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andrew11 [14]
3 years ago
8

A boy falls and hits his head with an impulse of 20N-s. On cement, the boy would hit for

Physics
1 answer:
ANTONII [103]3 years ago
5 0

Answer:

xjdifjfiiiffiigigififidghffigfigivig

Explanation:

bye

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Why is it important for scientist to do research?
sweet-ann [11.9K]

Answer:

Because if they dont research first they will be unprepared

Explanation:

3 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
Which instrument is launched into the atmosphere to collect pressure, temperature, humidity, wind speed, and other data?
koban [17]
C. Radiosonde is the answer

the above mentioned is not correct
5 0
3 years ago
A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?
kramer

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

Hope it helped u if yes mark me BRAINLIEST!

Tysm!

I would appreciate it!

3 0
3 years ago
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A track is traveling a. a speed of 25.0 m/s along a level road. A crate is resting on the bed of the truck, and the coefficient
Mice21 [21]

To solve this problem it is necessary to apply the concepts related to

conservation of energy, for this case manifested through work and kinetic energy.

W = \Delta KE

W = F*d

Where,

F= Force (Frictional at this case F_r = \mu N)

d= Distance

\Delta KE = \frac{1}{2} mv^2

Where,

m = mass

v = velocity

Equation both terms,

F*d = \frac{1}{2}mv^2

\mu mg *d = \frac{1}{2}mv^2

\mu g * d = \frac{1}{2}v^2

d = \frac{1}{2} \frac{v^2}{\mu g}

Replacing with our values we have that

d = \frac{1}{2} \frac{25^2}{0.65*9.8}

d = 49.05m

Therefore the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck is 49.05m

6 0
3 years ago
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