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4 years ago

When an object is getting closer to us , it emits redder light . True or False

Physics

2 answers:

Andrei [34K]4 years ago

8 0

True I think.............

ivanzaharov [21]4 years ago

7 0

True
............... thats true

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A 51.7-kg hiker ascends a 43.2-meter high hill at a constant speed of 1.20 m/s. Determine the work done in climbing the hill.

iren [92.7K]

W=m/d. w=51.7\1.20 W=43.083333333J

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2 years ago

A heavy log will float on water

posledela

Answer:

A it is less dense than water

Explanation:

The density, of a substance is its mass per unit volume

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4 years ago

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A 1 cm2 silicon solar cell has a saturation current of 10A and is illuminated with sunlight yielding a short-circuit photocurren

Yuki888 [10]

Answer:

The answer is "13% and 83%"

Explanation:

The highest power is generated for:

$\to \frac{dP}{dV_a}=0= I_s(e^{\frac{V_m}{V_t}} -1)-I_{Ph} +\frac{V_m}{V_t} I_s e^{\frac{V_m}{V_t}}$

Where voltage, $V_m$ , was its calculated maximum power wattage. This same following transcend national stable coins the power output:

$\to V_m=V_t \ In \frac{1+\frac{I_{ph}}{I_s}}{1+\frac{V_{m}}{V_t}}$

The accompanying consecutive values of $V_m$ are done utilizing iteration and an initial value of 0.5 V:

$\to V_m = 0.5,\ 0.542,\ 0.540 \ V$

Efficacy is equivalent to:

$\eta = |\frac{V_{m} I_{m}}{p_{in}}| =\frac{0.54 \times 0.024}{0.1}=1.3\%$

Currently, $I_m$ was calculated with the formula of voltage $V_m$ (4.6.1) as well as the sun is presumed to produce $100 \frac{mW}{cm^2}$ of power.

The fulfillment factor is equal to:

fill factor $= \frac{V_{m} I_{m}}{V_{oc}I_{sc}}= \frac{0.54 \times 0.024}{0.62 \times 0.025}=83\%$

where open circuit tension of equations (4.6.1) and I = 0. is calculated. The current is equal to the total current of the photo.

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3 years ago

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined

MrRa [10]

Incomplete question as the car's speed is missing.I have assumed car's speed as 6.0m/s.The complete question is here

An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 6.00 kN, and the radius of the circle is 15.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v 6.0m/s

Answer:

$F_{B}=-5755N$

Explanation:

Set up force equation

∑F=ma

∑F=W+FB

$\frac{mv^{2} }{R}=W+F_{B}\\ F_{B}=\frac{mv^{2} }{R}-W\\F_{B}=\frac{(W/g)v^{2} }{R}-W\\F_{B}=\frac{(6000N/9.8m/s^{2} )(6m/s)^{2} }{(15m)}-6000N\\F_{B}=-5755N$

The minus sign for downward direction

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3 years ago

Goal posts at the ends of football fields are padded as a safety measure for players who might run into them. How does thick pad

dimaraw [331]

When looking at it from a physics standpoint, the thick padding decreases the amount of force that a player experiences. Force equals mass times acceleration where acceleration is distance over time. The thick padding increases the amount of time the player makes contact with the post which decreases force.

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3 years ago

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