Answer:

They are reflected back at the same angle they came in

Explanation:

When parallel light rays hit a convex mirror they reflect outwards and travel directly away from an imaginary focal point (F). Each individual ray is still reflecting at the same angle as it hits that small part of the surface.

**Answer:**

x = 10.53 m

**Explanation:**

Let's analyze this problem a bit, the time that the cowboy must take to fall must be the time that the horse takes to arrive

Let's start by looking for the cowboy's time, which starts from rest and the point where the chair is is y = 0

y =y₀ + v₀ t - ½ g t²

0 = y₀ - ½ g t²

t =

we calculate

t = √(2 3.22 / 9.8)

t = 0.81 s

the horse goes at a constant speed

x = t

x = 13 0.81

x = 10.53 m

this is the distance where the horse should be when in cowboy it is left Cartesian

**Answer:**

The object fell from about 38.14 meters

**Explanation:**

We can use the formula for displacement under accelerated motion due to gravity to find the velocity of the object had 30 m before hitting the ground :

Now, knowing this velocity, we can find the time it took to fall from the initial position to 30 m before hitting the ground:

And now we can find what is the total distance covered in 1.5 s plus 1.29 seconds for this free falling object:

Answer:

a = dv/ dt = -9.20 × 10⁷t + 2.55 ×10⁵

The solutio to this problem uses the concept of calculus and motion

The acceleration of the bullet is simple the result of differenyiating the velocity function with respect to time.

Explanation:

**Explanation:**

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm