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5. In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which sl

pishuonlain [190]

Answer:

sorry about this question

5 0

2 years ago

If you put 120 volts of electricity through a pickle, the pickle will smoke and start glowing orange-yellow. The light is emitte

Alona [7]

Answer:

2.11eV

Explanation:

We know that speed of light is it's wavelength times frequency.

$\therefore f=v/\lambda\\=(3\times10^8m/s)/(589mm\times1m/1\times 10^9nm)\\=5.09\times10^1^4s^-1 \ or \ 5.09\times10^1^4Hz$

Planck's constant is $6.626\times 10^3^4Js$

The energy gap is calculated by multyplying the light's frequency by planck's constant:

$E_c=5.09\times10^1^4s^-^1\times 6.626\times10^-^3^4Js\\\\=3.37\times 10^-^1^9J \ \ \ \ \ \ #1eV=1.06\times 10^-^1^9J\\\\=2.11eV$

Hence, the energy gap is 2.11eV

4 0

3 years ago

A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I

Anuta_ua [19.1K]

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = $\frac{4\times \pi^2\times r^3}{GM}$ where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

8 0

3 years ago

A carrot hangs from the ceiling by a rope,

valentina_108 [34]

Answer: Diagram B

Explanation:

A free body diagram shows the forces acting on an object in a certain scenario.

In this scenario there are two forces acting on the carrot: the Tension force (Ft) from the rope that the carrot is hanging from and Gravitational force(Fg) which is pulling the carrot to the Earth.

The diagram depicting this is diagram B.

7 0

2 years ago

Read 2 more answers

A 150kg motorcycle starts from rest and accelerates at a constant rate along a distance of 350m. The applied force is 250N and t

notsponge [240]

A) The net force on the motorbike is 205.9 N

B) The acceleration of the motorbike is $1.37 m/s^2$

C) The final speed is 5.2 m/s

D) The elapsed time is 3.80 s

Explanation:

A)

There are two forces acting on the motorbike:

- The applied force, F = 250 N, forward

- The frictional force, $F_f$ , backward

The frictional force can be written as

$F_f = \mu mg$

where

$\mu=0.03$ is the coefficient of kinetic friction

$m=150 kg$ is the mass of the motorbike

$g=9.8 m/s^2$ is the acceleration of gravity

Therefore the net force is given by

$\sum F = F - F_f = F - \mu mg$

And substituting, we find

$\sum F=250 - (0.03)(150)(9.8)=205.9 N$

2)

The acceleration of the motorbike can be found by using Newton's second law, which states that the net force is equal to the product between mass and acceleration:

$\sum F = ma$