2\times 10^{-19}}

\lambda =216.06 nm\approx 220 nm

6 0
3 years ago
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