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Varvara68 [4.7K]
3 years ago
9

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk

. After 14sec, how far will it have traveled in the direction parallel to the edge of the sidewalk?
A) 0.005m
B) 0.01m
C) 1.0m
D) 9.9m​
Physics
2 answers:
ahrayia [7]3 years ago
6 0

Answer:

C) 1.0 m

Explanation:

The component of the velocity parallel to the sidewalk is:

vₓ = v cos θ

vₓ = 0.1 m/s cos 45°

vₓ = 0.0707 m/s

The distance traveled after 14 seconds is:

d = vₓ t

d = (0.0707 m/s) (14 s)

d = 0.99 m

Closest answer is C) 1.0 m.

DerKrebs [107]3 years ago
5 0

Answer:

C) 1.0m

Explanation:

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45 degrees relative to the edge of the sidewalk. After 14sec, it will have traveled 1.0m in the direction parallel to the edge of the sidewalk.

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For how long a time must a tow truck pull with a force of 550N on a stalled 1200kg car to give it a forward velocity of 2.0m/s?
horsena [70]

Answer:

4.36 seconds

Explanation:

According to the question;

  • Force is 550 N
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We are needed to find the time the car must the tow track pull the car.

  • From Newton's second law of motion;
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Rearranging the formula;

t = mv ÷ F

Thus;

Time = (1200 kg × 2.0 m/s²) ÷ 550 N

        = 4.36 seconds

Thus, the time needed to pull the car is 4.36 seconds

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3 years ago
As a United States citizen, the requirement to serve on juries if called upon is considered which of the following?
ratelena [41]

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if u are caught by the jews u die

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3 years ago
In a Rutherford scattering experiment a target nucleus has a diameter of 1.34×10-14 m. The incoming α particle has a mass of 6.6
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Answer:

E = 2.5 x 10⁻¹⁴ J

Explanation:

given,

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mass = 6.64 x 10⁻²⁷ kg

wavelength is equal to diameter

de broglie wavelength equal to diameter

         \lambda = \dfrac{h}{mv}

         1.33 \times 10^{-14}= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times v}

         v= \dfrac{6.626 \times 10^{-34}}{6.64 \times 10^{-27}\times 1.33 \times 10^{-14}}

              v = 7.5 x 10⁶ m/s

Kinetic energy is equal to

     E = \dfrac{1}{2}mv^2

     E = \dfrac{1}{2}\times 6.64 \times 10^{-27}\times (7.5\times 10^6)^2

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3 years ago
In the Bohr model of the hydrogen atom, an electron moves in a circular path around a proton. The speed of the electron is appro
blondinia [14]
In order to answer these questions, we need to know the charges on
the electron and proton, and then we need to know the electron's mass. 
I'm beginning to get the creepy feeling that, in return for the generous
5 points, you also want me to go and look these up so I can use them
in calculations ... go and collect my own straw to make the bricks with,
as it were. 

Ok, Rameses:

Elementary charge . . . . .  1.6 x 10⁻¹⁹  coulomb
                                        negative on the electron
                                        plussitive on the proton

Electron rest-mass . . . . .  9.11 x 10⁻³¹  kg


a).  The force between two charges is

      F  =  (9 x 10⁹) Q₁ Q₂ / R²

          =  (9 x 10⁹ m/farad) (-1.6 x 10⁻¹⁹C) (1.6 x 10⁻¹⁹C) / (5.35 x 10⁻¹¹m)²

          =     ( -2.304 x 10⁻²⁸) / (5.35 x 10⁻¹¹)²

          =          8.05 x 10⁻⁸  Newton .


b).  Centripetal acceleration  = 

                                               v² / r  .

                  A  =  (2.03 x 10⁶)² / (5.35 x 10⁻¹¹)

                     =      7.7 x 10²²  m/s² .

That's an enormous acceleration ... about  7.85 x 10²¹  G's !
More than enough to cause the poor electron to lose its lunch.

It would be so easy to check this work of mine ...
First I calculated the force, then I calculated the centripetal acceleration.
I didn't use either answer to find the other one, and I didn't use  "  F = MA "
either.

I could just take the ' F ' that I found, and the 'A' that I found, and the
electron mass that I looked up, and mash the numbers together to see
whether  F = M A .

I'm going to leave that step for you.   Good luck !
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