Answer:
6000 counts per secondExplanation:
If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;
2000 counts per second = 1 meter ... 1
In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;
x count per second = 3 meter ... 2
Solving the two expressions simultaneously for x we will have;
2000 counts per second = 1 meter
x counts per second = 3 meter
Cross multiply to get x
2000 * 3 = 1* x
6000 = x
This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample
Suppose an electron is trapped within a small region and the uncertainty in its position is 24.0 x 10-15 m. What is the minimum uncertainty in the electron's momentum
Answer:
Uncertainty in position (∆x) = 24 × 10⁻¹⁵ mUncertainty in momentum (∆P) = ?Planck's constant (h) = 6.26 × 10⁻³⁴ Js[tex]\longrightarrow \: \: \sf\Delta x .\Delta p = \dfrac{h}{4\pi} [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} {4 \times \frac{22}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34}} { \frac{88}{7} } [/tex]
[tex]\longrightarrow \: \: \sf24 \times {10}^{ - 15} .\Delta p = \dfrac{6.26 \times {10}^{ - 34} \times 7} { 8 } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 8 \times 24 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} } { 192 \times {10}^{ - 15} } [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ - 34} \times {10}^{15} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{43.82 \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 2} \times {10}^{ -19} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = \dfrac{4382 \times {10}^{ - 21} } { 192} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 22.822\times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \sf\Delta p = 2.2822 \times {10}^{1} \times {10}^{ - 21} [/tex]
[tex]\longrightarrow \: \: \underline{ \boxed{ \red{ \bf\Delta p = 2.2822 \times {10}^{ - 20} \: kg/ms}}}[/tex]
The kinetic energy of an object with a mass of 6.8 kg and a velocity of 5.0 m/s is J. (Report the answer to two significant figures.)
Answer:
[tex] \boxed{\sf Kinetic \ energy \ (KE) = 85 \ J} [/tex]
Given:
Mass (m) = 6.8 kg
Speed (v) = 5.0 m/s
To Find:
Kinetic energy (KE)
Explanation:
Formula:
[tex] \boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}[/tex]
Substituting values of m & v in the equation:
[tex] \sf \implies KE = \frac{1}{2} \times 6.8 \times {5}^{2} [/tex]
[tex] \sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 3.4 \times 25 [/tex]
[tex] \sf \implies KE =3.4 \times 25 [/tex]
[tex] \sf \implies KE = 85 \: J[/tex]
The kinetic energy of the object reported to two significant figures is: 85 Joules.
Given the following data:
Mass = 6.8 kg Velocity = 5.0 m/s.To find the kinetic energy of the object:
Kinetic energy refers to an energy that is possessed by a physical object or body due to its motion.
Mathematically, kinetic energy is calculated by using the formula;
[tex]K.E = \frac{1}{2} MV^2[/tex]
Where:
K.E is the kinetic energy. M is the mass of an object. V is the velocity of an object.Substituting the parameters into the formula, we have;
[tex]K.E = \frac{1}{2}[/tex] × [tex]6.8[/tex] × [tex]5^2[/tex]
[tex]K.E = 3.4[/tex] × [tex]25[/tex]
Kinetic energy = 85 Joules.
Therefore, the kinetic energy of the object is 85 Joules.
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A satellite of mass m circles a planet of mass M and radius R in an orbit at a height 2R above the surface of the planet. What minimum energy is required to change the orbit to one for which the height of the satellite is 3R above the surface of the planet
Answer:
ΔE = GMm/24R
Explanation:
centripetal acceleration a = V^2 / R = 2T/mr
T= kinetic energy
m= mass of satellite, r= radius of earth
= gravitational acceleration = GM / r^2
Now, solving for the kinetic energy:
T = GMm / 2r = -1/2 U,
where U is the potential energy
So the total energy is:
E = T+U = -GMm / 2r
Now we want to find the energy difference as r goes from one orbital radius to another:
ΔE = GMm/2 (1/R_1 - 1/R_2)
So in this case, R_1 is 3R (planet's radius + orbital altitude) and R_2 is 4R
ΔE = GMm/2R (1/3 - 1/4)
ΔE = GMm/24R
3. A very light bamboo fishing rod 3.0 m long is secured to a boat at the bottom end. It is
held in equilibrium by an 18 N horizontal force while a fish pulls on a fishing line
attached to the rod shown below. How much force F does the fishing line exert on the
rod? (3)
18 N
pivot
30°
1.8 m
3.0 in
The image in the attachment describes the situation of the fishing rod.
Answer: F = 10.8 N
Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.
Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:
τ = F.r.sin(θ)
F is the force acting on the object;
r is distance between where the torque is measured to where the force is applied;
θ is the angle between F and r;
For the fishing rod:
[tex]\tau_{1} = \tau_{2}[/tex]
[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]
Assuming part (1) is related to unknown force:
[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]
Replacing the corresponding values:
[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]
[tex]F = \frac{18*1.8}{3}[/tex]
F = 10.8
The fishing line exert on the the rod a force of 10.8N.
Two football teams, the Raiders and the 49ers are engaged in a tug-of-war. The Raiders are pulling with a force of 5000N. Which of the following is an accurate statement?
A. The tension in the rope depends on whether or not the teams are in equilibrium.
B. The 49ers are pulling with a force of more than 5000N because of course they’d be winning.
C. The 49ers are pulling with a force of 5000N.
D. The tension in the rope is 10,000N.
E. None of these statements are true.
Answer:
E. None of these statements are true.
Explanation:
We can't say the exact or approximate amount of tension on the rope, since we do know for sure from the statement who is winning.
for A, the tension on the rope does not depend on if both teams pull are in equilibrium.
for B, the 49ers would be pulling with a force more than 5000 N, if they were winning. The problem is that we can't say with all confidence that they'd be winning.
for C, we don't know how much tension exists on the rope, and its direction, so we can't work out how much tension the 49ers are pulling the rope with.
for D, just as for C above, we can't work out how much tension there is on the rope, since we do not know how much force the 49ers are pulling with.
we go with option E.
OBJECTI
1. The motion of a liquid inside a U-tube is an
example of what type of motion?
a. Simple Harmonic c. Random
b.Rectilinear
d. Circular
Answer:
option A
Explanation:
simple harmonic motion
Answer:
random motion I think not sure
If a train travels at a constant 18.0 m/s, how far would it move in one hour? In 1.00 minute? In 1.00 second?
Explanation:
Distance = speed × time
d = (18.0 m/s) (1 hr × 3600 s/hr)
d = 64,800 m
d = (18.0 m/s) (1 min × 60 s/min)
d = 1080 m
d = (18.0 m/s) (1 s)
d = 18.0 m
an electron travels at 0.3037 times the speed of light through a magnetic field and feels a force of 1.2498 pN. What is the magnetic field in teslas
Answer:
Explanation:
Charge on an electron (q) = 1.6 * 10 ^ -19 C
Velocity of electron (v) = 0.3037 * 300,000,000 = 91,110,000 m/sec
We know that, Force exerted on moving particle moving through a magnetic field :
[tex]F= q * v * B ( q,v\ and\ B\ are\ mutually\ perpendicular)[/tex]
1.2498 * 10 ^ -12 = 1.6 * 10^ -19 * 91110000 * B
B = 0.08573 T
A city of Punjab has a 15 percent chance of wet weather on any given day. What is the probability that it will take a week for it three wet weather on 3 separate days?
Answer: 0.0617
Explanation:
Given: The probability of wet weather on any given day in a city of Punjab : p=15%=0.15
Let X be a binomial variable that represents the number of days having wet weather.
Binomial probability formula : [tex]P(X=x)=^nC_xp^x(1-p)^x[/tex], where n= total outcomes, p = probability of success in each outcomes.
Here, n= 7 ( 1 week = 7 days)
The probability that it will take a week for it three wet weather on 3 separate days:
[tex]P(X=3)^=\ ^7C_3(0.15)^3(1-0.15)^{7-3}\\\\=\dfrac{7!}{3!(7-3)!}(0.15)^3(0.85)^4\\\\=\dfrac{7\times6\times5}{3\times2}\times 0.003375\times0.52200625\approx0.0617[/tex]
Hence, the required probability =0.0617
calculate the rate of loss of heat through a glass window of area 200 CM square and thickness 0.5 CM where temperature inside is 35 degree Celsius and outside is -5 degree Celsius conductivity of Glass is 2.2 into 10 to the power 3 cal per s per cm per k .
Answer:
The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass
Explanation:
If Superman really had x-ray vision at 0.12 nm wavelength and a 4.1 mm pupil diameter, at what maximum altitude could he distinguish villains from heroes, assuming that he needs to resolve points separated by 5.4 cm to do this?
Answer:
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
Explanation:
Given:
wavelength (λ) = 0.12 nm = 0.12 × 10⁻⁹ m
Pupil Diameter (d) = 4.1 mm = 4 × 10⁻³ m
Separation distance (D) = 5.4 cm = 0.054 m
Find:
Maximum altitude to see(L)
Computation:
Resolving power = 1.22(λ / d)
D / L = 1.22(λ / d)
0.054 / L = 1.22 [(0.12 × 10⁻⁹) / (4 × 10⁻³ m)]
0.054 / L = 1.22 [0.03 × 10⁻⁶]
L = 0.054 / 1.22 [0.03 × 10⁻⁶]
L = 0.054 / [0.0366 × 10⁻⁶]
L = 1.47 × 10⁶
Maximum altitude to see(L) = 1.47 × 10⁶ m (Approx)
A 60 mAs results in an exposure of 85 mGya, with all factors remaining the same, what would the new exposure be if 120 mAs is used?
Answer: d₂ = 170 mGya
Explanation:
the relationship between absonbed 'd' and exposure 'E' is given as;
D(Gv) = F . x (AS/xB)
F is a conversion coefficient depending on medium
so we can simply write
d₁/d₂ = x₁/x₂
Given that;
our x₁ = 60 mAs, x₂ = 120 mAs, d₁ = 85 mGya, d₂ = ?
from the given formula,
d₂ = (x₂d₁ / x₁)
now we substitute
d₂ = (120 × 85) / 60
d₂ = 170 mGya
∴ if 120 mAa is used, the new exposure will be 170 mGya
Determine usando ecuación de Bernoulli la Presión P1 necesaria para mantener la condición mostrada dentro del sistema mostrado en la figura, sabiendo que el aceite tiene un s.g =0.45 y el valor de d=90mm.
Answer:
PlROCA
Explanation:
A linear accelerator can be used to accelerate which of the following?
Question 3 options:
protons and electrons
protons and neutrons
protons only
protons, electrons, and neutrons
What did Edwin Hubble discover by observing absorption spectrum
blue light
red light
grey matter
ultra-violet light
Answer:
According to different sources Edwin Hubble observed red light of galaxies directly proportional to the distance of the galaxy from earth.
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a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with
Complete question:
a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of 955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.
Answer:
The initial speed of the block is 1.422 m/s
Explanation:
Given;
mass of the block, m = 2.0 kg
force constant of the spring, K = 955 N/m
compression of the spring, x = 4.6 cm = 0.046 m
Apply Hook's law to determine applied force on the spring;
F = Kx
F = (955 N/m)(0.046 m)
F = 43.93 N
Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;
F = ma
a = F / m
a = 43.93 / 2
a = 21.965 m/s²
Apply kinematic equation to determine the initial speed of the block;
v² = u² + 2ax
where;
v is the final speed of the block = 0
u is the initial speed of the block
x is the distance traveled by the block = compression of the spring
a is the block deceleration = -21.965 m/s²
0 = u² + 2(-21.965 )(0.046)
0 = u² - 2.021
u² = 2.021
u = √2.021
u = 1.422 m/s
Therefore, the initial speed of the block is 1.422 m/s
Convert 76.2 kilometers to meters?
Answer
76200meters
Explanation:
we know that 1km=1000meters
to convert km into meters we we divide km by meters
=76.2/1000
=76200meters
To celebrate a victory, a pitcher throws her glove straight upward with an initial speed of 5.0 m/s. How much time does it take for the glove to return to the pitcher
Answer:
The glove takes 1.02s to return to the pitchers hand.
Explanation:
Given;
initial velocity the pitcher's glove, u = 5 m/s
Apply kinematic equation
s = ut - ¹/₂gt²
where;
g is acceleration due to gravity = 9.8 m/s²
t is the time takes the glove to return to the pitchers hand
s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)
0 = ut - ¹/₂gt²
ut = ¹/₂gt²
u = ¹/₂gt
gt = 2u
t = (2u) / g
t = (2 x 5) / 9.8
t = 1.02 s
Therefore, the glove takes 1.02s to return to the pitchers hand.
A mass m = 0.6 kg is released from rest at the top edge of a hemispherical bowl with radius = 1.1 meters. The mass then slides without friction down the inner surface toward the bottom of the bowl. At a certain point of its path the mass achieves a speed v = 3.57 m/s. At this point, what angle \theta\:θ ( in degrees) does the mass make with the top of the bowl?
Answer:
The angle is [tex]\theta = 36.24 ^o[/tex]
Explanation:
From the question we are told that
The mass is [tex]m = 0.6 \ kg[/tex]
The radius is [tex]r = 1.1 \ m[/tex]
The speed is [tex]v = 3.57 \ m /s[/tex]
According to the law of energy conservation
The potential energy of the mass at the top is equal to the kinetic energy at the bottom i.e
[tex]m * g * h = \frac{1}{2} * m * v^2[/tex]
=> [tex]h = \frac{1}{2 g } * v^2[/tex]
Here h is the vertical distance traveled by the mass which is also mathematically represented as
[tex]h = r * sin (\theta )[/tex]
So
[tex]\theta = sin ^{-1} [ \frac{1}{2* g* r } * v^2][/tex]
substituting values
[tex]\theta = sin ^{-1} [ \frac{1}{2* 9.8* 1.1 } * (3.57)^2][/tex]
[tex]\theta = 36.24 ^o[/tex]
A long straight solenoid has 800 turns. When the current in the solenoid is 2.90 amperes the average flux through each turn is 3.25×10−3Wb.
A. What is the inductance of the coil?
B. What must be the magnitude fo the rate of change of the current (di/dt) in order for the self-induced emf to equal 7.50 mV?
Answer:
Explanation:
Relation between flux and inductance is as follows
φ = Li
where φ is flux associated with induction of inductance L when a current i flows through it
putting the values
3.25 x 10⁻³ x 800 = L x 2.9
L = .9 H
for induced emf in an induction , the relation is
emf induced = L di / dt
Putting the values
7.5 x 10⁻³ = .9 x di / dt
di / dt = 8.33 x 10⁻³ A / s
(a) The self inductance of the solenoid is 0.897 H.
(b) The magnitude of the rate of change of the current is 0.00836 A/s.
The given parameters;
number of turns, N = 800 turnscurrent in the solenoid, I = 2.9 flux through the solenoid, Ф = 3.25 x 10⁻³ WbThe self inductance of the solenoid is calculated as follows;
[tex]emf = \frac{d\phi}{dt}\\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\NBA = LI\\\\L = \frac{NBA}{I} \\\\L = \frac{N\phi}{I} \\\\L = \frac{800 \times 3.25\times 10^{-3}}{2.9} \\\\L = 0.897 \ H\\\\[/tex]
The magnitude of the rate of change of the current is calculated as follows;
[tex]emf = L \frac{di}{dt} \\\\\frac{di}{dt} \ = \frac{emf}{L} \\\\\frac{di}{dt} = \frac{7.5 \times 10^{-3}}{0.897} \\\\\frac{di}{dt} = 0.00836 \ A/s[/tex]
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PLEASE HELP ME WITH THIS PHYSICS QUESTION PLSSS!!!
Vf^2 = v0^2 + 2a (xf -x0)
Solve for v0
b. solve for Vo
[tex] vf ^{2} = vo^{2} + 2a(xf - xo) \\ vf ^{2} = vo ^{2} + 2axf - 2axo\\ vo ^{2} = vf ^{2} - 2axf + 2axo \\ vo = \sqrt{vf ^{2} - 2axf + 2axo } \\ vo = - \sqrt{vf^{2} - 2axf + 2axo } [/tex]
I hope I helped you ^_^
Calcular la resistencia de una varilla de grafito de 170 cm de longitud y 60 mm2. Resistividad grafito 3,5 10-5 Ωm
Answer:
R = 0.992 Ω
Explanation:
En esta pregunta, dada la información que contiene, debemos calcular la resistencia de la varilla de grafito.
Matemáticamente,
Resistencia = (resistividad * longitud) / Área De la pregunta;
Resistividad = 3,5 * 10 ^ -5 Ωm
longitud = 170 cm = 1,7 m
Área = 60 mm ^ 2 = 60/1000000 = 6 * 10 ^ -5 m ^ 2
Conectando estos valores a la ecuación anterior, tenemos;
Resistencia = (3.5 * 10 ^ -5 * 1.7) / (6 * 10 ^ -5) =
(3.5 * 1.7) / 6 = 0.992 Ω
A competitive diver leaves the diving board and falls toward the water with her body straight and rotating slowly. She pulls her arms and legs into a tight tuck position. What happens to her rotational kinetic energy
Answer: her rotational kinetic energy increases
A sinusoidal electromagnetic wave is propagating in a vacuum in the +z-direction. If at a particular instant and at a certain point in space the electric field is in the +x-direction and has a magnitude of 4.00 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time?
Answer:
B = 1.33 10⁻⁸ T , the magnetic field must be in the y + direction
Explanation:
In an electromagnetic wave the electric and magnetic fields are in phase
c = E / B
B = E / c
let's calculate
B = 4.00 / 3 10⁸
B = 1.33 10⁻⁸ T
To determine the direction we use that the electric and magnetic fields and the speed of the wave are perpendicular.
If the wave advances in the + Z direction and the electric field is in the + x direction, the magnetic field must be in the y + direction
You shine unpolarized light with intensity 52.0 W/m2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 15.0 W/m2. Find the intensity of the light that emerges from the first polarizer.
Answer:
The intensity of light from the first polarizer is [tex]I_1 = 26 W/m^2[/tex]
Explanation:
The intensity of the unpolarized light is [tex]I_o = 52.0 \ W/m^2[/tex]
Generally the intensity of light that emerges from the first polarized light is
[tex]I_1 = \frac{I_o}{2 }[/tex]
substituting values
[tex]I_1 = \frac{52. 0}{2 }[/tex]
[tex]I_1 = 26 W/m^2[/tex]
A polarized laser beam of intensity 285 W/m2 shines on an ideal polarizer. The angle between the polarization direction of the laser beam and the polarizing axis of the polarizer is 16.0 ∘. What is the intensity of the light that emerges from the polarizer?
Answer:
The intensity is [tex]I_1 = 263.35 \ W/m^2[/tex]
Explanation:
From the question we are told that
The intensity of the beam is [tex]I = 285\ W/m^2[/tex]
The angle is [tex]\theta = 16^o[/tex]
The intensity of the light that emerges from the polarizer is mathematically represented by Malus' law as
[tex]I_1 = I * cos^2 (\theta )[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
substituting values
[tex]I_1 = 285 * [cos(16)]^2[/tex]
[tex]I_1 = 263.35 \ W/m^2[/tex]
A 25g rock is rolling at a speed of 5 m/s. What is the kinetic energy of the rock?
Answer:
The answer is 312.5j
Explanation:
The kinetic energy (KE):
KE=1/2*m*v^2
M= mass of the object
v= velocity of the object
We have;
m=25g
v=5m/s
KE=1/2*25g*5^2m/s
KE =312.5j
An electron experiences a force of magnitude F when it is 5 cm from a very long, charged wire with linear charge density, lambda. If the charge density is doubled, at what distance from the wire will a proton experience a force of the same magnitude F?
Answer:
The distance of the proton is [tex]r_p =10 \ cm[/tex]
Explanation:
Generally the force experience by the electron is mathematically represented as
[tex]F_e = \frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e}[/tex]
Where [tex]\lambda _e[/tex] is the charge density of the charge wire before it is doubled
Also the force experience by the proton is mathematically represented as
[tex]F_p = \frac{q * \lambda_p }{ 2 \pi * \epsilon_o * r_p}[/tex]
Given that the charge density is doubled i.e [tex]\lambda_p = 2 \lambda_e[/tex] and that the the force are equal then
[tex]\frac{q * \lambda_e }{ 2 \pi * \epsilon_o * r_e} = \frac{q * 2 \lambda_e }{ 2 \pi * \epsilon_o * r_p}[/tex]
[tex]\frac{ \lambda_e }{ r_e} = \frac{ 2 \lambda_e }{ r_p}[/tex]
[tex]r_p * \lambda_e =2 \lambda_e * r_e[/tex]
[tex]r_p =2 r_e[/tex]
Now given from the question that [tex]r_e[/tex] the distance of the electron from the charged wire is 5 cm
Then
[tex]r_p =2 (5)[/tex]
[tex]r_p =10 \ cm[/tex]
In an adiabatic process:
a. the energy absorbed as heat equals the work done by the systemon its environment
b. the energy absorbed as heat equals the work done by theenvironment on the system
c. the work done by the environment on the system equals the changein internal energy
Answer:
c. the work done by the environment on the system equals the changein internal energy.
Explanation:
Adiabatic process:
When the boundary of a system is perfectly insulated, it means that the energy can not flow from the system and into the system ,these system is known as adiabatic system.
When the energy transfer in the system is zero ,then these type of process is known as adiabatic process.
From the first law of thermodynamics
Q= ΔU + W
Q=Heat transfer
ΔU=Change in internal energy
W=Work transfer
In adiabatic process , Q= 0
Therefore
0=ΔU +W
W=- ΔU
Negative sign indicates that ,the work done by the environment.
Therefore the correct option will be (c).
A concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.
Answer:
D.
Inverted and reduced
If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
What is a concave mirror?When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.
A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.
A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.
Hence option B is correct.
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