PHYSICS COLLEGE

The focal length of a concave mirror is 17.5 cm. An object is located 38.5 cm in front of this mirror. How far in front of the mirror is the image located?

Answers

Answer 1

Answer:

Answer:

Explanation:

object distance u = 38.5 cm ( negative )

focal length f = 17.5 cm ( negative )

mirror formula

1 / v + 1 / u = 1 / f

1 / v - 1 / 38.5 = - 1 / 17.5

1 / v = - 1 / 17.5 + 1 / 38.5

= - 0 .03116

v = - 1 / .03116 = - 32 cm

Image will be formed in front of the mirror at 32 cm distance .

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A 1,200 kg car travels at 20 m/s. what is it’s momentum ?

Answers

The momentum of the car is 24000 Kg•m/s

Momentum is defined as the product of mass and velocity. Mathematically, it can be expressed as:

Momentum = mass × velocity

With the above formula, we can obtain the momentum of the car as follow:

Mass = 1200 Kg

Velocity = 20 m/s

Momentum =?

Momentum = mass × velocity

Momentum = 1200 × 20

Momentum of car = 24000 Kg•m/s

Learn more about momentum:

brainly.com/question/250648

Answer:

24000 kg·m/s

Explanation:

Momentum is Mass x Velocity, so 1200 kg time 20 m/s = 24000 kg-ms/s

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. We replace one of the objects with another whose net charge is + 4Q. We move the +Q and +4Q charges to be 3 times as far apart as they were. What is the magnitude of the force on the +4Q charge ?A. F

B. 4F

C. 4F/3

D. 4F/9

E. F/3

Answers

Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

$F=(kQ^2)/(r^2)$ ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

$F'=(kQ* 4Q)/((3r)^2)\n\nF'=(4kQ^2)/(9r^2)$ .....(2)

Dividing equation (1) and (2), we get :

$(F)/(F')=((kQ^2)/(r^2))/((4kQ^2)/(9r^2))\n\n(F)/(F')=(kQ^2)/(r^2)* (9r^2)/(4kQ^2)\n\n(F)/(F')=(9)/(4)\n\nF'=(4F)/(9)$

Hence, the correct option is (d) i.e. " 4F/9"

Final answer:

The magnitude of the force on the +4Q charge, after replacing one of the original +Q charges and moving the charges three times farther apart, is calculated to be 4F/9 using Coulomb's Law. Therefore, the correct answer is D.

Explanation:

The magnitude of the electrostatic force between two charges can be described by Coulomb's Law, which states that F = k × (q1 × q2) / r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between the centers of the two charges. Originally, two objects each with charge +Q exert a force of magnitude F on each other. After one charge is replaced with a +4Q charge and they are moved to be three times as far apart, the force on the +4Q charge can be calculated using the modified version of Coulomb's Law that takes into account the new charges and distance.

Using the original scenario as a reference, where F = k × (Q × Q) / r^2, when the charge is replaced and the distance is tripled, the new force F' = k × (Q × 4Q) / (3r)^2 = 4kQ^2 / 9r^2. By comparing F' with F, we find that F' = (4/9)F. Thus, the magnitude of the force on the +4Q charge is 4F/9.

Neptunium. In the fall of 2002, scientists at Los Alamos National Laboratory determined that the critical mass of neptunium-237 is about 60 kg. The critical mass of a fissionable material is the minimum amount that must be brought together to start a nuclear chain reaction. Neptunium-237 has a density of 19.5 g/cm3. What would be the radius of a sphere of this material that has a critical mass?

Answers

To solve this problem it is necessary to apply the concepts related to density, such as the relationship between density and Volume.

The volume of a sphere can be expressed as

$V = (4)/(3) \pi r^3$

Here r is the radius of the sphere and V is the volume of Sphere

Using the expression of the density we know that

$\rho = (m)/(V) \rightarrow V = (m)/(\rho)$

The density is given as

$\rho = (19.5g/cm^3)((10^3kg/m^3)/(1g/cm^3))$

$\rho = 19.5*10^3kg/m^3$

Now replacing the mass given and the actual density we have that the volume is

$V = (60kg)/(19.5*10^3kg/m^3 )$

$V = 3.0769*10^(-3) m ^3$

The radius then is,

$V = (4)/(3) \pi r^3$

$r = \sqrt[3]{(3V)/(4\pi)}$

Replacing,

$r = \sqrt[3]{(3(3.0769*10^(-3)))/(4\pi)}$

The radius of a sphere made of this material that has a critical mass is 9.02 cm.

A fly lands on one wall of a room. The lower-left corner of the wall is selected as the origin of a two-dimensional Car- tesian coordinate system. If the fly is located at the point having coordinates (2.00, 1.00) m, (a) how far is it from the origin? (b) What is its location in polar coordinates?

Answers

Answer:

a) The fly is 2.24 m from the origin.

b) In polar coordinates, the position of the fly is (2.24 m, 26.7°).

Explanation:

Hi there!

The position vector of the fly is r = (2.00, 1.00)m. The distance from that point to the origin is the magnitude of the vector "r" (see figure).

a) Notice in the attached figure that the distance from the origin to the point where the fly is located is the hypotenuse of the triangle formed by r, the x-component of r (2.00 m) and the y-component ( 1.00 m). Then:

r² = (2.00 m)² + (1.00 m)²

r² = 5.00 m²

r = 2.24 m

The fly is 2.24 m from the origin.

b) To find the angle θ (see figure) we can use trigonometry:

cos θ = adjacent / hypotenuse

cos θ = 2.00 m / √5 m

θ = 26.7°

The same will be obtained if we use sin θ:

sin θ = opposite / hypotenuse

sin θ = 1.00 m / √5 m

θ = 26.7°

In polar coordinates, the position of the fly is (2.24 m, 26.7°).

If you drew magnetic field lines for this bar magnet, which statement would be true

Answers

Arrows point away from north and toward south.

and

Field lines loop around the magnet starting at the north pole and ending at the south pole.

field lines correct for future weiins

A 81.0 kg diver falls from rest into a swimming pool from a height of 4.70 m. It takes 1.84 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.