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pentagon [3]
3 years ago
13

Davien and Jackson are headed to a beach that is 50km from school. At noon, Davien leaves his house that is 10 km closer to the

beach than the school is and moves at 40 km/h. Jackson starts from school at 12:30 P.M. And moves at 100 km/h. What time does Davien reach the beach? * Answer Format: Time
Physics
1 answer:
snow_tiger [21]3 years ago
5 0

Distance between Davien's school and beach, D = 50 km.

Also, Davian's house is 10 km closer to the beach than the school.

So, distance between Davien's house and beach, d = 40 km.

Speed of Davian, s = 40 km/h.

Now, time taken is given by :

t =\dfrac{Distance}{Speed}\\\\t=\dfrac{40}{40}=1\ hour

Therefore, Davien reach the beach at 1 : 30 P.M .

Hence, this is the required solution.

You might be interested in
THIS MARCIN
nekit [7.7K]

Answer:

The image is formed at a ‘distance of 16.66 cm’ away from the lens as a diminished image of height 3.332 cm. The image formed is a real image.

Solution:

The given quantities are

Height of the object h = 5 cm

Object distance u = -25 cm

Focal length f = 10 cm

The object distance is the distance between the object position and the lens position. In order to find the position, size and nature of the image formed, we need to find the ‘image distance’ and ‘image height’.

The image distance is the distance between the position of convex lens and the position where the image is formed.

We know that the ‘focal length’ of a convex lens can be found using the below formula

1f=1v−1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

f

1

=

v

1

−

u

1

Here f is the focal length, v is the image distance which is known to us and u is the object distance.

The image height can be derived from the magnification equation, we know that

Magnification=h′h=vu\text {Magnification}=\frac{h^{\prime}}{h}=\frac{v}{u}Magnification=

h

h

′

=

u

v

Thus,

h′h=vu\frac{h^{\prime}}{h}=\frac{v}{u}

h

h

′

=

u

v

First consider the focal length equation to find the image distance and then we can find the image height from magnification relation. So,

1f=1v−1(−25)\frac{1}{f}=\frac{1}{v}-\frac{1}{(-25)}

f

1

=

v

1

−

(−25)

1

1v=1f+1(−25)=110−125\frac{1}{v}=\frac{1}{f}+\frac{1}{(-25)}=\frac{1}{10}-\frac{1}{25}

v

1

=

f

1

+

(−25)

1

=

10

1

−

25

1

1v=25−10250=15250\frac{1}{v}=\frac{25-10}{250}=\frac{15}{250}

v

1

=

250

25−10

=

250

15

v=25015=503=16.66 cmv=\frac{250}{15}=\frac{50}{3}=16.66\ \mathrm{cm}v=

15

250

=

3

50

=16.66 cm

Then using the magnification relation, we can get the image height as follows

h′5=−16.6625\frac{h^{\prime}}{5}=-\frac{16.66}{25}

5

h

′

=−

25

16.66

So, the image height will be

h′=−5×16.6625=−3.332 cmh^{\prime}=-5 \times \frac{16.66}{25}=-3.332\ \mathrm{cm}h

′

=−5×

25

16.66

=−3.332 cm

Thus the image is formed at a distance of 16.66 cm away from the lens as a diminished image of height 3.332 cm. The image formed is a ‘real image’.

5 0
2 years ago
As the distance from a charged particle, "q", increases, what happens to the electric potential?
mr Goodwill [35]

As the distance from a charged particle, "q", increases, the electric potential decreases.

<h3>Electric potential between particles</h3>

The electric potential between particles is the work done in moving a unit charge from infinity to a certain point against the electrical resistance of the field.

V = Kq/r

where;

  • K is Coulomb's constant
  • q is the magnitude of the charge
  • r is the distance between the charges

Thus, from the formula above, as the distance from a charged particle, "q", increases, the electric potential decreases.

Learn more about electric potential here: brainly.com/question/14306881

#SPJ1

6 0
1 year ago
in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's
GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}

Where

  • I₁ = Intensity at distance 1 (W/m²)
  • I₂ = Intensity at distance 2 (W/m²)
  • d₁ = distance 1 from a light source (m)
  • d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}

\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}

\frac{200 }{I_{2} } = \frac{36r^{2} }{r^{2}}

I_{2} = \frac{200} {36}

I₂ = 5.55 W/m²

Hence, the sun's intensity for a planet at a distance 6r from the sun is 5.55 W/m².

Learn more about intensity of light here:

brainly.com/question/13155277

#SPJ4

3 0
1 year ago
A student thinks that any real vibration must be damped. Is the student correct? If so, give convincing reasoning. If not, give
Stels [109]

Answer:

Yes the student is correct

Explanation:

The first law of thermodynamics states that energy can neither be created nor destroyed

The second law of thermodynamics states that the entropy (disorderliness) of an isolated system always increases

Therefore, whereby energy is not supplied to maintain the orderly oscillatory motion with constant amplitude, the amplitude of the system is bound to reduce with time that is the vibration of the system must be damped

6 0
3 years ago
An Object, Start from rest w Confront Aiceleration 8m/s2 along a
shutvik [7]

Answer:

A)   v = 40 m / s, B)   v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

         

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

         v = vo + a t

         v = 0 + 8 5

         v = 40 m / s

B) the average velocity can be found with the relation

         v_average = vf + vo / 2

         v-average = 0+ 40/2

          v_average = 20 m / s

6 0
2 years ago
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