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vaieri [72.5K]
3 years ago
12

A cabbie is trying to stop when he notices a fare is whistling them over. The

Physics
1 answer:
liberstina [14]3 years ago
7 0
  • K.E=18750J
  • Mass=m=2100kg
  • Velocity=v

\boxed{\sf K.E=\dfrac{1}{2}mv^2}

\\ \sf\longmapsto 18750=\dfrac{1}{2}2100v^2

\\ \sf\longmapsto 18750=1050v^2

\\ \sf\longmapsto v^2=\dfrac{18750}{1050}

\\ \sf\longmapsto v^2=17.85m^2

\\ \sf\longmapsto v=\sqrt{17.85}

\\ \sf\longmapsto v=4.1m/s

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Calculate the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.
Harrizon [31]

1.549×10-19lJ is the energy of a photon emitted when an electron in a hydrogen atom undergoes a transition from =7 to =1.

The equation E= hcE =hc, where h is Planck's constant and c is the speed of light, describes the inverse relationship between a photon's energy (E) and the wavelength of light ().

The Rydberg formula is used to determine the energy change.

Rydberg's original formula used wavelengths, but we may rewrite it using units of energy instead. The result is the following.

aaΔE=R(1n2f−1n2i) aa

were

2.17810-18lJ is the Rydberg constant.

The initial and ultimate energy levels are ni and nf.

As a change of pace from

n=5 to n=3 gives us

ΔE

=2.178×10-18lJ (132−152)

=2.178×10-18lJ (19−125)

=2.178×10-18lJ×25 - 9/25×9

=2.178×10-18lJ×16/225

=1.549×10-19lJ

Learn more about Rydberg formula here-

brainly.com/question/13185515

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8 0
2 years ago
The 0.15kg baseball has a speed of v=30 m/s just before it is struck by the bat. It then travels along the trajectory shown befo
notka56 [123]

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

The mass of the baseball, m = 0.15 kg

The speed at which it moves, v = 30 m/s

Time at which the baseball was in contact with the bat, t = 0.75 ms

t = 0.75/1000 s

t  = 0.00075 s

The impulsive force is given by the formula:

F=\frac{mv}{t}

Substitute m = 0.15 kg, v = 30, and t = 0.00075s into the formula above:

F=\frac{0.15 \times 30}{0.00075} \\\\F=6000N

The magnitude of the average impulsive force imparted to the ball if it is in contact with the bat is 6000 N

Learn more here: brainly.com/question/25892144

4 0
2 years ago
How many dots should be drawn in the lewis dot structure for an atom of sodium A)19 B)8 C)4 D)1
Stells [14]
Na is in the first column on the periodic table so therefore it would have 1 valence electron
D 1
6 0
3 years ago
What is the molar mass of a gas if 1.30g of the gas has a volume of 245mL at STP? ...?
padilas [110]
First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

PV= nRT
PV= mRT/MM

1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER
5 0
2 years ago
A 5.00 kg crate is suspended from the end of a short vertical rope of negligible mass. An upward force F(t) is applied to the en
Brut [27]

Answer:

75 N

Explanation:

In this problem, the position of the crate at time t is given by

y(t)=2.80t+0.61t^3

The velocity of the crate vs time is given by the derivative of the position, so it is:

v(t)=y'(t)=\frac{d}{dt}(2.80t+0.61t^3)=2.80+1.83t^2

Similarly, the acceleration of the crate vs time is given by the derivative of the velocity, so it is:

a(t)=v'(t)=\frac{d}{dt}(2.80+1.83t^2)=3.66t [m/s^2]

According to Newton's second law of motion, the force acting on the crate is equal to the product between mass and acceleration, so:

F(t)=ma(t)

where

m = 5.00 kg is the mass of the crate

At t = 4.10 s, the acceleration of the crate is

a(4.10)=3.66\cdot 4.10 =15.0 m/s^2

And therefore, the force on the crate is:

F=ma=(5.00)(15.0)=75 N

7 0
3 years ago
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