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MAVERICK [17]
3 years ago
5

Professor Frank Nabarro insists that all senior physics majors take his notorious physics aptitude test. The test is so tough th

at anyone not going on to a career in physics has no hope of passing, whereas 35% of the seniors who do go on to a career in physics still fail the test. Further, 70% of all senior physics majors in fact go on to a career in physics. Assuming that you fail the test, what is the probability that you will not go on to a career in physics? (Round your answer to four decimal places.)
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

0.5504

Explanation:

We can use Bay's theorem to solve this question

P( No physics career/fail) = P(No Physics career and fail)/P(Fail)

= P( Fail/ No physics career)×P( no physics career)/P( fail)

=\frac{0.3\times1}{(0.7\times0.35)+(0.3\times1)}

=0.5504

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A glass plate (n = 1.64) is covered with a thin, uniform layer of oil (n = 1.28). A light beam of variable wavelength from air i
Makovka662 [10]

Answer:

Explanation:

This is case of interference in thin films

for constructive interference in thin film the condition is

2μ t = (2n+1)λ/2    ;  μ is refractive index of oil , t is thickness of oil , λ is wave length of light .

2 x 1.28 x t = λ/2 , if n = 0

2 x 1.28 x t = 605 /2

t = 118.16 nm .

the minimum non-zero thickness of the oil film required = 118.16 nm.

8 0
3 years ago
if abus travelling at 20m/s is subject to steady decceleration of 5m/s².how long will it take yo come to rest.​
Likurg_2 [28]

Answer:

4 seconds

Explanation:

Given:

v₀ = 20 m/s

v = 0 m/s

a = -5 m/s²

Find: t

v = at + v₀

0 m/s = (-5 m/s²) t + 20 m/s

t = 4 s

6 0
3 years ago
A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?
stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, m_1 = 300\ g = 0.3\ kg

Initial speed of the cart, u_1=1.2\ m/s

Mass of the larger cart, m_2 = 2\ kg

Initial speed of the larger cart, u_2=0

After the collision,

Final speed of the smaller cart, v_1=-0.81\ m/s (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let v_2 is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1+m_2u_2-m_1v_1=m_2v_2

v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}

v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}

v_2=0.301\ m/s

So, the speed of the large cart after collision is 0.301 m/s.

4 0
3 years ago
A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the
Harlamova29_29 [7]

Answer:

Explanation:

Kinetic energy of block will be converted into heat energy by friction .

Heat energy produced = 1/2 m v²

= .5 x 4.8 x 1.2²

= 3.456 J

85% of energy is converted into heat energy , so heat energy produced

= .85 x 3.456 = 2.9376 J .

If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature

Q = msΔt

specific heat of iron s = 462 J / kg C

Putting the values ,

2.9376 = 4.8 x 462 x Δt

Δt = 13.24 x 10⁻⁴ ⁰C.

4 0
3 years ago
As scientists learn more what happens to scientific ideas?
blsea [12.9K]
They could change but most likely not
4 0
3 years ago
Read 2 more answers
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