Answer:
The two rays, CY and DM are diverging rays and when extended behind the mirror, they appear to intersect each other at point M'. Therefore, the properties of the images formed here are formed behind the mirror, between the pole and principal focus (f), the images are diminished and are virtual and erect.
Explanation:
<h2>Spherical Mirrors</h2>
- There are two kinds of spherical mirrors, concave and convex.
- The focal point (F) of a concave mirror is the point at which a parallel beam of light is "focussed" after reflection in the mirror. ...
- The focal length (f) and radius of curvature (R) are defined in the diagram at the right.
<h3>hope it helps and thanks for following </h3><h2>please give brainliest </h2>
Answer:
<h3>0.329m/s</h3>
Explanation:
According to law of conservation of momentum, the momentum of the object before collision is equal to that of the object after collision. Using the formula
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses of the object
u1 and u2 are the respective initial velocities
v is the final velocity
Given
m1 = 0.190kg
u1 = 1.5m/s
m2 = 0.305kg
u2 = -0.401m/s
Substitute
0.19(1.5)+(0.305)(-0.401) = (0.19+0.305)v
0.285 - 0.122305 = 0.495v
0.162695 = 0.495v
v = 0.162695 /0.495
v = 0.329m/s
<em>Hence their velocities after collision is 0.329m/s in the positive x direction</em>
<em></em>
Answer:
Correct answer: F = 214.56 N
Explanation:
Given:
V₀ = 0 m/s initial speed
V = 44.7 m/s speed after t = 29 seconds
m = 96 kg
F = ? horizontal force
Taking off an airplane is a uniformly accelerated motion to which the formula applies:
V = V₀ + a t and V₀ = 0 m/s
V = a t ⇒ a = V / t = 44.7 / 20 = 2.235 m/s²
a = 2.235 m/s²
the horizontal force is calculated using the formula:
F = m · a = 96 · 2.235 = 214.56 N
F = 214.56 N
God is with you!!!
Hi there!
To find the appropriate force needed to keep the block moving at a constant speed, we must use the dynamic friction force since the block would be in motion.
Recall:
The normal force of an object on an inclined plane is equivalent to the vertical component of its weight vector. However, the horizontal force applied contains a vertical component that contributes to this normal force.
We can plug in the known values to solve for one part of the normal force:
N = (1)(9.8)(cos30) + F(.5) = 8.49 + .5F
Now, we can plug this into the equation for the dynamic friction force:
Fd= (0.2)(8.49 + .5F) = 1.697 N + .1F
For a block to move with constant speed, the summation of forces must be equivalent to 0 N.
If a HORIZONTAL force is applied to the block, its horizontal component must be EQUIVALENT to the friction force. (∑F = 0 N). Thus:
Fcosθ = 1.697 + .1F
Solve for F:
Fcos(30) - .1F = 1.697
F(cos(30) - .1) = 1.697
F = 2.216 N
