Answer:
weight
Explanation:
Weight is the force of gravity with which a body is attracted towards the earth.
Calculate the man’s mass. (Use PE = m × g × h, where g = 9.8 N/kg.)
A man climbs a wall that has a height of 8.4 meters and gains a potential energy of 4,620 joules. His mass is about
kilograms
Answer:
mass=56.12kg
Explanation:
PE=mgh
4620=m×9.8×8.4
make m subject of the formula...
m =4620/(9.8×8.4)
m=4620/82.32
m=56.12kg
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction between the blocks is the same as the coefficient of static friction between the lower block and the horizontal surface. A horizontal force is applied to the upper block, and its magnitude is slowly increased. When the force reaches 52.1 N, the upper block just begins to slide. The force is then removed from the upper block, and the blocks are returned to their original configuration. What is the magnitude of the horizontal force that should be applied to the lower block, so that it just begins to slide out from under the upper block
Answer:
F = 156.3 N
Explanation:
Let's start with the top block, apply Newton's second law
F - fr = 0
F = fr
fr = 52.1 N
Now we can work with the bottom block
In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal
we apply Newton's second law
Y axis
N - W₁ -W₂ = 0
N = W₁ + W₂
as the two blocks are identical
N = 2W
X axis
F - fr₁ - fr₂ = 0
F = fr₁ + fr₂
indicates that the lower block is moving below block 1, therefore the upper friction force is
fr₁ = 52.1 N
fr₁ = μ N
a
s the normal in the lower block of twice the friction force is
fr₂ = μ 2N
fr₂ = 2 μ N
fr₂ = 2 fr₁
we substitute
F = fr₁ + 2 fr₁
F = 3 fr₁
F = 3 52.1
F = 156.3 N
Which statement correctly describes the Earth's magnetic field?
A. The geomagnetic south is consistent with a magnetic south.
B. The geomagnetic north is similar to the south pole of a magnet.
C. The Earth's magnetic poles are the same magnetic charge at the poles.
D. The Earth does not have a consistent pattern of magnetism.
Answer: B. The geomagnetic north is similar to the south pole of a magnet
Explanation:
A gas occupies 2 m^3 at 27°C at a pressure of 1 atmosphere. At a pressure of 2 atmospheres it occupies a volume of 1 m^2. What is its temperature at this new volume and pressure?
Answer:
27°C
Explanation:
We'll begin by converting 27 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Initial temperature (T₁) = 27 °C
Initial temperature (T₁) = 27 °C + 273
Initial temperature (T₁) = 300 K
Next, we shall determine the final temperature of the gas. This can be obtained as follow:
Initial volume (V₁) = 2 m³
Initial temperature (T₁) = 300 K
Initial pressure (P₁) = 1 atm
Final pressure (P₂) = 2 atm
Final volume (V₂) = 1 m³
Final temperature (T₂) =?
P₁V₁/T₁ = P₂V₂/T₂
1 × 2 / 300 = 2 × 1 / T₂
2/300 = 2/T₂
1/150 = 2/T₂
Cross multiply
T₂ = 150 × 2
T₂ = 300 K
Finally, we shall convert 300 K to celsius temperature. This can be obtained as follow:
T(°C) = T(K) – 273
T(K) = 300 K
T(°C) = 300 – 273
T(°C) = 27°C
Thus, the final temperature is 27°C
A 4.88 x 10-6 C charge moves 265 m/s
parallel (at 0°) to a magnetic field of
0.0579 T. What is the magnetic force
on the charge?
Answer:
[tex]F=0N[/tex]
Explanation:
From the question we are told that:
Charge [tex]Q=4.88 x 10-6 C[/tex]
Velocity [tex]v= 265m/s[/tex]
Angle [tex]\theta =0 \textdegree[/tex]
Magnetic field [tex]B=0.0579T[/tex]
Generally the equation for Force is mathematically given by
[tex]F=Q(\=v*\=B)[/tex]
[tex]F=qvBsin\theta[/tex]
Therefore
[tex]F=qvBsin0 \textdegree[/tex]
[tex]F=0N[/tex]
Answer:
0 newtons
Explanation:
When a moving charge is parallel to the magnetic field, it feels no Magnetic force at all.
Even if the Magnetic Field is 100,000,000 Tesla!
Una masa de Hidrogeno ocupa 0.2 L a 100°C. Determine su volumen a 0°C, si la presión se mantiene constante. Como la presión y la cantidad de materia se mantienen constantes, podemos aplicar la ley de Charles
Answer:
0,146 L
Explanation:
Según la ley de Charles; el volumen de una determinada masa de gas es directamente proporcional a su temperatura a presión constante.
A partir de los datos proporcionados;
V1 = 0,2 L
T1 = 100 ° C + 273 = 373 K
V2 =
T2 = 0 ° C + 273 = 273 K
V1 / T1 = V2 / T2
V1T2 = V2T1
V2 = V1T2 / T1
V2 = 0,2 × 273/373
V2 = 0,146 L
A car starts from rest. if rhe final velocity becomes 50m /s after 10 second calculate the acceleration of the car.
Answer:5m/s² well I am not pretty sure but hope it's help
Explanation:
u=0m
final velocity ,v=50m/s
t=10s
(v-u)/t=(50-0)/10=50/10=5m/s²
Which of the following could be used to create an open circuit?
A student drops a ball off the top of building and records that the ball takes 3.32s to reach the ground (g = 9.8 m/s^2). What is the ball speed just before hitting the ground?
Answer:
Explanation:
Here's what we know because it was given to us:
a = -9.8 m/s/s and
time = 3.32 seconds
Here's what we know because we rock physics:
v₀ = 0 (because the object was held still before it was dropped).
Here's the equation that ties all that info together in a single one-dimensional equation:
v = v₀ + at
Filling in and solving for v:
v = 0 + (-9.8)(3.32) and
v = -33m/s
The velocity is negative because the object is moving downwards and up is positive (but you knew that already too!)
1- Un avión vuela horizontalmente con una velocidad cuya magnitud es de 100 m/s y deja caer un proyectil desde una altura de 400 m con respecto al suelo. Determinar: a) El tiempo que transcurre antes de que el proyectil se impacte en el suelo. b) ¿Qué distancia horizontal recorre el proyectil después de iniciar su caída?
2- Se lanza horizontalmente un objeto con una velocidad de 80 m/s desde una altura de 160 m. Calcular: a) El tiempo que tarda en llegar al suelo b) La magnitud de la velocidad vertical que lleva a los 3 segundos. c) La distancia horizontal a la que cae el objeto a partir del punto desde que fue arrojada.
3- Se lanza un proyectil con una velocidad inicial de 80 m/s y una inclinación, sobre la horizontal, de 30°. Calcular: a) ¿Cuál es la altura máxima que alcanza la bala? b) ¿Cuáles son las posiciones, x y y, cuando han transcurrido 5 segundos?
4- Un joven patea una pelota contra un arco con una velocidad inicial de 20 m/s y con un ángulo de 45°. Determinar: a) cual es el alcance total
if the action force is 100N what will be the reaction force
Answer:
HONORS PHYSICS
Introduction
Matter & Energy
Math Review
Kinematics
Defining Motion
Graphing Motion
Kinematic Equations
Free Fall
Projectile Motion
Relative Velocity
Dynamics
Newton's 1st Law
Free Body Diagrams
Newton's 2nd Law
Static Equilibrium
Newton's 3rd Law
Friction
Ramps and Inclines
Atwood Machines
Momentum
Impulse & Momentum
Conservation Laws
Types of Collisions
Center of Mass
UCM & Gravity
Uniform Circular Motion
Gravity
Kepler's Laws
Rotational Motion
Rotational Kinematics
Torque
Angular Momentum
Rotational KE
Work, Energy & Power
Work
Hooke's Law
Power
Energy
Conservation of Energy
Fluid Mechanics
Density
Pressure
Buoyancy
Pascal's Principle
Fluid Continuity
Bernoulli's Principle
Thermal Physics
Temperature
Thermal Expansion
Heat
Phase Changes
Ideal Gas Law
Thermodynamics
Electrostatics
Electric Charges
Coulomb's Law
Electric Fields
Potential Difference
Capacitors
Current Electricity
Electric Current
Resistance
Ohm's Law
Circuits
Electric Meters
Circuit Analysis
Magnetism
Magnetic Fields
The Compass
Electromagnetism
Microelectronics
Silicon
P-N Junctions
Transistors
Digital Logic
Processing
Integration
Waves & Sound
Wave Characteristics
Wave Equation
Sound
Interference
Doppler Effect
Optics
Reflection
Refraction
Diffraction
EM Spectrum
Modern Physics
Wave-Particle Duality
Models of the Atom
M-E Equivalence
The Standard Model
Relativity
DYNAMICS
Newton's 1st Law
Free Body Diagrams
Newton's 2nd Law
Static Equilibrium
Newton's 3rd Law
Friction
Ramps and Inclines
Atwood Machines
Silly Beagle
Newtons's 3rd Law of Motion
Newton’s 3rd Law of Motion, commonly referred to as the Law of Action and Reaction, describes the phenomena by which all forces come in pairs. If Object 1 exerts a force on Object 2, then Object 2 must exert a force back on Object 1. Moreover, the force of Object 1 on Object 2 is equal in magnitude, or size, but opposite in direction to the force of Object 2 on Object 1. Written mathematically:
Newton's 3rd Law Equation
This has many implications, some of which aren’t immediately obvious. For example, if you punch the wall with your fist with a force of 100N, the wall imparts a force back on your fist of 100N (which is why it hurts!). Or try this. Push on the corner of your desk with your palm for a few seconds. Now look at your palm... see the indentation? That’s because the corner of the desk pushed back on your palm.
running tiger
Although this law surrounds your actions everyday, often times you may not even realize its effects. To run forward, a cat pushes with its legs backward on the ground, and the ground pushes the cat forward. How do you swim? If you want to swim forwards, which way do you push on the water? Backwards, that’s right. As you push backwards on the water, the reactionary force, the water pushing you, propels you forward. How do you jump up in the air? You push down on the ground, and it’s the reactionary force of the ground pushing on you that accelerates you skyward!
As you can see, then, forces always come in pairs. These pairs are known as action-reaction pairs. What are the action-reaction force pairs for a girl kicking a soccer ball? The girl’s foot applies a force on the ball, and the ball applies an equal and opposite force on the girl’s foot.
How does a rocket ship maneuver in space? The rocket propels hot expanding gas particles outward, so the gas particles in return push the rocket forward. Newton’s 3rd Law even applies to gravity. The Earth exerts a gravitational force on you (downward). You, therefore, must apply a gravitational force upward on the Earth!
How to calculate displacement, velocity, acceleration.
The formulas used to analyze the horizontal and vertical motion of projectiles launched at an angle invlove and use of which functions
Answer:
Explanation:
The formula to analyze motion in the vertical (or y) dimension is
[tex]A_y=Asin\theta[/tex] which says that the motion in the y-dimension is equal to the magnitude of the A vector times the sin of the angle.
The formula to analyze motion in the horizontal (or x) dimension is
[tex]A_x=Acos\theta[/tex] which says that the motion in the x-dimension is equal t the mignitude of the A vector times the cosine of the angle.
Many times this is used to find the upwards velocity and the horizontal velocity when an overall velocity is given with an angle of inclination.
Explain the energy transformations that occur when accelerating in a gasoline
vehicle.
The air also contains oxygen nitrogen and carbon dioxide what is the state of each of substance at -190°?
Answer:
oxygen - gas
nitrogen - liquid
carbon dioxide- gas
oxygen would be gas
nitrogen would be liquid
carbon dioxide would be gas
Interpret the following graphs by answering the questions below.
a. The figure below shows the Standing waves
How many nodes and antinodes?
Answer:
Nodes = 5 and antinodes = 4
Explanation:
Nodes are the points of zero amplitude and appear to be fixed. On the other hand, antinodes are points on a stationary wave that oscillates with maximum amplitude.
In this given standing wave, there are 5 points where the amplitude is 0. So, there are 5 nodes. Also, there are 4 points where the amplitude is maximum.
So, there are 5 nodes and 4 antinodes.
Convert:
1) 2kg into gram
2) 5200m into km
3) 20cm into m
HELPPP I NEED HELP ASAP NOW
Answer:
Your answers would be
4. A. sperm and testosterone.
7. C. prostate, penis, Testes
uterus, vagina, fallopian tubes
10. B. Protein
11. A. carbohydrate
12. B. amino acids (I'm not positive on this i haven't taken bio in years
27. D. Respiratory system
Explanation:
yeah
Two cars move at different velocities. Car A moves at a speed of 90 km/hr while Car B moves
at a speed of 75 km/hr . If both cars collided with a wall made up of the same material, which of
the two cars will create greater impact of collision. Both cars have the same masses.
(Serious Answers Please)
This type of collision is known as an Inelastic collision.
The speed rate at which car a used is more higher than that of car b because of its friction.
I hope this helps.
(Please help if you can, I need this last answer done by tonight.)
Use the universal law of gravitation to solve the following problem.
Hint: mass of the Earth is = 5.97 x 1024 kg
A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?
a. Write out the formula for this problem.
b. Plug in the values from this problem into the formula.
c. Solve the problem, writing out each step.
d. Correct answer
Answer:
a.
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
b.
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]
c.
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]
d. 1.144 × 10¹³ N
Explanation:
The universal law of gravitation is presented as follows;
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
The given mass of the scientific satellite, m = 1,300 kg
The height of the orbit of the satellite, r = 200 km above the Earth's surface
The length of the radius of the Earth, R = 6378 km
The mass of the Earth = 5.97 × 10²⁴ kg
a. The formula for the universal law of gravitation is presented as follows;
[tex]F =G \cdot \dfrac{M \cdot m}{r^{2}}[/tex]
Where;
M = The mass of the Earth = 5.97 × 10²⁴ kg
m = The mass of the satellite = 1,300 kg
r = The distance between the Earth and the satellite = R + r = 6,378 km + 200 km = 6,578 km
G = The Gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²
b. Plugging in the values from the problem into the formula gives;
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}}[/tex]
c. Solving gives;
[tex]F =6,378 \times 10^{-11} \times \dfrac{5.97 \times 10^{24} \times 1,300}{6,578^{2}} = \dfrac{9.519165 \times 10^{18}}{832117} \approx 1.144 \times 10^{13}[/tex]
The force acting between the Earth and the satellite, F ≈ 1.144 × 10¹³ N
d. 1.144 × 10¹³ N
11. The diagram to the right shows a harmonic of a
standing wave on a vibrating string.
Open the image for the rest :)
Answer:
(a) Fourth harmonic
(b) 2 wavelengths
(c) wavelength = 1 m
Explanation:
11.
(a) It is fourth harmonic.
(b) The distance between the two consecutive nodes is half of wavelength, so there are two wavelength.
(c) The distance between the two consecutive node or the antinode is half of wavelength.
So, wavelength = 2/2 = 1 m
A converging lens can be defined as __________
a lens that causes parallel light rays to bounce off the surface
a lens that allows parallel light rays to pass without changing direction
a lens that causes parallel light rays to separate from each other
a lens that causes parallel light rays to focus at a specific location
Answer:
a lens that causes parallel light rays to separate from each other
Answer:
a lens that causes parallel light rays to focus at a specific location
I took the test and got it right! :)
I need to know how to do it
Answer:
1.76km
Explanation:
Here,
1mi = 1.6km
1.10 miles = 1.6 x 1.10 km
= 1.76 km
sulfur and oxygen can react to form both sulfur dioxide and sulfur trioxide in sulfur dioxide there are 32.06 grams of sulfur and 32 grams of oxygen in sulfur dioxide there are 32.06 grams of sulfur are combined with 48 grams of oxygen
a. what is the ratio of the weights of oxygen that combine with 32.06 g of sulfur ?
b. How do these data illustrate the law of multiple proportions?
Answer:
a. 2:3
b. The data illustrates the law of multiple proportions by showing that the the masses of oxygen that reacts with a fixed mass of sulfur are in a ratio of small whole numbers
Explanation:
The weight of oxygen that combines with 32.06 grams of sulfur in sulfur dioxide = 32 grams
The weight of oxygen that combines with 32.06 grams of sulfur in sulfur trioxide = 48 grams
a. The ration of the weights of oxygen that combine with 32.06 g of sulfur = 32:48 = 2:3
b. The law of multiple proportions states that when two elements are able to interact chemically to form more than one compound, then the (different) weights of one of the element that combines with a fixed weight of the other element are in small whole number ratios
The data demonstrates the law of multiple proportions by showing that the ratios of the weights of oxygen that combine with a fixed weight of sulfur to form sulfur dioxide and sulfur trioxide is in the ratio of 2 to 3 which are small whole number ratios
A 52 kg child on a swing is travelling at 6 m/s . What is his gravitational potential energy if he has 1000 J of the mechanical energy?
Answer:
The correct answer is "64 J".
Explanation:
The given values are:
Mass,
m = 52 kg
Velocity,
v = 6 m/s
Mechanical energy,
= 1000 J
Now,
The gravitational potential energy will be:
⇒ [tex]P.E=1000-\frac{1}{2}mv^2[/tex]
[tex]=1000-\frac{1}{2}\times 52\times (6)^2[/tex]
[tex]=1000-26\times 36[/tex]
[tex]=1000-936[/tex]
[tex]=64 \ J[/tex]
When magma flows on the surface on the surface, it is already called lava
TRUE OR FALSE
Answer:
True
Explanation:
I guess you made a mistake on question.
but I understood what you wanted to say.
Hope this helps... :)
help....................
The surface tension of isopropyl alcohol in air has a value of 22.39 units and the surface tension of water in air is 72.86 units.
Answer:
The answer is "Option B and Option D"
Explanation:
Isopropyl alcohol has an air pressure of 22,39 units with such a water contact angle of 72,86 units.
The melting points of Isopropyl alcohol are higher than water because of the weaker forces between the previously studied.
The flatter meniscus is indeed the isopropyl ethanol so because forces between the particles are smaller.
QUESTION 30 A tennis ball moves back and forth 10 times in 5 sec. The frequency of its motion is
Answer:
so in 1 sec 2 times
so frequency = 2
Explanation:
The submarine emits a pulse of sound to detect other objects in the sea. The sped of sound in sea water is 1500m/s. An echo is received with a time delay of 0.50s after the original sound is emitted.
Calculate the distance between submarine and the other object
Answer:
d = 375 m
Explanation:
The speed of sound is constant in any medium, therefore we can use the uniform motion relationships
v = x / t
x = v t
In this case it indicates that the time since the sound is emitted and received is t = 0.50 s, in this time the sound traveled a round trip distance
x = 2d
2d = v t
d = v t/2
let's calculate
d = 1500 0.5 / 2
d = 375 m