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Dvinal [7]
3 years ago
7

A car crashes into a wall with an impulse of 40,000 kg.m/s. The collision took 2 seconds. How much force did the wall exert?

Physics
1 answer:
sesenic [268]3 years ago
5 0

Answer:

i hate physics lol but its 20,000N

Explanation:

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The outer layer of bone _____.
baherus [9]

Answer:

I believe its called the periosteum , its a thin , dense membrane that contains  nerves and blood vessels .

Hoped I helped-

Sleepy~

3 0
3 years ago
The cube with 2.00m wide and 2.00m long and 2.00m high has a weight of 960.00 N what pressure does it exert
andre [41]
2m X 2m = 4m ^2
960n / 4 = 240pa
3 0
3 years ago
Calculate the energy per photon and the energy per mole of photons for radiation of wavelength (a) 200 nm (ultraviolet), (b) 150
VMariaS [17]

Answer: a. E =9.9*EXP(-19)J

1 mole E= 596178J

b. E= 1.32*EXP(-15)J, 1 mole E=795MegaJ

c. E= 1.98*EXP(-23)J

1 mole E = 11.9J

Explanation: The Energy of a photon E, the wavelength are related by

E= h*c/wavelength

h is the Planck's constant 6.6*EXP(-34)J.s

c is speed of light 3*EXP(8)m/s

h*c=1.98*EXP(-25)

Now let's solve

a. E = h*c/wavelength

= h*c/(200*EXP(-9)m

=9.9*EXP(-19)J

1 mole of a photon contian 6.022*EXP(23)photons by advogadro

Now to get the energy of 1 mole of the photon we have

9.9*EXP(-19)*6.023*EXP(23)

=596178J

b. E=h*c/150*EXP(-12)m

=1.32*EXP(-15)J

1 mole will have

1.32*EXP(-15)*6.022*EXP(23)J

=795*EXP(6)J

c. E= h*c/1*EXP(-2)m

=1.98*EXP(-23)J

1 mole of the photon will have

1.98*EXP(-23)J *6.022*EXP(23)

= 11.9J.

You will notice that the longer the wavelength of the photon the lesser the Energy it as.

NOTE: EXP represent 10^

8 0
4 years ago
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in and the atom emits a photon of lig
PSYCHO15rus [73]
Give me some answer choices and i will be happy to help
4 0
4 years ago
jack be nimble jack be quick jack jumped over the candlestick with a velocity of 5.0 m/s at an angle of 30.0 degrees to horizont
Mrrafil [7]

Answer:

No

Explanation:

The vertical component of Jack's initial velocity is:

5.0

⋅

sin

30

∘

=

5.0

⋅

1

2

=

2.5

m/s

With gravitational acceleration

9.8

m/s

2

, he will reach the highest point of his trajectory after:

2.5

9.8

≈

0.255

s

The average vertical component of his velocity in that

0.255

s

will be:

1

2

⋅

2.5

=

1.25

m/s

So the highest point of his trajectory will be:

0.255

⋅

1.25

≈

0.32

m

So he will pass approximately

7

cm

above the top of the candle.

The horizontal component of his velocity will be a constant:

5.0

⋅

cos

30

∘

=

5.0

⋅

√

3

2

≈

4.33

m/s

So Jack's trajectory will be substantially longer than it is high and he will spend little time anywhere near above the candle.

4 0
3 years ago
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