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3 years ago

A 6 cm object is 8 cm from a convex lens that has a focal length of 2.7 cm. The image is 4 cm from the lens.

Physics

2 answers:

melomori [17]3 years ago

7 0

Answer:

3cm

Explanation:

MariettaO [177]3 years ago

5 0

Answer:

-3

Explanation:

EDGE 2020

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The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago

Scientific work is currently underway to determine whether weak oscillating magnetic fields can affect human health. For example

iragen [17]

Given Information:

Magnetic field = B = 1×10⁻³ T

Frequency = f = 72.5 Hz

Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m

Required Information:

Maximum Emf = ?

Answer:

Maximum Emf = 20.66×10⁻¹² volts

Explanation:

The maximum emf generated around the perimeter of a cell in a field is given by

Emf = BAωcos(ωt)

Where A is the area, B is the magnetic field and ω is frequency in rad/sec

For maximum emf cos(ωt) = 1

Emf = BAω

Area is given by

A = πr²

A = π(d/2)²

A = π(7.60×10⁻⁶/2)²

A = 45.36×10⁻¹² m²

We know that,

ω = 2πf

ω = 2π(72.5)

ω = 455.53 rad/sec

Finally, the emf is,

Emf = BAω

Emf = 1×10⁻³*45.36×10⁻¹²*455.53

Emf = 20.66×10⁻¹² volts

Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts

8 0

3 years ago

1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of

lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0

3 years ago

A 45.0-kg girl is standing on a 168-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat,

muminat

Answer:

The speed of the plank relative to the ice is:

$v_{p}=-0.33\: m/s$

Explanation:

Here we can use momentum conservation. Do not forget it is relative to the ice.

$m_{g}v_{g}+m_{p}v_{p}=0$ (1)

Where:

m(g) is the mass of the girl
m(p) is the mass of the plank
v(g) is the speed of the girl
v(p) is the speed of the plank

Now, as we have relative velocities, we have:

$v_{g/b}=v_{g}-v_{p}=1.55 \: m/s$ (2)

v(g/b) is the speed of the girl relative to the plank

Solving the system of equations (1) and (2)

$45v_{g}+168v_{p}=0$

$v_{g}-v_{p}=1.55$

$v_{p}=-0.33\: m/s$

I hope it helps you!

8 0

3 years ago

Suppose you wanted to be able to see astronauts on the moon. What is the smallest diameter of the objective lens required to res

LenaWriter [7]

Answer:

A:1.94

Explanation:

cause that the only one on there

5 0

3 years ago