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densk [106]
2 years ago
14

I need emergency help we only have 3 minutes left

Physics
1 answer:
jenyasd209 [6]2 years ago
3 0

Answer:

Option A

Explanation:

Ast the force is equal and the diayance is equal the beam is also balanced

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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
What energy is directly dependent upon velocity and mass?
erastova [34]

Answer:

Kinetic energy

Explanation:

3 0
3 years ago
Find the direction and magnitude of Ftot, the total force exerted on her by the others, given that the magnitudes F1 and F2 are
Mars2501 [29]

Answer:

 θ = 36°

Explanation:

given,

F₁ = 22.8 N

F₂ = 16.6 N

magnitude of force = ?

direction of force = ?

F = \sqrt{F_1^2 + F_2^2}

F = \sqrt{22.8^2 + 16.6^2}

F = \sqrt{795.4}

      F = 28.20 N

direction

\theta = tan^{-1}(\dfrac{F_2}{F_1})

\theta = tan^{-1}(\dfrac{16.6}{22.8})

\theta = tan^{-1}(0.728)

       θ = 36°

5 0
3 years ago
A 0.6 kg basketball moving 7.2 m/s to the right collides with a 0.04 kg tennis
KonstantinChe [14]

Answer:

B

Explanation:

4 0
3 years ago
Read 2 more answers
Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 30 mph. at the same tim
aleksandr82 [10.1K]

In triangle ABC , using Pythagorean theorem

BC = sqrt(AB² + AC²)

r = sqrt(y² + x²)                                eq-1

taking derivative both side relative to "t"

dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))

dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)

15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)

v₂ = 51.2 m/s

3 0
3 years ago
Read 2 more answers
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