Answers
Reason: Opposite angles
A = 37°
Reason: Sum of angels on a straight line = 180 so 180-90-53 = 37°
C = 90°
Reason: Right angles = 90°
D = 37°
Reason: Sum of angels on a straight line = 180 so 180-90-53 = 37°
Related Questions
A box with mass of 2 kg is pushed directly horizontally over a horizontal surface (with friction) at a constant speed of 10 m/s. The force of the push is 60 N. How much thermal energy is generated pushing the box a distance of 15 m
Answers
Answer:
E= 600 W
Explanation:
Given that
m = 2 kg
Speed , v= 10 m/s
Force , F= 60 N
Given that box is moving with constant velocity, it means that friction force will be 60 N.
f = 60 N
Therefore total energy generated
E= f x v
E= 60 x 10 = 600 W
E= 600 W
Thus the answer will be 600 W.
The marginal cost curve
(a) Lies below the ATC curve when the ATC curve slopes upward.
(b) Intersects the AFC and ATC curves at their respective minimum points.
(c) Lies above the AVC curve when the AVC curve slopes downward.
(d) Intersects the AFC and AVC curves at their respective minimum points.
(e) Intersects the AVC and ATC curves at their respective minimum points
Answers
Answer:
c
Explanation:
The marginal cost curve image has been attached from which we can clearly, indicate that
ATC = average total cost
AFC = average fixed cost
AVC = average variable cost.
From the graph we can indicate that the marginal cost curve
(c) Lies above the AVC curve when the AVC curve slopes downward.
The energy change in an endothermic reaction is: A. Internal B. External C. Negative D. Positive
Answers
Answer:
Positive
Explanation:
In an endothermic reaction, the products are at a higher energy than the reactants. This means that the enthalpy change of the reaction (∆H) is positive
a 2-n force is applied to a spring, and there is displacement of 0.4 m. how much would the spring be displaced if a 5-n force was applied?
Answers
Answer:1m
Explanation:
2n=0.4m
5n=?
5n×0.4/2n=1m
A physics student stands on a cliff overlooking a lake and decides to throw a softball to her friends in the water below. She throws the softball with a velocity of 23.5 m/s at an angle of 39.5∘ above the horizontal. When the softball leaves her hand, it is 11.5 m above the water. How far does the softball travel horizontally before it hits the water? Neglect any effects of air resistance when calculating the answer.
Answers
Answer:
66.86m
Explanation:
Velocity of ball thrown, u = 23.5 m/s
Initial height of the ball above the water, H = 11.5 m
Angle of projection, θ = 39.5°
Vertical components of veloclty = usinθ
Horizontal components of veloclty = ucosθ
The soft ball hits the water after time 't'
Considering the second equation of motion
S = ut + 1/2at^2........ 1
But since the ball went through motion under gravity ( free fall ) rather than linear motion, then equation 1 can be rewritten as:
H = ut +/- 1/2gt^2
H = - 11.5m
U = usinθ
θ = 39.5°
a = -g = -9.8m/s^2
- 11.5m = 23.5(sin39.5°)t + 1/2(-9.8)t^2
-11.5m = 23.5(0.6360)t - 4.9t^2
-11.5m = 14.946t - 4.9t^2
4.9t^2 -14.946t-11.5m = 0
Since the ball drifted horizontally
D = (Ucosθ)t
Where θ = 39.5°
U = 23.5m/s t=
Alternatively,
horizontal component of the velocity is 23.5 cos 39.5º = 18.1331 m/s
now how long does it take the ball to raise to a peak and fall to the water.
vertical component of velocity = 23.5 sin 39.5º = 14.947m/s
time to reach peak t = v/g = 11.947/9.8 = 1.5252 sec
peak reached above cliff top is
h = ½gt² = ½(9.8)(1.5252)²
= ½×22.797
= 11.3985m
now the ball has to fall 11.3985+ 11.5 = 22.8985m
time to fall from that height is
t = √(2h/g) = √(2• 22.8986/9.8) = 2.1617 sec
add up the two times to get time it is in the air, 2.1617 + 1.5252 = 3.6869
now haw far does the ball travel horizontally in that time
d = vt = 18.1331 ×3.6869= 66.856m
= 66.86m
A missile is moving 1350 m/s at a 25° angle it needs to hit a target 23,500 m away in a 55° direction in 10.2 seconds what is the magnitude of its final velocity
Answers
Answer:
3504 m/s
Explanation:
Let x be the horizontal component of distance
y - vertical component of distance
t-time
ax- horizontal component of acceleration
ay-Vertical component of acceleration
Vx-horizontal component of velocity
Vy-Vertical component of velocity
horizontally: x = V_x ×t + ½×a_x×t²
plugging the values we get
23500× cos 55º = 1350×cos25.0º × 10.20 + ½×a_x× (10.20)²
⇒ax = 19.2 m/s²
Moreover,
V'x = V_x + a_x×t = 1350×cos25.0º + 19.2×10.20= 1419 m/s
similarly in vertical direction:
y = V_y×t + ½×a_y×t²
23500×sin55º = 1350×sin25.0º×10.20s + ½×a_y×(10.20)²
⇒a_y = 258 m/s²
Also,
V'y = V_y + a_y×t = 1350×sin25.0º + 258×10.20 = 3204 m/s
Therefore
V = √(V'x² + V'y²) = 3504 m/s
therefore, magnitude of final velocity of missile=3504 m/s
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Brainliest
what is a hypothesis reffered to as after being verified by a large number or independent experiments
Answers
Answer:
The hypothesis may or may not be true and needs to be tested. It might be the answer to the problem. Hence, it must be tested thoroughly. When these predictions are tested again and again in independent scientific experiments and gets verified, the hypothesis is converted into a scientific theory.
Will mark as BRAINLIEST.......
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation 4x³+3x²-5x+2 , where x is in meters and t is in sec.
a)Find velocity of particle at i) t=2 sec ii) t=4 sec.
b) Find the acceleration of the particle at t=3 sec.
Answers
Explanation:
It is given that,
The Displacement x of particle moving in one dimension under the action of constant force is related to the time by equation as:
[tex]x=4t^3+3t^2-5t+2[/tex]
Where,
x is in meters and t is in sec
We know that,
Velocity,
[tex]v=\dfrac{dx}{dt}\\\\v=\dfrac{d(4t^3+3t^2-5t+2)}{dt}\\\\v=12t^2+6t-5[/tex]
(a) i. t = 2 s
[tex]v=12(2)^2+6(2)-5=55\ m/s[/tex]
At t = 4 s
[tex]v=12(4)^2+6(4)-5=211\ m/s[/tex]
(b) Acceleration,
[tex]a=\dfrac{dv}{dt}\\\\a=\dfrac{d(12t^2+6t-5)}{dt}\\\\a=24t+6[/tex]
Pu t = 3 s in above equation
So,
[tex]a=24(3)+6\\\\a=78\ m/s^2[/tex]
Hence, (a) (i) v = 55 m/s (ii) v = 211 m/s and (b) 78 m/s²
Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is today. When astronomers make such a statement, how can they know what was happening inside galaxies way back then
Answers
Answer:
First, as you may know, the light travels at a given velocity.
In vaccum, this velocity is c = 3x10^8 m/s.
And we know that:
distance = velocity*time
Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.
This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)
Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.
The astronomers could know what was happening inside galaxies way back then by the fact that;
they examine the spectra of galaxies (or the overall colors of galaxies) with the highest redshifts they can find
Astronomers Measure the wavelength of the light that is stretched, so the light is seen as 'shifted' towards the red part of the spectrum by using spectroscopy. This measure is also called redshift.
This invokes a ray of light through a triangular prism that splits the light into various components known as spectrum.
The way the astronomers could use this concept to know what was happening in the galaxies before is by examining the spectra of galaxies that have the highest redshifts.
Read more at; https://brainly.com/question/15995216
A block is attached to the end of a spring. The block is then displaced from its equilibrium position and released. Subsequently, the block moves back and forth on a frictionless surface without any losses due to friction. Which one of the following statements concerning the total mechanical energy of the block-spring system this situation is true?
1. The total mechanical energy is dependent on the maximum displacement during the motion.
2. The total mechanical energy is at its maximum when the block is at its equilibrium position
3. The total mechanical energy is constant as the block moves back and forth.
4. The total mechanical energy is only dependent on the spring constant and the mass of the block.
Answers
Answer:
The correct option is;
3. The total mechanical energy is constant as the block moves back and forth
Explanation:
The total mechanical energy is the sum of the potential and kinetic energies of the system
For a system that is isolated from the effects of external forces, but being acted upon by the internal conservative forces within the system, the total mechanical energy is constant
For a black and spring system, we have total mechanical energy, E = 1/2×K×A².
Where;
K = Constant
A = The amplitude of motion
Therefore, where there is no loss to friction, with A, remaining constant, the total mechanical energy will be constant.
Is there a way for us to control motion
Answers
Answer:
They are:
1) change position
2) distract yourself
3) Get fresh air
4) Face the direction you are going.
5) Drink water.
6) Play music.
7) Put your eyes on horizon.
Explanation:
Hope it helps.
A car is moving on straight highway with a speed of 108 km/h.
Answers
Answer:
5.3333 sec
Explanation:
initial speed: u = 108km/hr or 30 m/s
final speed: v = 0m/s
distance travelled: s = 80m
time the car took to stop: = t sec
[tex]v^{2} - u^{2}[/tex] = 2as,
a = ([tex]v^{2} - u^{2}[/tex])/2s
a = (0-900)/160
a = -5.625 [tex]ms^{-2}[/tex]
v = u + at,
t = (v - u)/a
t= (0 - 30)/(-5.625)
t = 5.3333 sec
Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a neutron and a proton meet (assume both to be almost stationary), they combine and form a deuteron, emitting a gamma ray whose energy is 2.2233 MeV. The masses of the proton and the deuteron are 1.007 276 467 u and 2.013 553 212 u, respectively. Based on this data, what is the mass of the neutron
Answers
Answer:
Explanation:
Energy of gamma ray = 2.2233 MeV
Let mass of neutron be n amu
mass defect of deuteron = 2.013553212 - ( 1.007 276 467 + n ) u .
in terms of energy this mass defect will be equal to energy of gamma ray
1 amu = 931 MeV
931 [ 2.013553212 - ( 1.007 276 467 + n ) ] = 2.2233
( 1.007 276 467 + n ) - 2.013553212 = .00238807733
n = 1.008664822 amu
so mass of neutron = 1.008664822 amu
A hammer is used to hit a nail into a board. Which statement is correct about the forces at play between the nail and the hammer? The nail exerts an equal force on the hammer in the same direction. The nail exerts a much smaller force on the hammer in the opposite direction. The nail exerts an equal force on the hammer in the opposite direction. The nail exerts a much smaller force on the hammer in the same direction.
Answers
Answer:
The nail exerts an equal force on the hammer in the opposite direction.
Explanation:
The Newtons third law states that there is an equal an opposite reaction for every action. When hammer pushes the nail, the nail will push the hammer back in opposite direction. When the hammer hits a nail then nail will exert the equal and opposite force to the hammer. These both objects will exert force on each other in opposite directions.
Hello, I am BrotherEye
Answer:
Answers are
1. "The nail exerts an equal force on the hammer in the opposite direction."
2. "500 N"
3. "The iron piece exerts a force of 1 N on the magnet in the opposite direction."
4. "When mass moves closer to the point of rotation, rotational inertia decreases."
5. "The skater spins slower because his rotational inertia has increased."
Explanation:
The acceleration due to gravity near Earth ... Select one: a. varies inversely with the distance from the center of Earth. by. varies inversely with the square of the distance from the center of Earth. c. is a constant that is independent of altitude d. varies directly with the distance from the center of Earth.
Answers
Answer:
b. varies inversely with the square of the distance from the center of Earth.
Explanation:
Comparing the Newton's law of universal gravitation and second law of motion;
from Newton's second law of motion,
F = ma ............. 1
from New ton's law of universal gravitation,
F = [tex]\frac{GMm}{r^{2} }[/tex] ........... 2
Equating 1 and 2, we have;
mg = [tex]\frac{GMm}{r^{2} }[/tex]
g = [tex]\frac{GM}{r^{2} }[/tex]
Therefore, the acceleration due to gravity near Earth, g, is inversely proportional to the square of the distance from the center of Earth.
Which value would complete the last cell?
(1 point)
3.0
100.0
25.0
4.0
Answers
Answer:
4.0
Explanation:
The following data were obtained from the question:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
Force is simply defined as the product of mass and acceleration. Mathematically, it is expressed as
Force (F) = mass (m) x acceleration (a)
F = ma
With the above formula, we can obtain th acceleration of the body as follow:
Force (F) = 20 N
Mass (m) = 5 kg
Acceleration (a) =.?
F = ma
20 = 5 x a
Divide both side by 5
a = 20/5
a = 4 m/s²
Therefore, the value that will complete the last cell in the question above is 4.
Can you solve this question please help me with this
Answers
Answer:
Explanation:
The velocity ratio of a wheel and axle is the ratio of the radius (R) of the wheel to the radius (r) of the axle. It is expressed as;
VR = R/r
Since radius = diameter/2
VR = (D/2)/(d/2)
VR = D/d
D is the diameter of the wheel and 'd' is the diameter of the axle.
Given VR = 3 and d = 5cm
3 = D/5
D = 15 cm
If the diameter of the wheel is 15cm, the radius of the wheel will be 15/2 = 7.5cm.
b) Workdone by the load = Load * distance moved by load
Given load = 60kg
Distance moved by load = 2π*radius of axle
Distance moved by load = 2π(0.025) = 0.157
workdone by load = 60* 0.157 = 9.42J
Effort = Workdone by load/distance moved by the wheel
Effort = 9.42/2π(0.075)
Effort = 9.42/0.471
Effort = 20kg
Hence the effort applied is 20kg
c) MA = Load/Effort
MA = 60/20
MA = 3
d) Efficiency = MA/VR * 100%
Efficiency = 3/3 * 100%
Efficiency = 100%
If the spring constant is 10 N/m and the spring is stretched 1 m, what is the Force?
Answers
Answer:
10N
Explanation:
Applying the Hooke law:
F = kx
F: Force
k: stiffness coefficient
x: stretched distance
F = 10N/m x 1m = 10N
A small cylinder is rolled along a ruler and completes two revolutions. The circumference is the distance around the outside of a circle. What is the circumference of the cylinder? A 4.4 cm B 5.2 cm C 8.8 cm D 10.2 cm
Answers
Answer; 4.4cm
Explanation: There is a ruler upon which the cylinder start from 1.4cm and reaches 10.2cm
distance traveled =10.2-1.4=8.8
since this cylinder is small so the linear distance can be approximately taked as rotational distance(as in case of point charge) so
2x2πxr =8.8
so the circumference will be 2πr=4.4cm
A dog is 10 m from a cat, whose speeds are 6 and 5 m / s, respectively. What time does the dog require to catch the cat?
Answers
Answer:
10 seconds
Explanation:
because the cat is moving one m/s slower than the dog, the dog has a relative speed of 1 m/s. 10 meters would take 10 seconds for the dog to cover
Forensic toxicologist analyze and identify drugs that are confiscated from criminals
True
False
Answers
Can someone explain how the weight of the block is 10.26N, with reference to an appropriate law of motion?
Answers
this process is called parellelogram method of resolving vectors.
PLS HELP ME Define Derived Quantities ?
Answers
Derived Quantities
Explanation: Those physical quantities which are derived from fundamental quantities are called derived quantities and their units are called derived units. e.g., velocity, acceleration, force, work etc.
Answer:
These are quantities calculated from two or more measurements
Explanation:
They can't me measured directly.
They can only be computed.
They are calculated in PHYSICAL SCIENCE.
hope it helps.
What did Bohr's model of the atom include that Rutherford's model did not have?
a nucleus
energy levels
electron clouds
smaller particles
Answers
Answer:
The correct option is energy levels
Explanation:
Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.
Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.
Answer:
energy levels
hope this helped!
Explanation:
Will mark as BRAINLIEST..... A balloon is ascending at the rate of 4.9 m/s. A packet is dropped from from the balloon when situated at a height of 100m. How long does it take the packet to reach the ground ? What is it's final velocity ?
Answers
Answer:
PFA
:-)
Explanation:
SHOW ADEQUATE WORKINGS IN THIS SECTION
12. Wale and Lekan of 55kg and 60kg respectively ran a race of 200m.
i. Calculate their work done in KJ
ii. If wale finished the race in 25secs while Lekan finished in 30secs, calculate their
power and who is more powerful out of these two. ( g = 10m/secs)
13. A machine has a velocity ratio of 5 and is 800
/0 efficient. If the machine is carrying a load
of 200kg, what will be effort applied?
Please help me answer anyone that you understand
Answers
Answer:
12 i. The work done by Wale = 107.910 kJ
The work done by Lekan = 117.720 kJ
Total work done = 225.36 kJ
ii. Wale's power = 4.3164 kW
Lekan's power = 3.924 kW
Wale has more power and is more powerful than Lekan
13. 313.92 N
Explanation:
i. The work done, W = Force, F × Distance moved by the force, D
The given parameters are
The mass of Wale = 55 kg
The mass of Lekan = 60 kg
The acceleration due to gravity, g =9.81 m/s²
The motion force of Wale and Lekan are;
Motion force of Wale = 9.81 × 55 = 539.55 N
Motion force of Lekan = 9.81 × 60 = 588.6 N
The work done by Wale = 539.55 × 200 = 107910 J = 107.910 kJ
The work done by Lekan= 588.6 × 200 = 117720 J = 117.720 kJ
107910 + 117720 =225630 J = 225.36 kJ
ii. Power = Work done/time
Wale finished the race in 25 s, therefore, his power = 107910/25 = 4316.4 W
Lekan finished the race in 30 s, therefore, his power = 117720/30 = 3924 W
Wale has more power and is more powerful than Lekan
13. The velocity ratio = 5
V. R. = Distance moved by effort/(Distance moved by load)
Efficiency = 80%
Work done by effort = x
Work done by machine = Efficiency × Work done by effort = 0.8 × x
Distance moved by effort, E = V. R. × Distance moved by load, D = 5 × D
Work done by effort = Force × Distance moved = 200×9.81× E
Work done by effort = 1962×E = 1962×E = 1962×5×D
Work done by machine = 1962 × D, when D = 1, we have;
0.8 × 1962×1 = 1569.6 J
Work done by effort = Force × Distance moved
Work done by effort = Force × 5×D = Force × 5 (D = 1)
From the principle of conservation of energy, we have;
Energy is neither created nor destroyed
Therefore
Work done by effort = Force × 5 = 1569.6 J
Force = 1569.6 /5 = 313.92 N.
A car is driving at 99 km/h, calculate the distance it travels in 70 minutes.
Give your answer in correct SI units rounded to 0 decimal places.
Answers
Answer:
The distance the car travels is 115500 m in S.I units
Explanation:
Distance d = vt where v = speed of the car and t = time taken to travel
Now v = 99 km/h. We now convert it to S.I units. So
v = 99 km/h = 99 × 1000 m/(1 × 3600 s)
v = 99000 m/3600 s
v = 27.5 m/s
The speed of the car is 27.5 m/s in S.I units
We now convert the time t = 70 minutes to seconds by multiplying it by 60.
So, t = 70 min = 70 × 60 s = 4200 s
The time taken to travel is 4200 s in S.I units
Now the distance, d = vt
d = 27.5 m/s × 4200 s
d = 115500 m
So, the distance the car travels is 115500 m in S.I units
Light travels at a speed of 2.998*108 m/s. Light takes approximately 3.25 minutes to travel from the Sun to reach a planet. Calculate the distance from the Sun to this planet in meters. Give your answer to 0 decimal places.
Answers
Answer:
585×10⁸ m
Explanation:
Distance = rate × time
d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)
d = 585×10⁸ m
In a Young's double-slit experiment, a set of parallel slits with a separation of 0.102 mm is illuminated by light having a wavelength of 575 nm and the interference pattern observed on a screen 3.50 m from the slits.(a) What is the difference in path lengths from the two slits to the location of a second order bright fringe on the screen?(b) What is the difference in path lengths from the two slits to the location of the second dark fringe on the screen, away from the center of the pattern?
Answers
Answer:
Rounded to three significant figures:
(a) [tex]2 \times 575\; \rm nm = 1150\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
(b) [tex]\displaystyle \left(1 + \frac{1}{2}\right) \times (575\;\rm nm) \approx 863\; \rm nm = 8.63\times 10^{-7}\; \rm m[/tex].
Explanation:
Consider a double-slit experiment where a wide beam of monochromatic light arrives at a filter with a double slit. On the other side of the filter, the two slits will appear like two point light sources that are in phase with each other. For each point on the screen, "path" refers to the length of the segment joining that point and each of the two slits. "Path difference" will thus refer to the difference between these two lengths.
Let [tex]k[/tex] denote a natural number ([tex]k \in \left\lbrace0,\, 1,\, 2,\, \dots\right\rbrace[/tex].) In a double-split experiment of a monochromatic light:
A maximum (a bright fringe) is produced when light from the two slits arrive while they were in-phase. That happens when the path difference is an integer multiple of wavelength. That is: [tex]\text{Path difference} = k\, \lambda[/tex].Similarly, a minimum (a dark fringe) is produced when light from the two slits arrive out of phase by exactly one-half of the cycle. For example, The first wave would be at peak while the second would be at a crest when they arrive at the screen. That happens when the path difference is an integer multiple of wavelength plus one-half of the wavelength: [tex]\displaystyle \text{Path difference} = \left(k + \frac{1}{2}\right)\cdot \lambda[/tex].MaximaThe path difference is at a minimum (zero) at the center of the screen between the two slits. That's the position of the first maximum- the central maximum, a bright fringe where [tex]k = 0[/tex] in [tex]\text{Path difference} = 0[/tex].
The path difference increases while moving on the screen away from the center. The first order maximum is at [tex]k = 1[/tex] where [tex]\text{Path difference} = \lambda[/tex].
Similarly, the second order maximum is at [tex]k = 2[/tex] where [tex]\text{Path difference} = 2\, \lambda[/tex]. For the light in this question, at the second order maximum: [tex]\text{Path difference} = 2\, \lambda = 2 \times 575\; \rm nm = 1.15\times 10^{-6}\; \rm m[/tex].
Central maximum: [tex]k = 0[/tex], such that [tex]\text{Path difference} = 0[/tex].First maximum: [tex]k = 1[/tex], such that [tex]\text{Path difference} = \lambda[/tex].Second maximum: [tex]k = 2[/tex], such that [tex]\text{Path difference} = 2\, \lambda[/tex].MinimaThe dark fringe closest to the center of the screen is the first minimum. [tex]\displaystyle \text{Path difference} = \left(0 + \frac{1}{2}\right)\cdot \lambda = \frac{1}{2}\, \lambda[/tex] at that point.
Add one wavelength to that path difference gives another dark fringe- the second minimum. [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex] at that point.
First minimum: [tex]k =0[/tex], such that [tex]\displaystyle \text{Path difference} = \frac{1}{2}\, \lambda[/tex].Second minimum: [tex]k =1[/tex], such that [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda[/tex].For the light in this question, at the second order minimum: [tex]\displaystyle \text{Path difference} = \left(1 + \frac{1}{2}\right)\cdot \lambda = \left(1 + \frac{1}{2}\right)\times (575\; \rm nm) \approx 8.63\times 10^{-7}\; \rm m[/tex].
what type of image is formed by a lens if m = -0.87
A. The image is larger than the object and the image is real
B. The image is smaller than the object and the object is real
C. The image is larger than the object and is virtual
D. The image is smaller than the object and is virtual
Answers
Answer:
B.
Explanation:
as there is a '-' sign before the magnification value, it forms a real image.
so the last two options get cancelled.
as the value of magnification is 0.87 ie. lesser than 1, we can say that the image is smaller.
The type of image is formed by a lens if m = -0.87 would be smaller than the object and the object is real
What is a lens?A lens is a transmissive optical tool that employs refraction to focus or disperse a light beam. The power of the lens is expressed in the dioptre which is the reciprocal of the focal length of the lens.
The image formation through the lens is calculated with the help of the lens formula given as follows
1/f = 1/v - 1/u
where f is the focal length of the lens
u is the distance of the object from the lens
v is the distance of the image formed from the
magnification (m) = size of image /size of the object
As given in the problem image is formed by a lens if m = -0.87
Thus, the negative sign represents that the image formed by the object is real, and the magnitude of the magnification is less than 1 this represents that size of the image is smaller than the size of the object.
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