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andrew-mc [135]
3 years ago
8

Plz help me im really confused. The questions are in the second pic and the vehicles are in the first the fetures are in the las

t 3 the check mars are just to see which data for that vehicle is safer. The check marks are just to see which data is safer for that car.

Physics
1 answer:
kirill115 [55]3 years ago
5 0

Answer:

1.ima say car 2 is safer than car 1.

2.airbags: airbags make it safer because if you get in a reck they will protect you againsed more harm then if you didnt have airbags.

breaking distance: it takes less distance to stop so if there was a reck up the streat you could stop and turn around so you wouldnt get into that wreck.

accrleration: it can make you go faster so incase someoe behind you is speading you could speadup and turn into a parking lot and you wouldent get hurt.

3. car 2 is safer because it has breaking distance and it can stop you way before you get into a wreck than car 1 that doesnt have that ability

Explanation: ima award myself brainliest. Dont ask why i just answerd my own question on here. Yay

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Do frictional forces act in the same direction or in the opposite direction to the applied force?
Evgesh-ka [11]

Frictional forces act in the direction opposite to the MOTION. That direction could be the same OR opposite to applied force.

-- If you push a loaded heavy wagon from behind, trying to get it going faster, friction is acting against you, opposite to your force.

-- If you push a loaded rolling heavy wagon from in front, trying to make it slow down, friction is acting with you, in the same direction as your force.

-- Opposite to the motion both times.

3 0
3 years ago
A blue-green photon (λ = 488 nm ) is absorbed by a free hydrogen atom, initially at rest. What is the recoil speed of the hydrog
Natalka [10]

Answer:

The recoil speed is 2.207\times 10^{4} m/s

Solution:

Wavelength of a blue-green photon, \lambda_{BG} = 488 nm = 488\times 10^{- 9} m

Now, the energy associated with the blue-green photon:

E_{BG} = \frac{hc}{\lambda_{BG}}

where

h = Planck's constant

C = speed of light ion vacuum

E_{BG} = \frac{6.626\times 10^{- 34}\times 3\times 10^{8}}{488\times 10^{- 9}}

E_{BG} = 4.07\times 10^{- 19} J

Also, we know that the recoil speed can be calculated by the KInetic energy which is equal to the Energy of the blue-green photon:

KE_{H} =\frac{1}{2}m_{p}v_{H}

where

v_{H} = velocity of Hydrogen atom

m_{p} = 1.67\times 10^{- 27} kg = mass of H-atom

Now,

KE_{H} =\frac{1}{2}m_{p}(v_{H})^{2}

4.07\times 10^{- 19} =\frac{1}{2}\times 1.67\times 10^{- 27}\times (v_{H})^{2}

v_{H} = \sqrt(4.87\times 10^{8}) = 2.207\times 10^{4} m/s

7 0
3 years ago
Can someone please help me answer 14 and 15??
ziro4ka [17]
14. The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth's atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.

15.
About 5.4 billion years from now, the sun will have exhausted all of its hydrogen. The sun's core will get really hot and dense, thus shrinking; however, the outer region of the sun will expand and grow. ... Even if the expanding dying sun doesn't reach Earth, the sun's high temperatures will completely burn the planet.
7 0
4 years ago
The nature-nurture question asks:
beks73 [17]
The answer is D. How do your environment and your genes interact to make you who you are
6 0
3 years ago
Read 2 more answers
In the Bohr model of the hydrogen atom, an electron in the 7th excited state moves at a speed of 4.47 104 m/s in a circular path
tekilochka [14]

Answer:

I = 4.38 x 10⁻⁷ A

Explanation:

Given,

speed = 4.47 x 10⁴ m/s

radius of the circular path, r = 2.59 x 10⁻⁹ m

effective current = ?

The current represented by the orbiting electron is equal to

I = q \times f

f is the frequency

q is charge of electron

we know,

f = \dfrac{v}{2\pi r}

f = \dfrac{4.47\times 10^{4}}{2\pi\times 2.59\times 10^{-9}}

f = 2.74 x 10¹² Hz

now,

I = 1.6 \times 10^{-19}\times 2.74\times 10^{12}

I = 4.38 x 10⁻⁷ A

Hence, the effective current associated with the orbiting electron is equal to I = 4.38 x 10⁻⁷ A

7 0
4 years ago
Read 2 more answers
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