The distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
<h3>What is concave mirror?</h3>
A concave mirror has a reflective surface that is curved inward and away from the light source.
Concave mirrors reflect light inward to one focal point and it usually form real and virtual images.
<h3>
Object distance of the concave mirror</h3>
Apply mirrors formula as shown below;
1/f = 1/v + 1/u
where;
- f is the focal length of the mirror
- v is the object distance
- u is the image distance
when image height = object height, magnification = 1
u/v = 1
v = u
Substitute the given parameters and solve for the distance of the object from the mirror's vertex
1/f = 1/v + 1/v
1/f = 2/v
v = 2f
v = 2(19.5 cm)
v = 39 cm
Thus, the distance of an object from the mirror's vertex if the image is real and has the same height as the object is 39 cm.
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A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.
<h3>Why does the first ball comes to rest after collision ?</h3>
Let m be the mass of the two identical balls.
u1 = velocity before the collision of ball 1
u2 = 0 = velocity of second ball that is at rest
v1 and v2 are the velocities of the balls after the collision.
From the conservation of momentum,
∴ mu1 + mu2 = mv1 + mv2
∴ mu1 = mv1 + mv2
∴ u1 = v1 + v2
In an elastic collision, the kinetic energy of the system before and after collision remains same.
∴ 
∴ 
∴ 
₁
₂ = 0
- It is impossible for the mass to be zero.
- Because the second ball moves, velocity v2 cannot be zero.
- As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>
An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.
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Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative
8 ∙ 10^-4 / 2 ∙ 10^2 = (8/2) ∙ ((10^-4)*(10^-2)) = <span>4 ∙ 10^-6</span>
No, the object's displacement and distance travelled will be equal, but since the initial position is unknown, the object's position might not match up with its displacement and distance travelled.
We cannot assert that the displacement or distance equals the position because the initial position is not provided. We could reach a different conclusion if the starting position had been zero because the distance from zero is equal to the position.
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