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irinina [24]
3 years ago
7

E= (500.0lm) 4 (100)

Physics
1 answer:
drek231 [11]3 years ago
7 0

Answer:

didn't understand your question

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Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at t
TEA [102]

Answer: The second force applied to the doorknob

Explanation:

The formulae for torque is simple the product of the applied force and the perpendicular distance.

The greater the perpendicular force, the greater the torque assuming a constant value of force.

Applying the force at the doorknob gives for a greater distance between the force and the turning point compared to applying the force at the midpoint of the door ( which is at a shorter distance)

6 0
3 years ago
A car starts from rest at a stop light and reaches 20m/s in 3.5s. Determine the acceleration of the car
olga55 [171]

The car accelerated at around ~5.7 m/s

8 0
3 years ago
What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s
Sholpan [36]
<span>On the scale the only external forces are the man's weight acting downwards and the normal force which the scale exerts back to support his weight. So F = Ma = mg + Fs The normal force Fs (which is actually the reading on the scale) = Ma + Mg But a = 0 So Fs = Mg which is just his weight. Fs = 75 * 9.8 = 735N</span>
7 0
3 years ago
_____ is the frictional force needed to slow an object in motion
s2008m [1.1K]

Answer:

<u>Drag force</u> is the frictional force needed to slow an object in motion

Explanation:

8 0
3 years ago
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Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen
mash [69]
Since energy cannot be created nor destroyed, the change in energy of the electron must be equal to the energy of the emitted photon.

The energy of the emitted photon is given by:
E=hf
where
h is the Planck constant
f is the photon frequency
Substituting f=8.88 \cdot 10^{14}Hz, we find
E=hf=(6.6 \cdot 10^{-34} Js)(8.88 \cdot 10^{14} Hz)=5.86 \cdot 10^{-19} J

This is the energy given to the emitted photon; it means this is also equal to the energy lost by the electron in the transition, so the variation of energy of the electron will have a negative sign (because the electron is losing energy by decaying from an excited state, with higher energy, to the ground state, with lower energy)
\Delta E= -5.86 \cdot 10^{-19} J
6 0
3 years ago
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