The distance of the object to the lens will be a) 7.2 cm
What is focal length?The distance between the convex lens or a concave mirror and the focal point of a lens or mirror is called the focal length. It is the point where parallel rays of light meet or converge.
magnification (m) = h(image )/ h(object ) = image distance / object distance
given
h(image ) = 2.5 * h(object)
m = 2.5 * h(object) / h(object ) = 2.5
2.5 = image distance / object distance
2.5 = v / u
v = 2.5 u
using lens formula
given
focal length (f) = + 12cm
1/v - 1/u = 1/focal length
1/2.5 u - 1/u = 1/12
u = - 7.2 cm
correct option a)7.2 cm
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Please help 23 also 29 the answer choices are reflection or refraction your fraction absorption
Answer:
23. Option B. Hertz
29. Refraction.
Explanation:
23. Determination of the unit of measurement of frequency.
Frequency is simply defined as the number of oscillation made in one second. Mathematically, frequency can be represented as:
Frequency = 1/period
f = 1/T
Period is measured in seconds.
Thus, the unit of frequency becomes
f = 1/T
f = 1/s = s¯¹ (Hertz)
Therefore, the unit of frequency is Hertz.
29. When a wave enters a new medium, the speed of the wave is uttered. This leads to the bending of the wave. When this occurs, we can say refraction has taken place.
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.70 ss apart. The speed of sound in air is 343 m/sm/s, and in concrete is 3000 m/sm/s. Part A How far away did the impact occur
Answer:
271.095 m
Explanation:
✓ Let speed of sound in air that was given as (343 m/s) be represented as (Vi)
✓( speed of sound in concrete that was given as (3000 m/s ) be debited as (Vc)
✓ Let the distance travelled by the sound = s
✓duration of Time that exist between heard of sounds = 0.70s
But we know that
Time = (Distance / Speed)
✓Time it takes the sound to travel through air= s/vi = s/343
✓Time it takes the sound to travel through concrete= s/vc = s/3000
✓ (s/343) - (s/3000) = 0.70
Finding LCM and simplify
[(3000s - 343s)]/1029000 = 0.70
2657s /1029000 = 0.70
Making " s" subject of the formula
s= (1029000 × 0.70)/2657
s=720300/ 2657
s= 271.095 m
Hence, The impact took place at a distance of 271.095 m away from the person.
what the answer for this question ?
Answer:
Explanation:
It is better to say “It is correct.” The word THIS is a demonstrative pronoun. The word THIS should be used to point out or to indicate the things. The word CORRECT is an abstract noun therefore it can not be pointed out this is why it is better to say IT IS CORRECT.
The energy of motion is called...?
A uniform metre rule of mass 100g balance the 40cm mark when a mass x is placed at the 20cm mark
what is the value of x
Answer:
X = 50 g
Explanation:
Please see attached photo for explanation.
From the attached photo,
Anticlock–wise moment = X × 20
Clockwise moment = 100 × 10
Anticlock–wise moment = clockwise moment
X × 20 = 100 × 10
X × 20 = 1000
Divide both side by 20
X = 1000 / 20
X = 50 g
Therefore, the value of X is 50 g
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave?
SOLVE IT COMPLETELY
The speed of a wave is 40 m/s. If the wavelength is 80 centimeters, what is the frequency of the wave ?
Answer:-Given:-Velocity (V) = 40 m/s
Wavelength [tex] (\lambda) [/tex] = 80 cm = 0.8 m
To Find:-The frequency (F) of the wave.
Solution:-We know,
[tex] \bf V \: = \: F \: × \: \lambda [/tex]
40 = F × 0.8
F = [tex] \frac{40}{0.8} [/tex]
F = 50
The frequency of the wave is 50 Hz. [Answer]19
What is the mass of a ball that is traveling 30 m/s and has 4.5 J of kinetic energy?
ave a GPF of 35512
Answer: The mass of ball is 10 grams.
Explanation:
Given : Kinetic energy = 4.5 J
Velocity = 30 m/s
The formula for Kinetic energy is as follows.
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
K.E = Kinetic energy
m = mass
v = velocity
Substitute the values into above formula as follows.
[tex]K.E = \frac{1}{2}mv^{2}\\4.5 J = \frac{1}{2} \times m \times (30 m/s)^{2}\\m = \frac{4.5 J \times 2}{900 m^2/s^2} (1 Js^{2} = kg m^{2})\\= 0.01 kg (1 kg = 1000 g)\\= 10 g[/tex]
Thus, we can conclude that the mass of ball is 10 grams.
A 100-n object and a 50-n object are placed on scales a and b respectively inside an elevator ascending with constant velocity 3.0m/s which statement below correctly describes the readings on the scales inside the elevator
Answer: b
Explanation:
The reading of the scale of the elevator ascending with constant velocity is 150 N.
Reading of the scale
The reading of the scale on the elevaor is calculated by applying Newton's second law of motion;
R = m(a + g)
R = ma + mg
R = F + W
where;
a is the acceleration of the objectsAt constant velocity, the acceleration of the object is zero (0).
R = 0 + 100 + 50
R = 150 N
Thus, the reading of the scale of the elevator ascending with constant velocity is 150 N.
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A ball is rolling away from you at a constant speed, hits a wall, and then suddenly stops moving. Imagine a position-versus-time graph showing the ball’s motion. Describe the line on the graph.
Answer:
It would show a "straight line".
Explanation:
Newton's First Law of Motion states that an object at rest will remain at rest and an object in motion will remain in motion (at a constant velocity) unless acted upon by an unbalanced force. Newton's first law of motion is also called the law of inertia.
In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of 58 N to the input piston, which has a radius r1. As a result, the output plunger, which has a radius r2, applies a force to the car. The ratio r2/r1 has a value of 8.2. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.
Answer:
The correct answer is "3899.92 N".
Explanation:
The given values are:
Force,
[tex]F_{app}=58 N[/tex]
Ratio,
[tex]\frac{R_2}{R_1}=8.2[/tex]
As we know,
Area, [tex]A=\pi r^2[/tex]
or,
⇒ [tex]\frac{F_2}{F_1} =\frac{A_2}{A_1}[/tex]
On substituting the value of "A", we get
⇒ [tex]\frac{F_2}{F_1} =\frac{\pi r_2^2}{\pi r_1^2}[/tex]
⇒ [tex]\frac{F_2}{F_1} =\frac{r_2^2}{r_1^2}[/tex]
On applying cross-multiplication, we get
⇒ [tex]F_2=F_1\times (\frac{r_2}{r_1} )^2[/tex]
On substituting the given values, we get
⇒ [tex]=58\times (8.2)^2[/tex]
⇒ [tex]=58\times 67.2[/tex]
⇒ [tex]=3899.92 \ N[/tex]