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LenaWriter [7]
3 years ago
5

A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a rac

e.
After running 45 meters, how far is she from the beginning of the track?


10 meters

35 meters

45 meters

55 meters

Physics
2 answers:
liberstina [14]3 years ago
6 0
55
yes it is this 55 yes yes yes
shusha [124]3 years ago
3 0

Answer : The correct option is (D).

Explanation :

Given that,

A track begins at 0 meters and has a total distance of 100 meters. Juliet starts at the 10-meter mark while practicing for a race.

We have to find her position after she runs 45 meters.

From the attached figure,

Let A is the position of Juliet. O is the initial point such that OA = 10 m, AB = 45 m and OP = 100 m.

So, using simple mathematics, it is clear that the position of Juliet after running 45 meters will be 55 m. It is OB in the figure.

So, the correct option is (D) " 55 meters ".

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Complete the ray diagram and label incident ray, refracted ray, angle of incidence, and angle of refraction ​
Novay_Z [31]

Answer:

Solution

verified

Verified by Toppr

(a) The labelled diagram is shown.

(b) The refractive index of diamond is 2.42. Refractive index of diamond is the ratio of the speed of light in air to the speed of light in diamond.i.e.,

μ=  

Speedoflightindiamond

Speedoflightinair

​

 

and, the ratio of these velocities is 2.42. i.e., This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air. In other words, the speed of light in diamond is  

1/2.42

times the speed of light in vacuum.

Explanation:

a) Draw and label the diagram given :

  (i) Incident ray

  (ii) Refracted ray

  (iii) Emergent ray

  (iv) Angle of reflection

  (v) Angle of deviation

  (v) Angle of emergence

(b) The refractive index of diamond is 2.42. What is the meaning of this statement in relation to speed of light?

4 0
3 years ago
Prove that the weight of an object on moon is 1/6th of that on earth​
Elena L [17]

Answer:

The mass of moon is 1/100 times and its radius 1/4 times that of earth. As a result, the gravitational attraction on the moon is about one sixth when compared to earth. Hence, the weight of an object on the moon is 1/6th its weight on the earth.

4 0
3 years ago
A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

Electric field, E = 82 kN/C

a) Charge density,σ = ?

  using expression of charge density

 σ = E x ε₀

ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

 σ = 82 x 10³ x 8.85 x 10⁻¹²

 σ = 725.7 x 10⁻⁹ C/m²

 σ = 725.7 nC/m²

change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

   Q = σ  A

   Q = 725.7 x 10⁻⁹ x 0.55²

   Q = 219.52 nC

Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

7 0
4 years ago
A car is rounding a 100-m-radius curve at 25 m/s.What is the minimum possible coefficient of static friction between the tires a
Crazy boy [7]

Answer:

The minimum possible coefficient of static friction between the tires and the ground is 0.64.

Explanation:

if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :

Fc = f

m×(v^2)/(R) = μ×m×g

    (v^2)/(R) = g×μ

               μ = (v^2)/(R×g)

                  =  ((25)^2)/((100)×(9.8))

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Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.

4 0
3 years ago
I have two of these so if you think this is easy then another free ten points in my profile!
Agata [3.3K]

Answer:

z

Explanation:

x repersents a new moon and the others repersent quarter moons

(x is a new moon because new moons are often the phase when the moon is close to earths sun)

7 0
3 years ago
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