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Anika [276]
2 years ago
4

A concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face o

f the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.
Physics
1 answer:
Trava [24]2 years ago
5 0

Answer:

D.

Inverted and reduced

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A cylinder is sliced in half along its diagonal. Determine the location of the center of mass and the inertia properties relativ
Black_prince [1.1K]

Answer:

See the explanation below

Explanation:

To better understand this problem, a cylinder sketch is attached before and after the cut, we see that after the cut, the shape of this resembles that of a right triangle.

We can find, the centroid in the xy plane, knowing that the centroid for a triangle is located a third of its base.

In the z axis there is no displacement of the centroid.

3 0
2 years ago
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
OlgaM077 [116]

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F  mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

8 0
2 years ago
A flute player hears four beats per second when she compares her note to an 880 Hz tuning fork (note A). She can match the frequ
ludmilkaskok [199]

Answer:

884Hz

Explanation:

Beats is the absolute difference between two frequencies therefore

Beats = f1-f2

4=f1-880

F1=880+4

F1=884Hz

7 0
2 years ago
Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i
SCORPION-xisa [38]

Answer:

The value is  r =  5.077 \  m

Explanation:

From the question we are told that

   The  Coulomb constant is  k =  9.0 *10^{9} \  N\cdot  m^2  /C^2

   The  charge on the electron/proton  is  e =  1.6*10^{-19} \  C

    The  mass of proton m_{proton} =  1.67*10^{-27} \  kg

    The  mass of  electron is  m_{electron } =  9.11 *10^{-31} \ kg

Generally for the electron to be held up by the force gravity

   Then    

       Electric force on the electron  =  The  gravitational Force

i.e  

            m_{electron} *  g  = \frac{ k *  e^2  }{r^2 }

         \frac{9*10^9 *  (1.60 *10^{-19})^2  }{r^2 }  =     9.11 *10^{-31 }  *  9.81

         r =  \sqrt{25.78}

         r =  5.077  \  m

7 0
2 years ago
A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f
lys-0071 [83]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 1.94 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 6 N

d = 0.4 m

\theta = 36 degree

so we will have

W = 6 \times 0.4 cos36

W = 1.94 J

7 0
2 years ago
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