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kirill115 [55]
3 years ago
14

Sherlock Holmes examines a clue by holding his magnifying glass (with

Physics
1 answer:
Alborosie3 years ago
8 0

Answer:

Distance: -30.0 cm; image is virtual, upright, enlarged

Explanation:

We can find the distance of the image using the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where:

f = 15.0 cm is the focal length of the lens (positive for a converging lens)

p = 10.0 cm is the distance of the object from the lens

q is the distance of the image from the lens

Solving for q,

\frac{1}{q}=\frac{1}{f}-\frac{1}{q}=\frac{1}{15.0}-\frac{1}{10.0}=-0.033\\q=\frac{1}{0.033}=-30.0 cm

The negative sign tells us that the image is virtual (on the same side of the object, and it cannot be projected on a screen).

The magnification can be found as

M=-\frac{q}{p}=-\frac{-30}{10}=3

The magnification gives us the ratio of the size of the image to that of the object: since here |M| = 3, this means that the image is 3 times larger than the object.

Also, the fact that the magnification is positive tells us that the image is upright.

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As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh
marshall27 [118]

Answer:

The resultant velocity is  v_t=10 knots

Explanation:

Apply the law of conservation of momentum

     M_L *v_L + M_f * V_f = (M_L + M_f) v_t

Where M_L is the mass of the Luxury Liner = 40,000 ton

            v_L is the velocity of Luxury Liner = 20 knots due west

            M_f mass of freighter = 60,000

           v_f is the velocity of freighter = 10 knots due north

Apply the law of conservation of momentum toward the the west direction

         v_f = 0 \ knots

So the equation would be

              M_L *v_L = (M_L + M_f) v_t

Substituting values

            40000*20 = (40000+ 60000)v_t_w

Where v_t_w the final velocity due west

Making v_t_w the subject

          v_t_w = \frac{40,000* 20}{(40000 + 60000)}

                = 8 \ knots

Apply the law of conservation of momentum toward the the north direction          

          v_L = 0 \ knots

So the equation would be

           M_f *v_f = (M_L + M_f) v_t_n

Where v_t_n the final velocity due north

     Making v_t_n the subject

          v_t_n = \frac{60,000* 10}{(40000 + 60000)}

                = 6 \ knots

The resultant velocity is

       v_t = \sqrt{v_t_w^2 + v_t_n^2}

            = \sqrt{8^2 +6^2}

           v_t=10 knots

8 0
3 years ago
(ASAP) would it be 125 m/s2 to calculate for her speeding up?
serg [7]

Answer:

0\:\mathrm{ m/s^2}

Explanation:

Recall the formula for acceleration:

\displaystyle\\a=\frac{v_f-v_i}{\Delta t}, where v_f is final velocity, v_i is initial velocity, and \Delta t is elapsed time (change in velocity over this amount of time).

Let's look at our time vs velocity graph. At t=0 seconds, V=25 m/s. So her initial velocity is 25 m/s.

We want to find the acceleration during the first 5 seconds of motion. Well, looking at our graph, at t=5 seconds, isn't our velocity still 25 m/s? Therefore, final velocity is 25 m/s (for this period of 5 seconds).

We are only looking from t=0 seconds to t=5 seconds which is a total period of 5 seconds. Therefore, elapsed time is 5 seconds.

Substituting values in our formula, we have:

\displaystyle a=\frac{25-25}{5}=\frac{0}{5}=\boxed{0\:\mathrm{m/s^2}}

Alternative:

Without even worrying about plugging in numbers, let's think about what acceleration actually is! Acceleration is the change in velocity over a certain period of time. If we are not changing our velocity at all, we aren't accelerating! In the graph, we can see that we have a straight line from t=0 seconds to t=5 seconds, the interval we are worried about. This indicates that our velocity is staying the same! At t=0 seconds, we have a velocity of 25 m/s and that velocity stays the same until t=5 seconds. Even though we are moving, we haven't changed velocity, which means our average acceleration is zero!

8 0
2 years ago
Pascal has 96 miles remaining to complete his cycling trip. If he reduced his current speed by 4 miles per hour, the remainder o
Verdich [7]

Answer:

V = 20 miles /sec

Explanation:

We have remaining distance   =  d  = 96 miles

Lets call  Pascal velocity  V in miles per hour

Now if he increases his velocity by  50 % (equivalent to multiply by 1.5 ) he will need a time t₁ to arrive then as V = d/t

1.5* V  = d/ t₁      ⇒   1.5 * V  =  96 /t₁

And in the case of reducing his velocity

(V / 4) = d/ (t₁ + 16 )     ⇒  V * (t₁ + 16 ) = 4*d     ⇒ V*t₁ + 16*V = 384

So we a 2 equation system with two uknown variables

1.5*V = 96/t₁      (1)

V*t₁  + 16*V = 384     (2)

We solve  from equation    (1)      t₁  = 64/V

And by substitution   in equation (2)

V * (64/V) + 16* V = 384

64  + 16 *V  = 384         ⇒   16*V = 320      ⇒  V= 320/16

V = 20 miles /sec

6 0
3 years ago
Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i
Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

L = L_i_n_i + L_3_s

L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt

L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s

Moment if inertia is

I = 2ml^2

I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2

Angular speed

ω = L/I

= 38 / 5.75\\\\=6.61

The speed of each ball is

V = ωL

= 6.61\times0.5\\\\= 3.31m/s

7 0
3 years ago
Find the slit separation of a double-slit arrangement that will produce interference fringes 0.018 rad apart on a distant screen
inessss [21]

Answer:

1.64 * 10^(-5) m

Explanation:

Parameters given:

Angular separation, θ = 0.018 rad

Wavelength, λ = 589 nm = 5.89 * 10^(-7) m

The angular separation when there are 2 slots is given as

θ = λ/2d

where d = separation between slits

d = λ/2θ

d = (589 * 10^(-9))/(2 * 0.018)

d = 1.64 * 10^(-5) m

5 0
3 years ago
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