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

Discuss the relationship between acceleration and velocity. Are these scalar or vector quantities?

pickupchik [31]

Acceleration is just speeding up, and velocity is speeding up in a certain direction. If you are changing your velocity you accelerating, as you are changing direction..

5 0

3 years ago

The force on an object is F⃗ =−17j⃗ . For the vector v⃗ =2i⃗ +3j⃗ , find: (a) The component of F⃗ parallel to v⃗

Igoryamba

Answer:

(a) $\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{103}{13}\,j$ , (b) $\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j$ , (c) $W = -51$

Explanation:

The statement is incomplete:

The force on an object is $\vec F = -17\,j$ . For the vector $\vec v = 2\,i +3\,j$ . Find: (a) The component of $\vec F$ parallel to $\vec v$ , (b) The component of $\vec F$ perpendicular to $\vec v$ , and (c) The work $W$ , done by force $\vec F$ through displacement $\vec v$ .

(a) The component of $\vec F$ parallel to $\vec v$ is determined by the following expression:

$\vec F_{\parallel} = (\vec F \bullet \hat {v} )\cdot \hat{v}$

Where $\hat{v}$ is the unit vector of $\vec v$ , which is determined by the following expression:

$\hat{v} = \frac{\vec v}{\|\vec v \|}$

Where $\|\vec v\|$ is the norm of $\vec v$ , whose value can be found by Pythagorean Theorem.

Then, if $\vec F = -17\,j$ and $\vec v = 2\,i +3\,j$ , then:

$\|\vec v\| =\sqrt{2^{2}+3^{3}}$

$\|\vec v\|=\sqrt{13}$

$\hat{v} = \frac{1}{\sqrt{13}} \cdot(2\,i + 3\,j)$

$\hat{v} = \frac{2}{\sqrt{13}}\,i+ \frac{3}{\sqrt{13}}\,j$

$\vec F \bullet \hat{v} = (0)\cdot \left(\frac{2}{\sqrt{13}} \right)+(-17)\cdot \left(\frac{3}{\sqrt{13}} \right)$

$\vec F \bullet \hat{v} = -\frac{51}{\sqrt{13}}$

$\vec F_{\parallel} = \left(-\frac{51}{\sqrt{13}} \right)\cdot \left(\frac{2}{\sqrt{13}}\,i+\frac{3}{\sqrt{13}}\,j \right)$

$\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j$

(b) Parallel and perpendicular components are orthogonal to each other and the perpendicular component can be found by using the following vectorial subtraction:

$\vec F_{\perp} = \vec F - \vec F_{\parallel}$

Given that $\vec F = -17\,j$ and $\vec F_{\parallel} = -\frac{102}{13}\,i-\frac{153}{13}\,j$ , the component of $\vec F$ perpendicular to $\vec v$ is:

$\vec F_{\perp} = -17\,j -\left(-\frac{102}{13}\,i-\frac{153}{13}\,j \right)$

$\vec F_{\perp} = \frac{102}{13}\,i + \left(\frac{153}{13}-17 \right)\,j$

$\vec F_{\perp} = \frac{102}{13}\,i -\frac{68}{13}\,j$

(c) The work done by $\vec F$ through displacement $\vec v$ is:

$W = \vec F \bullet \vec v$

$W = (0)\cdot (2)+(-17)\cdot (3)$

$W = -51$

8 0

3 years ago

An 80,000 kg plane is flying at 900 km/h at a height of 10 kilometers. What is its total mechanical energy? Group of answer choi

Romashka-Z-Leto [24]

The mechanical energy of the given flying object is 10,340 MJ. So the correct choice is 10,340 MJ.

Mechanical energy: The mechanical energy of an object is the sum of its potential energy and kinetic energy.

The formula for the mechanical energy E is:

E=K+U

where K is the kinetic energy and U is the potential energy.

Kinetic energy: The energy possessed by an object due to the motion.

The kinetic energy is given by the formula,

K=(1/2)*(mv^2)

where m is the mass of the object and v is the velocity of the object.

Potential energy: The energy possessed by an object due to its relative position from a given point.

The potential energy is given by the formula,

U=m*g*h

where g is the acceleration due to gravity and h is the height of the object from the ground.

Substitute K=(1/2)*(mv^2) and U=m*g*h in the formula of the mechanical energy to get the following formula,

E=(1/2)*(mv^2)+m*g*h

Given that m=80000 kg, h=10 km, and v=900 km/h, and the value of g is 9.8 m/s^(-2), the value of the total mechanical energy is,

E=(1/2)*(80,000 kg)(900 km/h )^2+(80,000 kg)*(9.8 m/s^(-2))*(10 km)

Note: 1 km= 1000 m and 1 km/h = 5/18 m/s. 1 MJ = 10^(6) J

E=(1/2)*(80,000 kg)*(900*(5/18) m/s )^2+(80000 kg)*(9.8 m/s^(-2))*(10000 m)

E=(1/2)*(80,000 kg)*(250 m/s)^2+7840*10^(6) J

E=2500*10^(6) J+7840*10^(6) J

E=10340*10^(6) J

E=10340 MJ

Learn more about Potential energy here:

brainly.com/question/15647581

#SPJ4

4 0

2 years ago

Darya [45]

Mass=0.25kg
Force=95N

$\\ \bull\sf\dashrightarrow F=ma$

$\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}$

$\\ \bull\sf\dashrightarrow a=\dfrac{95}{0.25}$

$\\ \bull\sf\dashrightarrow a=380m/s^2$

3 0

3 years ago

Ou are given a 25.3 µf capacitor that is connected to a 13.0 v dc power supply. what will be the charge that is stored on this c

Black_prince [1.1K]

charge stored in the capacitor=3.29 x 10⁻⁴ C

Explanation:

we use the formula

Q= C V

Q= charge

C= capacitor=25.3 μF= 25.3 x 10⁻⁶ F

V= voltage= 13 V

Q=(25.3 x 10⁻⁶ ) (13)

Q= 3.29 x 10⁻⁴ C

5 0

3 years ago