Answer:
[tex]d_{i}=465 cm=4.65 m[/tex]
Explanation:
Using the thin lens equation, we have:
[tex]\frac{1}{f}=\frac{1}{d_{o}}+\frac{1}{d_{i}}[/tex]
Where:
f is the focal length
d(o) is the distance of the object
d(i) is the distance of the image
[tex]\frac{1}{15}=\frac{1}{15.5}+\frac{1}{d_{i}}[/tex]
[tex]\frac{1}{15}-\frac{1}{15.5}=\frac{1}{d_{i}}[/tex]
[tex]d_{i}=465 cm=4.65 m[/tex]
Therefore, the screen is at 465 cm.
I hope it helps you!
A car travels through a valley at constant speed, though not at constant velocity. Explain how this is possible.
Answer:
Explanation:
An object following a circular path can be covering the same distance along the circle's circumference with every passing minute, giving it a constant speed. Since a change in either speed OR direction means a change in velocity, the object's velocity is not constant.
velocity is a vector so therefore direction affects it being constant.
An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.
Velocity is a vector so therefore direction affects it being constant.
What is velocity?When an item is moving, its velocity is the rate at which its direction is changing as seen from a certain point of view and as measured by a specific unit of time
An object following a circular path can be covering the same distance along the circle's circumference with every passing minute, giving it a constant speed. Since a change in either speed OR direction means a change in velocity, the object's velocity is not constant. An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.
Velocity is a vector so therefore direction affects it being constant.
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A football player kicks a ball with a mass of 0.5 kg. The average acceleration of the football was 15 m/s/s. How much force did the kicker apply to the football?
A. 15.5 N
B. 7.5 N
C. 14.5 N
D. 30 N
Answer:
Explanation:
F = m*a = (0.5)*15 = 7.5 N
4. What is the electric field strength (E) at a distance of 0.50 m from a 1.00×10°C charge?
Answer:
3.6*10¹⁰N/C
Explanation:
The formula for Electric field strength is expressed as:
E = kQ/r²
k is the coulombs constant = 9*10^9Nm²/C²
Q is the charge = 1.00×10°C
r is the distance = 0.50m
Substitute the parameters into the formula as shown:
E = kQ/r²
E = 9*10^9(1)/0.5²
E = 9*10^9/0.25
E = 36*10^9
E = 3.6*10¹⁰N/C
Hence the electric field strength is 3.6*10¹⁰N/C
4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going when it hit the ground?
Answer:
v= 17.15 m/s
Explanation:
mass of the book=0.1 Kg
height above ground, h= 15 m
Using conservation of energy
Potential energy is converted into kinetic energy
[tex]mgh = \frac{1}{2}mv^2[/tex]
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 15}[/tex]
[tex]v=\sqrt{294}[/tex]
v= 17.15 m/s
Hence, the book will hit the ground at the speed of 17.15 m/s.
a 45 kg boy sits on a horse on a carousel 5.0 m from the center of the circle. he makes a revolution every 8.0 s.
calculate his speed.
what is centripetal force acting on the boy?
For every complete revolution the boy makes around the center of the carousel, he travels a distance of 2π (5.0 m) = 10π m, which gives a linear speed of
v = (10π m) / (8.0 s) ≈ 3.927 m/s
Then his centripetal acceleration would be
a = v ² / (5.0 m) ≈ 3.084 m/s²
so that the centripetal force exerted on him has magnitude
F = (45 kg) a ≈ 138.791 N ≈ 140 N
(rounded to 2 significant digits)
please help thank you
which of the following statements BEST describes the difference between an atom and an ion ?
Answer:
well the correct answer is
d. An atom contains equal numbers of protons and electrons whereas an ion contains unequal numbers of protons and electrons .Explanation:
A charged atom is known as an ion, well it can be negative as well as positive charge.
if atom has more protons than electrons then it get positively charged and known as cationif the atom has more electrons that the number of protons then the atom get negatively charged and known as anionWhy does liquid water boil away when exposed to the Martian atmosphere?
a. the surface of Mars is much hotter than the surface of Earth.
b. it is quickly absorbed into the dry soil.
c. radiation from the Sun passes through Mars' atmosphere and boils the water.
d. the low air pressure lowers the boiling point
Answer: d. the low air pressure lowers the boiling point
Explanation:
Note that there cannot be less than 6.1 millibars of liquid water. This fact is important because the Martian surface atmospheric pressure hovers just below that value. Any water that could form from melting water on a warm afternoon will easily vanish into the desiccated Martian atmosphere.
A 151 kg crate is pulled along a level surface by an engine. The coefficient of kinetic friction between the crate and the surface is 0.399. How much power must the engine deliver to move the crate at a constant speed of 6.04 m/s
Answer:
3566.26 Watts
Explanation:
From the question given above, the following data were obtained:
Mass (m) of crate = 151 kg
Coefficient of kinetic friction (μ) = 0.399
Velocity (v) = 6.04 m/s
Power (P) =?
Next, we shall determine the normal reaction. This can be obtained as follow:
Mass (m) of crate = 151 kg
Acceleration due to gravity (g) = 9.8 m/s²
Normal reaction (R) = mg
Normal reaction (R) = 151 × 9.8
Normal reaction (R) = 1479.8 N
Next, we shall determine the force applied to move the crate. This can be obtained as follow:
Coefficient of kinetic friction (μ) = 0.399
Normal reaction (R) = 1479.8 N
Force (F) =?
F = μR
F = 0.399 × 1479.8
F = 590.44 N
Finally, we shall determine the power used to move the crate as follow:
Velocity (v) = 6.04 m/s
Force (F) applied = 590.44 N
Power (P) =?
P = F × v
P = 590.44 × 6.04
P = 3566.26 Watts
Therefore, the power used to move the crate is 3566.26 Watts.
Why is 1/2kx^2=gym not a linear equation
describe the importance of the neutron in a atomic nuclei
Neutrons are required for the stability of nuclei, with the exception of the single-proton hydrogen nucleus. Neutrons are produced copiously in nuclear fission and fusion. They are a primary contributor to the nucleosynthesis of chemical elements within stars through fission, fusion, and neutron capture processes.
Hope it helps!
PLEASE HELP i’m giving 32 points!!
1.) How many grams of
potassium chloride (KCI)
can be dissolved at 80°C?
2.) At 50°C, how much potassium chlorate (KCIO3) can be dissolved in 300 grams of water?
Answer: 1 How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of K2Cr2O7 is added to 100 g H2O at. 0 °C. With constant stirring, to what temp-. 2 34 °C? 4. How many grams of KCl will dissolve in 1 liter of H2O at 50 °C? 5. 58.0 g of ... A saturated solution of KClO3 was made with 300 g of H2O at. 34 °C.
Explanation:
Help! Help!
Alcohol abuse has...
A. only physiological aspects.
B. only psychological aspects.
C. physiological and psychological aspects.
Answer
I feel the answer is C because it could cause mental and physical trauma
Explanation:
What is the force needed to accelerate a 10 kg object at 2 m/s??
Answer:
5 newtons
Explanation:
Just divide them
short essay your teacher have provided you
Deshaun Watson launches a football at a speed of 24.7 ms and an angle of 33° above the horizontal How far down
the football field does the football land? What is the max height the football reaches during flight?
Show work
Answer:
9.23m
Explanation:
Max height = u²sin²theta/2g
u is the speed = 34.7m/s
theta is the angle of elevation = 33°
g is the acceleration due to gravity = 9.8m/s²
Substitute into the formula
Max height = 24.7²sin²33/2(9.8)
Max height = 610.09sin²33/2(9.8)
Max height = 610.09(0.29663)/19.6
Max height = 180.97/19.6
Max Height = 9.23m
Hence the max height the football reaches during flight is 9.23m
A 9.0 × 10 3 kg satellite orbits the earth at the distance of 2.56 × 10 7 m from Earth's surface. What is its period?
Answer:
1.6537 * 10^9
Explanation:
Given that :
Distance (r) = 2.56 × 10^7 m
Using the relation from Kepler's third law of motion :
T2 = [ 4* pi^2/ G (M1 + M2) ]r^3
M1 = mass of earth = 5.97 *10^24
M2 = Mass of satellite = 9.0 × 10^3 kg
G = Gravitational constant = 6.67 *10^-11
T2 = [ (4 x 3.14^2) / 6.67 x 10^-11 ( 9 x 10^3 + 5.97 x 10^24)]/ (2.56 x 10^7)^3
= 1.6537 * 10^9
What 2 factors do you need in order to calculate speed?
Answer:
Distance and time.
Explanation:
Speed=Distance/time
The two factors which we need in order to calculate the speed of an object are the distance covered by the object and the time taken to cover that distance.
What is Speed?
Speed is the rate of change of position of an object in any direction. Speed is a scalar quantity as it has only magnitude and no direction. It is measured as the ratio of the distance covered by an object to the time taken in which the distance was covered by that object.
Speed has the dimension of distance covered by the time taken. Thus, the SI unit of speed is the combination of the basic units of distance and the basic unit of Time. Thus, the SI unit of speed is meter per second (m/s).
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le Which describes how light waves can interact with wood?
(14 Points)
O A. absorption and reflection
O B. diffraction and transmission
0 C. reflection and refraction
O D. transmission and refraction
Answer: A ,absorption and reflection
Explanation
Absorption and reflection are the two phenomena which describe how the light waves can interact with the wood. Thus, the correct option is A.
What is Absorption and reflection of light?Light absorption is a process through which light is absorbed and converted into different form of energy. An example of absorption of light is the process of photosynthesis which takes place in plants. However, light absorption does not occur exclusively in the plants, but in all the creatures and inorganic substances present on this planet.
When a ray of light approaches a smooth polished surface such as a metal surface and the light ray bounces back, it is called as the reflection of light. The incident light ray which lands on the surface is reflected off the surface. The ray that bounces back from the surface is called the reflected ray.
Therefore, the correct option is A.
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A satellite of mass m orbits a moon of mass M in uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon. What is the gravitational field strength of the moon at this new distance
The satellite is moved to a new circular orbit that is 2R from the center of the moon, then the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.
What is gravity?It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another.
As given in the problem A satellite of mass m orbits a moon of mass M in a uniform circular motion with a constant tangential speed of v. The gravitational field strength at a distance R from the center of the moon is gR. The satellite is moved to a new circular orbit that is 2R from the center of the moon.
The gravitational field strength is inversely proportional to the square of the distance from the center of the planet.
Thus, the gravitational field strength of the moon at this new distance would be one-fourth of the initial gravitational field.
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Victor covers 210 km by car at a speed of 70 km/hr. find the time taken to cover this distance.
Answer:
3 hrs
Explanation:
the distance covered by victor= 210 km
speed= 70 km/hrs
so, 70×3= 210
so the answer is 3 hrs
BTW im a small kid so don't just right away say the explanation sucks and the subject physics is not yet started in my grade.
edit: don't give me brainless answer plz.
A woman does 236 J of work
dragging her trash can 24.4 m to
the curb, using a force of 18.9 N.
At what angle was her force
directed?
Answer:
workdone = force × distance 236J = 18.9cos(o) × 24.4236/24.4 = 18.9cos(o)(0.5117)cos^-1 = (o)59.21°A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the fraction of the length of the rod above water
Answer:
[tex]\frac{y}{L}[/tex] = 0.66
Hence, the fraction of the length of the rod above water = [tex]\frac{y}{L}[/tex] = 0.66
and fraction of the length of the rod submerged in water = 1 - [tex]\frac{y}{L}[/tex] = 1 - 0.66 = 0.34
Explanation:
Data given:
Density of the rod = 5/9 of the density of the water.
Let's denote density of Water with w
And density of rod with r
So,
r = 5/9 x w
Required:
Fraction of the length of the rod above water.
Let's denote total length of the rod with L
and length of the rod above with = y
Let's denote the density of rod = r
And density of water = w
So, the required is:
Fraction of the length of the rod above water = y/L
y/L = ?
In order to find this, we first need to find out the all type of forces acting upon the rod.
We know that, a body will come to equilibrium if the net torque acting upon a body is zero.
As, we know
F = ma
Density = m/v
m = Density x volume
Volume = Area x length = X ( L-y)
So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:
F = mg
F = (Density x volume) x g
g = gravitational acceleration
F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)
where,
X (L-y) = volume
w = density of water.
Another force acting upon it is:
F = mg
F2 = X x L x r x g
Now, the torques acting upon the body:
T1 + T2 = 0
F1 ( y + [tex](\frac{L-y}{2})[/tex] ) g sinФ - F2 x ([tex]\frac{L}{2}[/tex]) x gsinФ = 0
plug in the equations of F1 and F2 into the above equation and after simplification, we get:
[tex](L^{2} - y^{2} ) . w[/tex] = [tex]L^{2}[/tex] . r
where, w is the density of water and r is the density of rod.
As we know that,
r = 5/9 x w
So,
[tex](L^{2} - y^{2} ) . w[/tex] = [tex]L^{2}[/tex] . 5/9 x w
Hence,
[tex](L^{2} - y^{2} )[/tex] = [tex]\frac{5L^{2} }{9}[/tex]
[tex]\frac{L^{2} - y^{2} }{L^{2} }[/tex] = [tex]\frac{5}{9}[/tex]
Taking [tex]L^{2}[/tex] common and solving for [tex]\frac{y}{L}[/tex], we will get
[tex]\frac{y}{L}[/tex] = 0.66
Hence, the fraction of the length of the rod above water = [tex]\frac{y}{L}[/tex] = 0.66
and fraction of the length of the rod submerged in water = 1 - [tex]\frac{y}{L}[/tex] = 1 - 0.66 = 0.34
Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble Constant were 700 km/s/Mpc, what would this imply about the age of our universe?
Answer:
The current value of the Hubble's constant = 73 km/sec/Mpc.
t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc
Explanation:
The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.
Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.
Hence,
v is directly proportional to d
where, v = apparent velocity
d = distance
if we equate velocity and distance then there comes Hubble constant.
v = [tex]H_{0}[/tex] x d
[tex]H_{0}[/tex] = 73 km/sec/Mpc
where, Mpc = Mega Parsec = 1 Mpc = 3.086 x [tex]10^{19}[/tex] km
We can use Hubble constant to tell the age of universe.
t = d/v
t = d/( [tex]H_{0}[/tex] xd)
t = 1/[tex]H_{0}[/tex]
Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.
Now, if we hypothetically change the value of Hubble constant,
from [tex]H_{0}[/tex] = 73 km/sec/Mpc to [tex]H_{0}[/tex] = 700 km/sec/Mpc
then the age of universe will be:
t = 1/[tex]H_{0}[/tex]
first convert the units of new [tex]H_{0}[/tex] into 1/s
[tex]H_{0}[/tex] = (700) x (/3.08 x [tex]10^{19}[/tex] )
[tex]H_{0}[/tex] = 227.27 x[tex]10^{-19}[/tex] = 2.27 x [tex]10^{-21}[/tex] 1/s
So,
Age of universe will be:
t = 1/[tex]H_{0}[/tex] = 1/2.27x[tex]10^{-21}[/tex] 1/s
t = 2.27 x [tex]10^{21}[/tex] s
t = 71.9 trillion years
t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc
Given value:
Hubble Constant,
[tex]H_o = \frac{700 \ km/s}{Mpc}[/tex]We know,
[tex]Mpc = 3.086\times 10^{19} \ km[/tex]By substituting the value of "Mpc" in Hubble constant, we get
→ [tex]H_o = \frac{700}{3.086\times 10^{19}}[/tex]
[tex]= 227\times 10^{-19} \ 1/s[/tex]
[tex]= 2.27\times 10^{-21} \ 1/s[/tex]
The Hubble's time will be:
→ [tex]H_o = \frac{1}{t}[/tex]
or,
→ [tex]t = \frac{1}{H_o}[/tex]
[tex]= \frac{1}{2.27\times 10^{-21}}[/tex]
[tex]= 4.4\times 10^{20} \ seconds[/tex]
Thus the above approach is right.
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A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
the following ATTENTION：using the energy/work formulae only: 3.The kinetic energy of the Rock half way down ？ 4.the speed of the Rock half way down？
5.The speed of the Rock as it hits the ground？
Answer:
See the answers below.
Explanation:
In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].
[tex]E_{A}=E_{B}[/tex]
The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.
So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.
[tex]E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]
At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.
[tex]E_{B}=m*g*h+\frac{1}{2}*m*v^{2}[/tex]
Therefore we will have the following equation:
[tex](6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s][/tex]
The kinetic energy can be easily calculated by means of the kinetic energy equation.
[tex]KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J][/tex]
In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.
[tex]E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s][/tex]
Energy that travels in waves through matter
A mechanical wave is a disturbance in matter that transfers energy through the matter. The matter through which a mechanical wave travels is called the medium
or media in plural.
Mechanical wave are waves that travels through a medium or matter.
What are mechanical wave?A wave is a disturbance that travels or propagates from the place where it was created.
Waves transfer energy from one place to another, but they do not necessarily transfer any mass.
There are two types of waves;
mechanical wave and electromagnetic waves
electromagnetic waves are waves that that do not need any means of medium for propagation.
Examples of electromagnetic wave includes; light wave , radio wave e.t.c
Mechanical waves are waves that need medium for propagation. Example of mechanical waves includes ; sound wave , water wave e.t.c
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Stand in a doorway so your toes and nose are against the doorway. 4. Grab a weight in each hand and hold your arms out from your body on either side of the wall. 5. Try to stand on your tip toes. What happens
A 3kg plastic tank that has a volume of 0.2m ^3 Is filled with liquid water . Assuming the density of water is 1000kg/m^3 determine the weight of the combined system
Answer:
The weight of the combined system is 1989.8 N.
Explanation:
The weight, W, of an object can be expressed as;
W = mg
where m is the mass of the object, and g is the acceleration due to gravity.
Weight of the combined system = weight of tank + weight of water
mass of the plastic tank = 3 kg
weight of the plastic tank = m x g
= 3 x 9.8
= 29.4 N
Weight of the plastic tank is 29.4 N
But,
density = [tex]\frac{mass}{volume}[/tex]
mass = density x volume
volume of water = 0.2 [tex]m^{3}[/tex], density of water = 1000 kg/[tex]m^{3}[/tex].
mass = 1000 x 0.2
= 200
mass of water = 200 kg
weight of water = 200 x 9.8
= 1960 N
Thus,
combined weight of the system = 29.4 + 1960
= 1989.8 N
The weight of the combined system is 1989.8 N.
The wavelength of a particular color of yellow light is 579 nm. The energy of this wavelength of light is
Answer:
3.44× 10⁻¹⁹Joules
Explanation:
Energy of the wavelength is expressed using the formula:
E = hc/λ
h is the Planck constant
c is the velocity of light
λ is the wavelength
Given
h = 6.63 × 10^-34 m² kg / s
c = 3×10⁸ m/s
λ = 579nm = 579 × 10⁻⁹m
λ = 5.79× 10⁻⁷m
Substitute the given values into the formula
E = hc/λ
E = (6.63 × 10⁻³⁴× 3×10⁸)/5.79× 10⁻⁷
E = 19.89× 10⁻³⁴⁺⁸/5.79× 10⁻⁷
E = 19.89× 10⁻²⁶/5.79× 10⁻⁷
E = 3.44× 10⁻²⁶⁺⁷
E = 3.44× 10⁻¹⁹Joules
Hence the energy of this wavelength of light is 3.44× 10⁻¹⁹Joules
( I will give a brainliest )
What must be changed, temperature or heat energy, during condensation?
Answer:
The answer is temperature lol
Explanation:
:)
Helppp!!!!!!!!!!!!?!!??