Answer:
5.0cm
Explanation:
To get the height of the image, we will use the magnification formula as shown:
m = Hi/H = v/u
Hi is the image height
H is the object height
v is the image distance
u is the object distance
First, we need to get the image distance v
Using the mirror formula;
1/f = 1/v+ 1/u
1/-6 = 1/v + 1/12
1/v = -1/12 - 1/6
1/v = -1-2/12
1/v = -3/12
v = 12/-3cm
v = -4cm
Next is to get the image height Hi
Using the expression;
Hi/H = v/u
Hi/15 = 4/12
Hi/15 = 1/3
3Hi = 15
Hi = 15/3
Hi = 5.0cm
Hence the image height is 5.0cm
Answer:
D. 5.0 cm
Explanation:
got it correct on the test
Object A has a mass of 5 kg and a velocity of 6 m/s to the east while Object 1 point B has a mass of 12 kg and velocity 0.6 m/s also to the east. What is the momentum of the system? (Let east be positive)
help plss I got family
Answer:
Momentum of system = 37.2 Kgm/s.
Explanation:
Given the following data;
Mass A = 5 kgVelocity A = 6 m/sMass B = 12 kgVelocity B = 0.6 m/sTo find the momentum of the system;
Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.
Mathematically, momentum is given by the formula;
Momentum = mass * velocity
For object A;
Momentum A = 5 * 6
Momentum A = 30 Kgm/s
For object B;
Momentum B = 12 * 0.6
Momentum B = 7.2 Kgm/s
Next, we would determine the momentum of this system using the formula;
Momentum of system = Momentum A + Momentum B
Substituting the values into the formula, we have;
Momentum of system = 30 + 7.2
Momentum of system = 37.2 Kgm/s.
the force of attraction between the earth and objects on it is called..
Answer: gravitational pull? gravity
Explanation:
86,400 seconds into day
Answer:
60×60×24=86,400
Explanation:
i.e. 86,400 secs is 1 day
1 hour = 60 minutes
1 minute = 60 seconds
1 day = 24 hours
60 min. × 60 sec. × 24 hr.
60 × 60 × 24 = 86,400 seconds
∴ 86,400 seconds = 1 day
Clois what is the weight of a body in the earth, if its weig is 5Nin moon?
Explanation:
because the moon has less mass than earth, the force due to gravity at the lunar surface is only about 1/6 that on earthso,the weight of a body on earth is 6×5N =30N
What is the centripetal acceleration of a point on the perimeter of a bicycle wheel diameter 70.0 cm when the bike is moving 8.0 m/s? (160 m/s)
It’s 180 m/s^2 dude. I think I have you in my class lol.
The centripetal acceleration of an object is due to the changing velocity in a circular path and the centripetal acceleration of the bicycle is 182.85 m/s².
What is Centripetal acceleration?Centripetal acceleration of an object can be defined as the property of the motion of an object which is traversing a circular path. Any object which is moving in a circular path and has an acceleration vector pointed towards the center of that circular path is known as Centripetal acceleration.
The centripetal acceleration of an object can be calculated by the formula:
ac = v²/ r
where, ac = centripetal acceleration,
v = velocity of the object,
r = radius of the circular path
The centripetal acceleration of the object will be:
ac = (8)²/ r
radius = diameter/ 2
radius = 70/ 2
radius = 35cm or 0.35 m
ac = 64/ 0.35
ac = 182.85 m/s²
Therefore, the centripetal acceleration of the bicycle will be 182.85 m/s².
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Given: F = k· m. g
Solve for "k"
Answer:
[tex]F = kmg \\ k = \frac{F}{mg} [/tex]
Explanation:
F = k . m . g
=> F = k . mg
[tex] = > k = \frac{F}{mg} (ans)[/tex]
An ant moves towards the plane mirror with speed of 2 m/s & the mirror is moved towards the ant with the same speed. What is the relative velocity between the ant and its image?
We know
[tex]\boxed{\sf Relative\:velocity(V_{AB})=V_A-V_B}[/tex]
[tex]\\ \sf\longmapsto V_{AB}=2-2[/tex]
[tex]\\ \sf\longmapsto V_{AB}=0m/s[/tex]
The vector difference between the velocities of two bodies : the velocity of a body with respect to another regarded as being at rest compare relative motion
[tex]Relative velocity $\left(\mathrm{V}_{\mathrm{AB}}\right)=\mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}$$$\begin{aligned}&\longmapsto \mathrm{V}_{\mathrm{AB}}=2-2 \\&\longmapsto \mathrm{V}_{\mathrm{AB}}=0 \mathrm{~m} / \mathrm{s}\end{aligned}$$[/tex]
What is relative velocity and its unit?The relative velocity of an object with respect to another is the velocity with which one object moves with respect to another object. The unit of velocity can be referred to as the ratio of unit of distance and that of time. The SI unit of Relative velocity is meter per second.
What is absolute velocity?The concept of absolute velocity is mainly used in turbomachinery design and defines the velocity of a fluid particle in relation to the surrounding, stationary environment. Together with the relative velocity (w) and the circumferential speed (u), it forms the velocity triangle.
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Examine the motion map. One animal is an antelope that is already running. The other is a cheetah that starts running after the antelope passes it. Does A or B represent the motion of the cheetah?
Answer:
A although am not seeing the cheetah
como previene un diafragma el embarazo
Answer:
Un diafragma es una pieza de silicona o de goma reutilizable que cubre el cuello del útero. El diafragma se coloca dentro de la vagina con espermicida antes de tener relaciones sexuales para prevenir un embarazo. El diafragma es un dispositivo anticonceptivo que impide que los espermatozoides ingresen en el útero.
A skipper on a boat notices wave crests passing his anchor chain every 5.6 s . He estimates the distance between wave crests to be 16 m . He also correctly estimates the speed of the waves. Find this speed.
Answer:
v = 2.85 m/s
Explanation:
Given that,
A skipper on a boat notices wave crests passing his anchor chain every 5.6 s.
The distance between wave crests to be 16 m.
We need to find the speed of the waves. The speed of a wave can be calculated by the formula as follows :
[tex]v=f\lambda\\\\v=\dfrac{\lambda}{T}\\\\v=\dfrac{16}{5.6}\\\\v=2.85\ m/s[/tex]
So, the speed of the wave is 2.85 m/s.
Assuming Faraday constant to be 96500c/mol and relative atomic mass of copper 63,calculate the mass of copper liberated by 2A current in 5min.ans 0.196gm
Answer: The mass of copper liberated is 0.196 g.
Explanation:
The oxidation half-reaction of copper follows:
[tex]Cu\rightarrow Cu^{2+}+2e^-[/tex]
Calculating the theoretical mass deposited by using Faraday's law, which is:
[tex]m=\frac{M\times I\times t(s)}{n\times F}[/tex] ......(1)
where,
m = actual mass deposited = ? g
M = molar mass of metal = 63 g/mol
I = average current = 2 A
t = time period in seconds = 5 min = 300 s (Conversion factor: 1 min = 60 sec)
n = number of electrons exchanged = 2
F = Faraday's constant = 96500 C/mol
Putting values in equation 1, we get:
[tex]m=\frac{63 g/mol\times 2A\times 300s}{2\times 96500 C/mol}\\\\m=0.196g[/tex]
Hence, the mass of copper liberated is 0.196 g.
Question 7 of 25
Which two chemical equations are balanced?
O A. 2PBrg + 3Cl2 - 2PC13 + 3Br2
O B. 2Na+ MgCl2
2NaCl + Mg
-
C. 2LIOH + 2H2S - Li2S + 2H20
D. 2AgNO3 + NaCl - NaNO3 + 2AGCI
Answer:
2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂
2Na + MgCl₂ → 2NaCl + Mg
Explanation:
A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation
Among the given chemical reactions, we have;
2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂
In the above reaction;
The number of phosphorus, P, on either side of the equation = 2
The number of bromine atoms, Br, on either side of the equation = 6
The number of chlorine atoms, Cl, on either side of the equation = 6
Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced
The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg
In the above reaction, there are two sodium atoms, Na, one magnesium atom and two chlorine atoms on both sides of the reaction, therefore, the reaction is balanced
Please help me with these I might need more than only 1 person to answer
Explanation:
a) copper
b) olive oil
Hope it helps✌✌
factor that affect gravitation
1) Two major factors, mass and distance, affect the strength of gravitational force on an object.
2) Most common factors that affect gravity are mass of the body, distance from centers, shape of bodies,etc. Gravity is a binding force, always acting to bind any material closure and closure inwards. If we think this way, then gravity is maximum at its centre and decreases slowly away from its centre.
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In a lunar experiment, a 950-g aluminum (920 J/(°Ckg)) sphere is dropped from the space probe while is 75 m above the Lunar ground. If the sphere’s temperature increased by 0.11°C when it hits the ground, what percentage of the initial mechanical energy was absorbed as thermal energy by the aluminum sphere?
Answer:
13.759 % of the initial mechanical energy is lost as thermal energy.
Explanation:
By the First Law of Thermodynamics we know that increase in internal energy of the object ([tex]U[/tex]), in joules, is equal to the lost amount of the change in gravitational potential energy ([tex]U_{g}[/tex]), in joules:
[tex]\frac{x}{100} \cdot \Delta U_{g} = \Delta U[/tex] (1)
Where [tex]x[/tex] is the percentage of the energy loss, no unit.
By definition of the gravitational potential energy and internal energy, we expand this equation:
[tex]\frac{x\cdot m \cdot g \cdot h}{100} = m\cdot c\cdot \Delta T[/tex] (1b)
Where:
[tex]m[/tex] - Mass of the object, in kilograms.
[tex]g[/tex] - Gravitational acceleration, in meters per square second.
[tex]h[/tex] - Initial height of the object above the lunar ground, in meters.
[tex]c[/tex] - Specific heat of aluminium, in joules per degree Celsius-kilogram.
[tex]\Delta T[/tex] - Temperature increase due to collision, in degree Celsius.
If we know that [tex]m = 0.95\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h = 75\,m[/tex], [tex]c = 920\,\frac{J}{kg\cdot ^{\circ}C}[/tex] and [tex]\Delta T = 0.11\,^{\circ}C[/tex], then the percentage of energy loss due to collision is:
[tex]x = \frac{100\cdot c\cdot \Delta T}{g\cdot h}[/tex]
[tex]x = \frac{100\cdot \left(920\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (0.11\,^{\circ}C)}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (75\,m)}[/tex]
[tex]x = 13.759\,\%[/tex]
13.759 % of the initial mechanical energy is lost as thermal energy.
Difference between gravitational force and frictional
Explanation:
gravitational force or gravity is the force of attraction between objects that have mass.
frictional force is the forec that opposes motion (movement) when two surface are in contact. (acts in the opposite direction of motion)
so to sum it up,
- gravtional force is the force of attraction between objects that have mass.
- objects on the earth have weight because of gravitational force between them and the earth.
frictional force
- oppose motion
- slow down and stop moving objects and;
- produce heat
hope it helps :)
What would be the acceleration in a body moving with uniform velocity and why?
Explanation:
The derivative of a constant term is always 0. So the acceleration of the body would be zero. Hence, the acceleration of a body moving with uniform velocity will always be zero.
hope it helps you
How do we aquire knowledge in science?
Answer:
By studying.
By paying attention to what the teacher is saying.
That's the way to receive knowledge.
HAVE A GREAT DAY
Explanation:
Please give me that little crown...
AND REMEMBER TO SMILE
A ball is dropped from the roof of a 25-m-tall building. What is the velocity of the object when it touches the ground? Suppose the ball is a perfect golf ball and it bounces such that the ve locity as it leaves the ground has the same magnitude but the op posite direction as the velocity with which it reached the ground How high will the ball bounce? Now suppose instead that the ball bounces back to a height of 20 m. What was the velocity with which it left the ground?
Answer:
a) [tex]h=25m[/tex]
b) [tex]v=19.8m/sec[/tex]
Explanation:
From the question we are told that:
Height [tex]h=25m[/tex]
Bounce Height [tex]h'=20m[/tex]
Generally the Kinematic equation is mathematically given by
[tex]V=\sqrt{2gh}\\\\V=\sqrt{2*9.81*25}[/tex]
[tex]V=22.1m/sec[/tex]
Therefore Height
[tex]h=\frac{V^2}{2g}\\\\h=\frac{22.1^2}{2*9.81}[/tex]
[tex]h=25m[/tex]
b)
Generally the Kinematic equation is mathematically given by
[tex]v^2=2ah[/tex]
[tex]v^2=2*9.8*20[/tex]
[tex]v=\sqrt{2*9.8*20}[/tex]
[tex]v=19.8m/sec[/tex]
define Archemedics principle?
THIS IS YOUR ANSWER :
Archimedes’ principle, physical law of buoyancy, discovered by the ancient Greek mathematician and inventor Archimedes, stating that any body completely or partially submerged in a fluid (gas or liquid) at rest is acted upon by an upward, or buoyant, force, the magnitude of which is equal to the weight of the fluid displaced by the body.
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Phương trình chuyển động thẳng đều của một chất điểm có dạng: x = 2t – 10. (x: km, t: h). Quãng đường đi được của chất điểm sau 2h là bao nhiêu?
Answer:
Distance cover in 2 hour = 6 kilometer
Explanation:
Given equation:
x = 2t - 10
where
x = kilometer
t = hour
Find:
Distance cover in 2 hour
Computation:
T = 2
So,
x = 2t - 10
x = 2(2) - 10
x = 4 - 10
x = -6
Distance cover in 2 hour = 6 kilometer
Answer:
The distance is 6 km.
Explanation:
The equation of uniform linear motion of a particle has the form: x = 2t – 10. (x: km, t: h). What is the distance traveled by the particle after 2 hours?
x = 2t - 10
distance traveled after t = 2 hours
Substitute t = 2 in the given expression
x = 2 x 2 - 10
x = 4 - 10
x = - 6 km
So, the distance is 6 km.
Define derived unit with example
Answer:
A derived unit is a SI unit of measurement comprised of a combination of the seven base units. Like SI unit of force is the derived unit, newton or N where N=s21×m×kg. There are some commonly used derived units which includes: 1. Pressure = AreaForce=m2N.
Explanation:
D 2. Which of the following is an example of a specific goal?
O I want to lose weight.
I want to get stronger.
O I want to be able to walk one mile in 15 minutes.
I want to improve my speed.
I want to improve my speed
Answer:
D
Explanation:
the scientist who gives the definition of work
Answer:
Work, in physics, measure of energy transfer that occurs when an object is moved over a distance by an external force at least part of which is applied in the direction of the displacement. ... To express this concept mathematically, the work W is equal to the force f times the distance d, or W = fd.
A large atomic bomb produces around 8.00x1015J of energy. How much matter does that bomb turn into energy?
A)2.67x10^79kg
B)2.67x10^7kg
C)7.20x10^32kg
D)0.0889kg
Answer:
D)0.0889kg
Explanation:
If a large atomic bomb produces around 8.00x10¹⁵J of energy, then from Einstein's energy law i.e E=mc² we can calculate that the amount of matter is 0.0889kg. The correct option is D.
What is Einstein's energy law?
The Einstein Energy Law, often referred to as Einstein's famous equation relates the energy of a particle (E) to its mass (m) and the speed of light (c). The equation is as follows:
E = mc²
This equation shows that energy (E) and mass (m) are equivalent and interchangeable and that a small amount of mass can be converted into a large amount of energy. This is the basis for understanding nuclear reactions and the energy released in processes like nuclear fusion and fission.
Here in the question,
To calculate the mass that is converted into energy in an 8.00x10¹⁵J nuclear reaction, we can rearrange the equation to solve for m:
m = E / c²
where c = 3 x 10⁸ m/s is the speed of light in a vacuum.
Substituting the values, we get:
m = 8.00x10¹⁵ J / (3 x 10⁸ m/s)²
m = 0.0889kg
Therefore, a large atomic bomb that produces around 8.00x10¹⁵ J of energy converts approximately 0.0889kg of matter into energy. This is a very small amount of matter, but because c² is such a large number, a huge amount of energy can be released from a small amount of matter in a nuclear reaction.
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(9x10^2) + (1x10^2)
10x10^4
10x10^3
1x10^4
1x10^3
Answer:
10 x 10⁴
I hope it's helps you
Answer:
Yes
Explanation:
A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult. Assuming the acceleration is constant, how far did it travel during those 2.0 s?
First find Acceleration
Initial velocity=u=0m/sFinal velocity=v=42m/sTime=t=2sDistance=sAcceleration=a[tex]\boxed{\sf Acceleration=\dfrac{v-u}{t}}[/tex]
[tex]\\ \sf\longmapsto Acceleration=\dfrac{42-0}{2}[/tex]
[tex]\\ \sf\longmapsto Acceleration=\dfrac{42}{2}[/tex]
[tex]\\ \sf\longmapsto Acceleration=21m/s^2[/tex]
Using second equation of kinematics
[tex]\boxed{\sf s=ut+\dfrac{1}{2}at^2}[/tex]
[tex]\\ \sf\longmapsto s=0(2)+\dfrac{1}{2}(21)(2)^2[/tex]
[tex]\\ \sf\longmapsto s=21(2)[/tex]
[tex]\\ \sf\longmapsto s=42m[/tex]
Tick (3) the correct statement about electrostatic charges.
(a) Earthing causes positive charges to flow from the object to the ground.
(b) Similar types of electric charges attract one another.
(c) An electroscope is used to determine the presence of electrostatic charges.
Answer:
similar type of electric charges attract one another
I think this is a coorect staement
Answer:
C ) An electroscope is used to determine the presence of electrostatic charges
make a list of principle of lever.
Answer:
The force applied to make the object move
Answer:
input work = output work
E * ED = L* LD
Explanation:
The principle of lever is that input work is always equal to output work .
1. Una carga Q1 = + 12 μC se coloca a una distancia r = 0.024 m desde una carga Q2 = + 16 μC. a) Determina la magnitud de la fuerza electrostática que actúa sobre las dos cargas, Q1 y Q2. b) ¿Es la fuerza la atracción o repulsión? 2. Determina la intensidad del campo eléctrico a una distancia radial de r = 48 mm desde una carga de Q = 24 μC. 3. Una carga Q1 = 24 mC se coloca a una distancia r = 0.032 m desde una carga Q2 = - 12 μC. a. Determina la cantidad de energía potencial eléctrica que tiene la carga Q1. b. Determina el potencial eléctrico en la posición de Q2.
Answer:
1. a. 3,000 N
b. Repulsión
2. 46.875 × 10⁶ N/C
3. a. 81,000 J
b. 6.75 × 10⁹ V
Explanation:
1. Los parámetros dados son;
Q₁ = +12 μC, Q₂ = +16 μC
La distancia entre las cargas, r = 0.024
La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;
[tex]F = k \times \dfrac{Q_1 \cdot Q_2}{r^2}[/tex]
Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²
Por lo tanto, obtenemos;
F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000
La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N
(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.
2) La intensidad de un campo eléctrico, E, se da como sigue;
[tex]E = \dfrac{k \cdot Q}{r^2}[/tex]
La magnitud de la carga, Q = 24 μC
La distancia donde se mide el campo, r = 48 mm = 0.048 m
Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C
La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C
3. La magnitud de las cargas son;
Q₁ = 24 mC
Q₂ = -12 μC
La distancia entre las cargas, r = 0.032 m
un. El potencial eléctrico de una carga, [tex]U_E[/tex] , se da de la siguiente manera;
[tex]U_E = k \times \dfrac{Q_1 \cdot Q_2}{r}[/tex]
Por lo tanto;
[tex]U_E[/tex] = 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000
La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J
b. El potencial eléctrico de Q₁ en Q₂, V₁ = [tex]k \times \dfrac{Q_1 }{r}[/tex]
Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹
El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V