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alexdok [17]
3 years ago
9

1. in which of the following cases is the wavelength of the emitted radiation greatest? electron jumps from third orbit to secon

d orbit. electron jumps from third orbit to first orbit. electron jumps from fourth orbit to first orbit. electron jumps from fifth orbit to second orbit.
Physics
1 answer:
Yuki888 [10]3 years ago
7 0
The more energy orbits the radiation jumps the more energy it has. So if the frequency stays the same each time then the wavelength will get longer if there is more energy.

In this case the situation in which the radiation jumps the most energy orbits is when: the electron jumps from the fourth orbit to the first orbit. This will emit the longest wavelength
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If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w
Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

Thank you for your question. Please don't hesitate to ask in Brainly your queries. 
5 0
3 years ago
Why are the orbits of planets only nearly circular and not perfectly circular?
Galina-37 [17]

Explanation:

this is my answer this is helpful for you

7 0
2 years ago
A car enters a level, unbanked semi-circular hairpin turn of 100 m radius at a speed of 28 m/s. The coefficient of friction betw
meriva

Answer:

As  28m/s = 28m/s

Explanation:

r = the radius of the curve

m =  the mass of the car

μ = the coefficient of kinetic friction

N = normal reaction

When rounding the curve, the centripetal acceleration is

a = \frac{v^{2}}{r}

therefore

\mu mg = m \frac{v^{2}}{r} \\\\ \mu =  \frac{v^{2}}{rg}

v = \sqrt{\mu rg}

\mu = \sqrt{0.8 \times 100\times9.8} \\\\= 28m/s

As  28m/s = 28m/s

8 0
3 years ago
- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7
frez [133]

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b=   -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+  7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

6 0
3 years ago
A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n
Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

N sin\theta = \frac{mv^2}{R}

N cos\theta = mg

so we have

v = \sqrt{Rg tan\theta}

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0
3 years ago
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