All categories

3 months ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

Physics

1 answer:

kicyunya [3.2K]3 months ago

3 0

Answer:

The magnification is calculated to be 3

Explanation:

It has been provided that

The lens' focal length is f = 15 cm

The object distance is u = -10 cm

Using the lens formula:

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v signifies image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Thus, the magnification of the lens is found to be 3.

You might be interested in

An 800-N billboard worker stands on a 4.0-m scaffold weighing 500 N and supported by vertical ropes at each end. How far would t

Maru [3345]

Answer:

2.5 m

Explanation:

Billboard worker's weight = 800 N

Number of ropes = 2

Length of scaffold = 4 m

Weight of scaffold = 500 N

Tension present in rope = 550 N

The total torques will be

$-800(4-x)-500\times 2+550\times 4=0\\\Rightarrow -800(4-x)=500\times 2-550\times 4\\\Rightarrow -800(4-x)=-1200\\\Rightarrow -x=\dfrac{1200}{800}-4\\\Rightarrow -x=-2.5\\\Rightarrow x=2.5\ m$

The worker is positioned at 2.5 m

7 0

3 months ago

A person shooting at a target from a distance of 450 metres finds that the sound of the bullet hitting the target comes 1 / 2 se

Sav [3153]

Answer:

1350 m/s

Clarification:

Bullet speed

The bullet travels 450 m

Sound travels a distance of 450 m

Using the equation S= V × t

==> t= S/V

Thus, the time for the bullet t1=450/vb

and the sound's travel time t2=450/vs

Given that there's a 1/2 sec interval from when the shot is fired to the moment the shooter hears the sound

==> t1+t2= 1/2 sec

==> 450/vb+450/vs=0.5sec ---- (1)

At a distance 'x' from both the gun and target, it takes 3 seconds for a person to hear the bullet sound from firing to impact.

Firing sound duration

Distance =x m

Speed of sound = V

time =T1

==> T1 = x/vs

During this time period, the bullet covers 450 m and the sound of impact travels a distance 'x'

The time taken for sound = 450/vb

The time it takes sound to travel distance 'x'= x/vs

therefore let T2= 450/vb + x/vs

However, all this occurs within 3 seconds, i.e., T = 3 sec

because firing takes place before hitting the target, implying the strike sound is heard in time T = T2-T1= 450/vb + x/vs -x/vs

Making T= 3sec

==> 3= 450/vb

==> vb= 1350 m/s

From equations 1 and 2

applying the same principle, in 3 seconds the observer sees the bullet travel 450 m and perceives the sound

3 0

2 months ago

13–82. the 8-kg sack slides down the smooth ramp. if it has a speed of 1.5 m> s when y = 0.2 m, determine the normal reaction

Softa [3030]

The second question necessitates a figure to provide an answer. For the initial question
The acceleration of the sack is
1.5² - 0² = 2a(0.2)
a = 5.63 m/s²
The ramp's reaction force is
F = 8 kg (5.63 m/s²)
F = 45 N
Differentiate the kinematic equation with respect to time to find the velocity's rate of increase.

4 0

3 months ago

A hard rubber rod with an electric potential energy of 5.2 × 10–3 J has a charge of 4.0 µC at the tip. What is the electric pote

Keith_Richards [3271]

1) The electric potential energy can be defined as the product of the electric potential and the associated charge:
$U=q V$
where
q refers to the charge
V denotes the electric potential

In this scenario, the charge on the rod is $q=4.0 \mu C = 4.0 \cdot 10^{-6}C$ , and the potential energy is $U=5.2 \cdot 10^{-3} J$ , thus we may rearrange the earlier formula to find the electric potential at the tip:
$V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3}J}{4.0 \cdot 10^{-6} C}=1300 V=1.3 \cdot 10^3 V$

2) Using this same formula, if the charge changes to $q=2.0 \mu C = 2.0 \cdot 10^{-6} C$ , the resulting electric potential will be:
$V= \frac{U}{q}= \frac{5.2 \cdot 10^{-3} J}{2.0 \cdot 10^{-6}C}=2600 V = 2.6 \cdot 10^{3}V$

8 0

3 months ago

Read 2 more answers

A very tall building has a height H0 on a cool spring day when the temperature is T0. You decide to use the building as a sort o

ValentinkaMS [3465]

Response:

The temperature is $T = \frac{h}{H_O \alpha_{steel} } + T_O$