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Lana71 [14]
3 years ago
11

What kind of image is formed by an object that is placed beyond the center of curvature on the principal axis of a concave mirro

r?
Physics
2 answers:
Delicious77 [7]3 years ago
8 0
As it is outside the focal point it must be real.Real images must be inverted.
As it is beyond the centre of curvature it is also beyond 2F which means that the image is inside the centre of curvature ( between F and 2F from the mirror ) As the image is closer to the mirror than the object it must be diminished in size.

Hope this helps you :)
bogdanovich [222]3 years ago
6 0

The answer for PLATO users is D

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Normal force is exerted _____to the surface of an object.
Elanso [62]

The normal force is always (underline, bold) is always perpendicular to the surface an object is sitting on. If the object is on an inclined plane, then the normal will not be vertical but it will be perpendicular to the angle of the incline.

The diagram below (left) shows a normal force (GH) that is not vertical, but it is perpendicular to the surface. The object on the right is the more usual normal a mass on a table top.

The vertical line on the right is the normal and it points up.

3 0
3 years ago
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<h3>B. True</h3>

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3 years ago
John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0
cestrela7 [59]

Answer:

(a) The horizontal ground reaction force  F_{g,x}=325\, N

(b)  The vertical ground reaction force F_{g,y}=696\, N

(c)   The resultant ground reaction force F_g=768\, N

Explanation:  

Given

John mass , m = 65 kg

Horizontal acceleration , a_x= 5.0 \frac{m}{s^{2}}

Vertical acceleration , a_y=0.9 \frac{m}{s^{2}}

(a) Using Newton's 2nd law in horizontal direction

F_{g,x}=ma_x

=>F_{g,x}=65\times 5\, N=325\, N

Thus the horizontal ground reaction force  F_{g,x}=325\, N

(b) Using Newton's 2nd law in vertical direction

F_{g,y}-mg=ma_y

=>F_{g,y}=mg+ma_y

=>F_{g,y}=65\times (9.81+0.9)\, N=696\, N

Thus the vertical ground reaction force F_{g,y}=696\, N

(c)  Resultant ground reaction force is

F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}

=>F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N

=>F_g=768\, N

Thus  the resultant ground reaction force F_g=768\, N

5 0
3 years ago
A research submarine has a 30-cm-diameter window that is 8.1 cm thick. The manufacturer says the window can withstand forces up
malfutka [58]

The pressure at a certain depth underwater is:

P = ρgh

P = pressure, ρ = sea water density, g = gravitational acceleration near Earth, h = depth

The pressure exerted on the submarine window is:

P = F/A

P = pressure, F = force, A = area

The area of the circular submarine window is:

A = π(d/2)²

A = area, d = diameter

Set the expressions for the pressure equal to each other:

F/A = ρgh

Substitute A:

F/(π(d/2)²) = ρgh

Isolate h:

h = F/(ρgπ(d/2)²)

Given values:

F = 1.1×10⁶N

ρ = 1030kg/m³ (pulled from a Google search)

g = 9.81m/s²

d = 30×10⁻²m

Plug in and solve for h:

h = 1.1×10⁶/(1030(9.81)π(30×10⁻²/2)²)

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5 0
3 years ago
A private aviation helicopter's main rotor blades rotate at approximately
Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity \omega a helicopter's main rotor blades:

\omega=450 rpm=450 \frac{rev}{min}

However, we are asked to express this \omega in the International Systrm (SI) units. In this sense, the SI unit for time is second (s):

\omega=450 \frac{rev}{min} \frac{1 min}{60 s}

\omega=7.5 \frac{rev}{s}

4 0
3 years ago
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