A baseball is thrown a distance of 20 meters what is its speed if it takes 0.5 seconds to cover the distance?

Kaylis [27]

(1) The image of an object placed further from the lens than the focal point will be upside down and smaller than the object.

(2) When light rays reflect, they bounce back.

(3) Images formed by a concave lens will look magnified.

(4) When light rays enter a different medium, they bend.

<h3>1.0 Object placed further from the lens than the focal point</h3>

The image of an object placed further from the lens than the focal point will be diminished and inverted.

Thus, the correct answer will be "upside down and smaller than the object".

<h3>2.0 What is reflection of light?</h3>

The ability of light to bounce back when it strike a hard surface is known as refection.

<h3>3.0 Image formed by concave lens</h3>

A concave lens is diverging lens is usually virtual, erect and magnified.

<h3>4.0 Refraction of light</h3>

The change in speed of light when it travels from medium to another medium is known as refraction. Refraction is also, the ability of light to bend around obstacles.

Learn more about reflection and refraction of light here: brainly.com/question/1191238

4 0

2 years ago

Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo

Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567 x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0

2 years ago

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$

4 0

3 years ago

A 0.0600-kilogram ball traveling at 60.0 meters

Vikki [24]

The momentum of ball is given by:

$\Delta Q=mv \\ \Delta Q=6*10^{-2}*60 \\ \Delta Q=3.6kg*m/s$

Since both have the same momentum, we have:

$\Delta Q=MV \\ 3.6=10^{-2}V \\ \boxed {V=360m/s}$

Number 3

If you notice any mistake in my english, please let me know, because i am not native.

7 0

3 years ago

(6) Module 4 Study Guide

velikii [3]

We have the meats Arby’s we beat them kids

3 0

3 years ago