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17 days ago

Rohit placed a pencil perpendicular to principal axis in front of a converging

Physics

1 answer:

serg [1.1K]17 days ago

3 0

Response:

45cm

Clarification:

A converging mirror is generally termed a concave mirror. The focal length and the image distance for a concave mirror are both expressed as positive values.

Using the mirror formula to derive the object distance;

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

Where f denotes the focal length, u indicates the object distance, and v represents the image distance.

Given f = 30cm, and v = 2u (The formed image is double the size of the pencil)

Plugging these values into the formula to solve for u yields;

$\frac{1}{30} = \frac{1}{u} + \frac{1}{2u}\\\frac{1}{30} = \frac{2+1}{2u}\\\frac{1}{30} = \frac{3}{2u}\\$

By cross-multiplying, we obtain;

2u = 90

Dividing both sides by 2;

2u/2 = 90/2

u = 45cm

The object's distance from the mirror measures 45cm

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A small mass m is tied to a string of length L and is whirled in vertical circular motion. The speed of the mass v is such that

serg [1198]

Response:

(mv^2/R)/(mg)=1/2

v^2=R/2g

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6 days ago

A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [913]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0

13 days ago

1. Susie wondered if the height of a hole punched in the side of a quart-size milk carton would affect how far from the containe

Sav [1105]

Hypothesis: The liquid will project far.
Independent Variable: Height of the hole.
Dependent Variable: Distance of the squirt.
Constant: All other factors aside from the independent variable, such as the liquid volume.
Control: None that I recognize.
Number of groups: 4
Trials per group: 4

7 0

2 days ago

Read 2 more answers

Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom

ValentinkaMS [1149]

Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

A. Your weight including the saddle

C. Value of the rope's strength

B. Your weight

D. Value of the centripetal force

3 0

15 days ago

A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b

Softa [913]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a. Equation 1

E(r) = kq/r² for a<r<b. Equation 2

E(r) = 0 for b<r<c. Equation 3

E(r) = kq/r² for r>c. Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b, noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a. Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0

10 days ago