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RideAnS [48]
3 years ago
5

please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0
\frac{total distance}{ total time}

\frac{72m}{36s} = 2m/s

Same thing for the rest 
 2) 5km/h
 3)360km/h

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What is the maximum value of the magnetic field at a distance of 2.5 m from a light bulb that radiates 100 W of single-frequency
Anvisha [2.4K]

Answer:

1.04\times 10^{-7} T

Explanation:

IP  = Power of the bulb = 100 W

r  = distance from the bulb = 2.5 m

I = Intensity of light at the location

Intensity of the light at the location is given as

I = \frac{P}{4\pi r^{2}}

I = \frac{100}{4(3.14) (2.5)^{2}}

I = 1.28 W/m²

B_{o} = maximum magnetic field

Intensity is given as

I = \frac{B_{o}^{2}c}{2\mu _{o}}

1.28 = \frac{B_{o}^{2}(3\times 10^{8})}{2(12.56\times 10^{-7})}

B_{o} = 1.04\times 10^{-7} T

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3 years ago
What happened to the maximum height of consecutive swings
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Answer:

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Explanation:

3 0
3 years ago
A swimmer is capable of swimming 0.42 m/s in still water. part a if she aims her body directly across a 66-m-wide river whose cu
satela [25.4K]
<span>If the swimmer is swimming perpendicular to the current, it will take her 66m / 0.42 m/s = 157.14 seconds to cross the river. At the same time, the current will be taking her downstream at a rate of 0.32 m/s. So, when she reaches the opposite bank, her total downstream distance traveled will have been 0.32*157.14 = 50.28 meters.</span>
3 0
3 years ago
A machine can never be 100% efficient because some work is always lost due to .
klasskru [66]
Friction
Hope it helped
5 0
3 years ago
Read 2 more answers
nert xenon actually forms many compounds, especially with highly electronegative fluorine. The ΔH o f values for xenon difluorid
loris [4]

Answer:

For Xenon fluoride, the average bond energy is 132kj/mol

For tetraflouride,the average bond energy is 150.5kj/mol.

For hexaflouride, the average bond energy is 146.5 kj/mol

Explanation:

For xenon fluoride

105/2 = 52.5

For F-F

159/2 = 79.5

Average bond energy of Xe-F = 79.5 + 52.5 = 132kj/mole

For tetraflouride

284/4 = 71

For F-F

159/2 = 79.5

Average bond energy = 79.5 + 71 = 150.5kj/mol

For hexaflouride

402/6 = 67

F-F = 159/2 = 79.5

Average bond energy = 67 + 79.5 = 146.5kj/ mol

3 0
3 years ago
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