All categories

3 years ago

Which of the following is accurate in describing converging and diverging lenses?

Physics

2 answers:

kenny6666 [7]3 years ago

3 0

B. Both of these types of lenses have the ability to produce real images.

koban [17]3 years ago

3 0

Answer:

A. Both of these types of lenses have the ability to produce upright images.

Explanation:

Let's analyze each statement proposed:

A. Both of these types of lenses have the ability to produce upright images. --> TRUE. In fact, depending on the position of the object relative to the lens, both converging and diverging lenses can produce an image which is upright.

B. Both of these types of lenses have the ability to produce real images. --> FALSE. In fact, diverging lenses can produce virtual images only, because the rays of light are not focused into a single point.

C. Parallel rays converge into a single point at the focal point of a diverging lens, whereas parallel rays diverge at the focal point of a converging lens and spread out. --> FALSE. It is actually the opposite: in a diverging lens, parallel rays diverge and spread out, while in a converging lens, parallel rays converge into a single point (the focal point)

D. Diverging lenses will produce images behind the lens, and converging lenses will produce images in front of the lens. --> FALSE. It's actually the opposite: the image produced by a diverging lens is in front of the lens, while the image produced by a converging lens is behind the lens.

You might be interested in

Suppose a cup of coï¬ee is at 100 degrees Celsius at time t = 0, it is at 70 degrees at t = 10 minutes, and it is at 50 degrees

Alexandra [31]

Answer:

T ambient = 10 degrees

Explanation:

Using Newton's Law of Cooling:

T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1

Ti = 100

We have two points to evaluate the above equation as follows:

T = 70 @ t = 10 using Eq 1

70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2

T = 50 @ t = 20 using Eq 1

50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3

Solving the above Eq 2 and Eq 3 simultaneously:

Using Eq 2:

(70 - Tamb) / (100 - Tamb) = e^(-10k)

Squaring both sides we get:

((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4

Substitute Eq 4 into Eq 3

50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2

After simplification:

50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)

5000 - 50*Tamb = 4900 - 40*Tamb

Tamb = 100 / 10 = 10 degrees

6 0

4 years ago

500km is equal to how many millimeters

Sindrei [870]

Answer:

500000000

if you can give me brainliest that would be great

7 0

4 years ago

A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L

⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒ P = 24624000 N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax we use σmax

⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

7 0

3 years ago

A projectile is launched with an initial speed of 60.0 m/s at an angle of 30.0° above the horizontal. The projectile lands on a

vodka [1.7K]

Answer:

51.96 m/s^-1

Explanation:

a) see the attachment

b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be

v_ox=v_o*cosФ

=60*cos (30)

= 51.96 m/s^-1

3 0

3 years ago

Read 2 more answers

Please help me find the answer

Hoochie [10]

Answer:

26b) 66.7%

27) 500 N

Explanation:

26.a) In a two pulley system, the load is attached to one of the pulleys. The other pulley is attached to a fixed surface, as well as one end of the rope. The other end of the rope goes around moving pulley, then around the fixed pulley.

26.b) Mechanical advantage is the ratio between the forces:

MA = load force / effort force

Efficiency is the ratio between the work:

e = work done on load / work done by effort

Work is force times distance.

e = (F load × d load) / (F effort × d effort)

Rearranging:

e = (F load / F effort) × (d load / d effort)

e = MA × (d load / d effort)

In a two pulley system, the load moves half the distance of the effort. So the efficiency is:

e = (4/3) × (1/2)

e = 2/3

e = 66.7%

27) In a three pulley system, the load moves a third of the distance of the effort.

e = (F load / F effort) × (d load / d effort)

0.40 = (600 N / F) × (1/3)

F = 500 N

8 0

3 years ago