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mario62 [17]
3 years ago
10

A car traveling initially at 7.35 m/s acceler-

Physics
1 answer:
Flura [38]3 years ago
4 0

Answer:

v_f=9,07~m/s

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes uniformly in time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

v_f=v_o+at

The car initially travels at vo=7.35 m/s and accelerates at a rate of a=0.824~m/s^2 during t=2.09 s.

The final velocity is:

v_f=7.35+0.824*2.09

\mathbf{v_f=9,07~m/s}

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Find the wavelength (in nm) of a 2.52 eV photon. a. 492.34127 nm.b. 585.88611 nm.c. 541.5754 nm.d. 418.49008 nm.
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Answer:

The correct option is (a).

Explanation:

Given that,

The energy of photon, E = 2.52 eV

We need to find the wavelength of the photon in nm. The formula for the energy of a photon is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}\\\\\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{2.52\times 1.6\times 10^{-19}}\\\\=4.93\times 10^{-7}\ m\\\\=493\ nm

The nearest option is a) i.e. 492.34127 nm.

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3 years ago
You are driving a passenger car on a two-lane highway. unless otherwise posted, the speed limit is
riadik2000 [5.3K]
The speed limit is: 55 mph
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Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant
Elena L [17]

Answer:

19 m/s

Explanation:

The complete question requires the final speed to be calculated.

Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.

For this question, first the unknown acceleration must be calculated and used to determine the final velocity

Step 1: Calculate the acceleration

a=\frac{v_{2}-v{1}}{t_{1}}

a=\frac{11-7}{8}

a=\frac{4}{8}

a=0.5 m/s^{2}

Step 2: Calculate the velocity using the acceleration calculated above

a=\frac{v_{3}-v{2}}{t_{2}}

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Why was Thompson's Plum pudding model a significant departure from previous models?
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Answer:

Thomson's cathode-ray tube experiments led him to develop the plum-pudding model, which stated that each atom had positively charged particles spread throughout its negatively charged matter. Reword the statement so it is true. ... More alpha particles were deflected than he expected.

Explanation:

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3 years ago
Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to
stiks02 [169]

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

\mu_{sB}=0.126

\mu_{sC}=0.168

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

\sum \tau_{A}=0

N(3m)-W(1.5m)=0

When solving for N we get:

N=\frac{W(1.5m)}{3m}

N=\frac{(1962N)(1.5m)}{3m}

N=981N

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

\sum F_{y}=0

-F_{By}+N_{c}=0

F_{By}=N_{c}

Next, the forces in x.

\sum F_{x}=0

-f_{sB}-f_{sC}+P_{x}=0

We can find the x-component of force P like this:

P_{x}=360N(\frac{4}{5})=288N

and finally the torques about C.

\sum \tau_{C}=0

f_{sB}(1.75m)-P_{x}(0.75m)=0

f_{sB}=\frac{288N(0.75m)}{1.75m}

f_{sB}=123.43N

With the static friction force in point B we can find the coefficient of static friction in B:

\mu_{sB}=\frac{f_{sB}}{N}

\mu_{sB}=\frac{123.43N}{981N}

\mu_{sB}=0.126

And now we can find the friction force in C.

f_{sC}=P_{x}-f_{xB}

f_{sC}=288N-123.43N=164.57N

f_{sC}=N_{c}\mu_{sC}

and now we can use this to find static friction coefficient in point C.

\mu_{sC}=\frac{f_{sC}}{N}

\mu_{sC}=\frac{164.57N}{981N}

\mu_{sB}=0.168

3 0
3 years ago
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