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.
Learn more about concave mirror here: brainly.com/question/27841226
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Experiments and investigations must be B. Repeatable.
The net force acting on the object perpendicular to the table is
∑ F[perp] = F[normal] - mg = 0
where mg is the weight of the object. Then
F[normal] = mg = (15 kg) (9.8 m/s²) = 147 N
The maximum magnitude of static friction is then
0.40 F[normal] = 58.8 N
which means the applied 40 N force is not enough to make the object start to move. So the object has zero acceleration and does not move.
Answer:
= 1647.92 N
Explanation:
Given:
length of the arm, r = 0.465 m
distance of forearm from elbow, r' = 2.15 cm = 0.0215 m
Mass of the forearm, M = 2.45 kg
Mass of the object, m = 6.55 kg
Let, the Force by bicep be,
under the motionless condition, the net moment about elbow is zero
thus,
× 0.0215 - Mg × (0.465/2) - mg × (0.465) = 0
or
× 0.0215 - 2.45 × 9.8 × (0.465/2) - 6.55 × 9.8 × (0.465) = 0
or
× 0.0215 - 5.582 - 29.848 = 0
or
= 1647.92 N
hence, the force exerted by the elbow is 1647.92 N
