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andreev551 [17]
3 years ago
14

You can carry a 50 N weight up a flight of stairs that is 3 m high in 10 s. Then you carry 40 N up two flights of stairs in 20 s

. Finally, you carry 30 N up three flights in 45 s. Is the job that took the most power the same as the job that took the most energy? Explain.
Physics
1 answer:
d1i1m1o1n [39]3 years ago
5 0
You can do 50 and 10 and carry the 30 and it would be in the same power so you’ll have the same energy
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A 1100 kg car is traveling around a flat 82.3 m radius curve. The coefficient of static friction between the car tires and the r
Novay_Z [31]

Answer:

The maximum speed of car will be 20.5m/sec

Explanation:

We have given mass of car = 1100 kg

Radius of curve = 82.3 m

Static friction \mu _s=0.521

We have to find the maximum speed of car

We know that at maximum speed centripetal force will be equal to frictional force m\frac{v^2}{r}=\mu _srg

v=\sqrt{\mu _srg}=\sqrt{0.521\times 82.3\times 9.8}=20.5m/sec

So the maximum speed of car will be 20.5m/sec

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A force is something that ___ an object to ____
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Answer:

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Put the phase of mitosis or mitosis in order from first to last
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The phases of mitosis in order from first to last are: Interphase, prophase, metaphase, anaphase, and telophase & cytokinesis.
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You measure its length as it passes. by how many millimeters do you determine the rod has contracted?
LUCKY_DIMON [66]
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4 0
4 years ago
4. There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from 320 nm to 400 nm. It is not
vampirchik [111]

The ranges of frequency are:

UVA: [7.50-9.38]\cdot 10^{14} Hz

UVB: [9.38-10.71]\cdot 10^{14} Hz

Explanation:

The relationship between frequency and wavelength for an electromagnetic wave is the following:

f = \frac{c}{\lambda}

where

f is the frequency

c=3.0\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the UVA, the range of wavelength is 320 nm - 400 nm, so

\lambda_1 = 320 \cdot 10^{-9} m\\\lambda_2 = 400 \cdot 10^{-9} m

So the corresponding frequencies are

f_1 = \frac{3\cdot 10^8}{320\cdot 10^{-9}}=9.38\cdot 10^{14} Hz\\f_2= \frac{3\cdot 10^8}{400\cdot 10^{-9}}=7.50\cdot 10^{14} Hz

For the UVB, the range of wavelength is 280 nm - 320 nm, so

\lambda_1 = 280 \cdot 10^{-9} m\\\lambda_2 = 320 \cdot 10^{-9} m

So the corresponding frequencies are

f_1 = \frac{3\cdot 10^8}{280\cdot 10^{-9}}=1.07\cdot 10^{15} Hz\\f_2= \frac{3\cdot 10^8}{320\cdot 10^{-9}}=9.38\cdot 10^{14} Hz

So the ranges of frequency are:

UVA: [7.50-9.38]\cdot 10^{14} Hz

UVB: [9.38-10.71]\cdot 10^{14} Hz

Learn more about waves, frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

3 0
3 years ago
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