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steposvetlana [31]
2 years ago
7

Which best describes the image of a concave mirror when

Physics
1 answer:
mr_godi [17]2 years ago
5 0

Answer:

when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted

Explanation:

Let's use the constructor equation to describe the image of a concave mirror

    1 / f = 1 / p + 1q

where f is the focal length, p and q the distance to the object and the image, respectively

     1 /q = 1/f - 1/p

tell us that the image is between the focal and twice the focal, let's calculate the position of the image

for both ends

case 1, distance to the object       p = f

     

       1 / q = 1 / f -1 / f

       1 / q = 0

       q = ∞

    the image is in infinity

case2, distance to object p = 2f

     1 / q = 1 / f - 1 / 2f

     1 / q = 1 / 2f

     q = 2f

the image is twice the focal length, the object and the image are at the same point

therefore the image when the object goes from the focal length to twice the focal length the image goes from infinity to twice the focal length, this image is real and inverted

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Abnormal protrusion of the eye out of the orbit is known as
kifflom [539]

Answer:

Exophthalmos

Explanation:

Exophthalmos is a disorder which can be either bilateral or unilateral. Sometimes it is also known by other names like Exophthalmus, Excophthamia, Exobitism.

It is basically the bulging of eye anterior out of orbit which if left unattended may result in eye openings even while sleeping consequently resulting in comeal dryness and damage which ultimately may lead to blindness.

It is commonly caused by trauma or swelling of eye surrounding tissues resulting from trauma.

7 0
3 years ago
What creates an ionic bond?
natta225 [31]
Ionic bonding is the bonding between a positive metal with a negative nonmetal (metals are always positive while non metals are opposite). The meeting of a metal with a non metal creates an ionic bond.
6 0
3 years ago
Read 2 more answers
Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o
Dominik [7]

Answer:

The center of mass for the object is  y_c = 1.063 \  m from the origin

Explanation:

From the question we are told that

   The mass of the first object is  m_1 =  1.99 \  kg

   The position of first object with respect to origin y_1 =  2.99 \ m

   The mass of the second object is  m_2 =  2.96 \  kg

   The position of second object with respect to origin y_2 =  2.57 \ m

   The mass of the third object is  m_3 =  2.43  \  kg

   The position of third object with respect to origin y_3 =  0 \ m

   The mass of the fourth object is  m_3 =  3.96  \  kg

   The position of fourth object with respect to origin y_3 =  -0.502  \ m

Generally the center of mass of the object along the x-axis is  zero  because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

    y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}

=>y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96  + 2.43 + 3.96}

=>y_c = 1.063 \  m

3 0
3 years ago
From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
2 years ago
Where do you feel that you are traveling at the fastest speed when on the swing?
il63 [147K]

Answer:

C

Explanation:

I think it's C, because at that point, you are going fastest. Sorry if im wrong, hope this helps.

7 0
3 years ago
Read 2 more answers
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