Answer:
C, B, and D are ones i think i know that can occur during a physical change
Explanation:
Ex 2) A cannon ball is shot straight up into the air with an initial velocity of 25 m/s[Up).
What is the maximum height of the cannonball?
Explanation:
The final velocity at the cannon ball's maximum height is zero ([tex]v_y = 0[/tex]). We can use the equation
[tex]v_y^2 = 0 = v_{0y}^2 - 2gy_{max}[/tex]
[tex]\Rightarrow y_{max} = \dfrac{v_{0y}^2}{2g}[/tex]
[tex]y_{max} = \dfrac{(25\:\text{m/s})^2}{2(9.8\:\text{m/s}^2)}[/tex]
[tex]\:\:\:\:\:\:\:=31.9\:\text{m}[/tex]
(a) Define the term ginning? (b) Name the simple device used for spinning?
A simple device used for spinning is a hand spindle also known as takli.
hope it helps
What was the initial speed of a car if its speed is 40 m/s after 5 seconds of
accelerating at 4 m/s27
A. 10 m/s
B. 60 m/s
C. 20 m/s
D. 25 m/s
Reset Selection
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Answer:
[tex]\boxed {\boxed {\sf C. \ 20 \ m/s }}[/tex]
Explanation:
We are asked to find the initial speed of a car.
We are given the final speed, the time, and the acceleration, so we will use the following kinematic equation:
[tex]v_f=v_i+at[/tex]
We know the final speed is 40 meters per second, the acceleration is 4 meters per second squared, and the time is 5 seconds.
[tex]v_f[/tex]= 40 m/s t= 5 s a= 4 m/s²Substitute the values into the formula.
[tex]40 \ m/s = v_i+(4 \ m/s^2 * 5 \ s)[/tex]
Multiply inside the parentheses.
[tex]40 \ m/s =v_i+ 20 \ m/s[/tex]
We are solving for the initial speed, so we must isolate the variable [tex]v_i[/tex].
20 meters per second is being added to [tex]v_i[/tex]. The inverse operation of addition is subtraction. Subtract 20 m/s from both sides of the equation.
[tex]40 \ m/s - 20 \ m/s = v_i + 20 \ m/s - 20 \ m/s[/tex]
[tex]40 \ m/s - 20 m/ s = v_i[/tex]
[tex]20 \ m/s=v_i[/tex]
The initial speed of the car is 20 meters per second and choice C is correct.
Describe what causes the planets to stay in orbit around the Sun
Answer:
Gravitional pull or gravity
Explanation:
Its that simple
A 3000 kg car stops at a red light, and is rear-ended by a 5000 kg truck traveling at 20m/s. In the collision, the two cars stick together. What is the final speed of the two cars just after the collision in m/s (Numeric Answers only)
Explanation:
this is actually not as simple as it sounds here.
quite some energy is lost in the deformation of the bodies of car and truck, and it also needs more energy to get a standing object going than to accelerate an already moving object.
but assuming the simple described circumstances, then the energy and impulse of the moving truck of 5000 kg is transferred to a new combined system of car and truck of now 5000 + 3000 = 8000 kg.
so, the 20m/s inertia energy of the truck is now distributed to the truck/car combination.
since the same energy has to move now more mass, it is clear that the combined speed will be lower.
20×5000 = x×8000
20×5 = x×8
x = 100/8 = 12.5 m/s
that is the resulting speed of the combined truck/car object.
Which of the following formulas describes the change in momentum of an
object?
A. change in momentum = force x time over which force is applied
B. change in momentum = acceleration distance over which
acceleration is applied
C. change in momentum = force x distance over which force is
applied
O D. change in momentum = acceleration time over which
acceleration is applied
*Sorry for the bad quality picture!*
A frictionless pendulum with a mass of 0.4 kg and a length of 2.1 m starts at point A, at an angle 0 of 60°. As it swings downward, it passes through point B, which is 30 degrees from equilibrium. What is the kinetic energy of the pendulum at point B?
A) 3.9 J
B) 3.0 J
C) 1.1 J
D) 4.1 J
The conservation of mechanical energy allows finding the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
The conservation of mechanical energy is a theorem of greater importance in physics and ordinary life, it states that if there is no friction force the total mechanistic energy remains constant at all points.
Mechanical energy is the sum of kinetic energy plus all potential energies. In the attachment we see a diagram of the pendulum's movement at the two points of interest.
They indicate that the pendulum is released from an initial angle of θ₁ = 60º, let's find the mechanical energy at that point.
Em₀ = U = m g h
Where the height is measured from the lowest point of the movement.
h = L - L cos tea1 = L (1 cos tea1)
The second point of interest occurs for θ₂ = 30º.
At this point part of the energy is indica and part gravitational potential.
[tex]Em_f[/tex] = K + U₂
[tex]Em_f[/tex] = ½ m v² + m g h ’
There is no friction in the system, therefore mechanical energy is conserved.
Em₀ = Em₀_f
mg L (1 - cos θ₁) = ½ m v² + m g L (1 - cos θ₂)
v² = 2g L (cos θ₂ - cos θ₁)
Let's calculate.
v² = 2 9.8 2.1 (cos 30 - cos 60)
v² = 41.16 0.366
v = 3.88 m / s
In conclusion using the conservation of mechanical energy we can find the result for the speed of the pendulum when it is at 30º is:
The speed is: 3.88 m / s
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Answer:
3.0 J
Explanation:
Just took the test
HELP! I need some assistance
Answer:
I'm a Filipino hahahah I'm not
What will be the current through a resistance of 50Ω if the applied voltage across the resistance is 117V?
Formulas: V=IR, I=V/R, R=V/I
Note: No need to write the unit of your answer.
2.34 A
Explanation:
[tex]V = IR \Rightarrow I = \dfrac{V}{R} = \dfrac{117\:\text{V}}{50\:Ω} = 2.34\:\text{A}[/tex]
A car with a mass of 1500 kg is pulled by a rope that is horizontal to the ground. The tension in the rope is 2000 N and a friction force of 350 N opposes the car's motion. What is the magnitude of the car's acceleration?
Answer:
Explanation:
Assuming the ground is level as well.
F = ma
a = F/m
a = (2000 - 350) / 1500
a = 1.1 m/s²
The acceleration of the car is 1.1 m/s².
To calculate the magnitude of the acceleration of the car, we use the formula below.
F-F' = ma................ Equation 1Where:
F = Tension in the ropeF' = Friction forcem = mass of the cara = acceleration of the carMake a the subject of the equation
a = (F-F')/m................. Equation 2From the equation,
Given:
F = 2000 NF' = 350 Nm = 1500 kgSubstitute these values into equation 2
a = (2000-350)/1500a = 1650/1500a = 1.1 m/s²Hence, the acceleration of the car is 1.1 m/s².
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A spring of spring constant k = 200 N m−1 is slowly extended from an extension of 3.0 cm to an extension of 5.0 cm. Calculate the work done by the extending force. 60
Answer:
31
Explanation:
No need
The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force on the rocket if it is now accelerating at 12mls²??
From your previous question
Mass=30kgAcceleration=12m/s^2[tex]\\ \sf\longmapsto F=ma[/tex]
[tex]\\ \sf\longmapsto F=30(12)[/tex]
[tex]\\ \sf\longmapsto F=360N[/tex]
Which term describes energy stored in the bonds between atoms?
A. Nuclear energy
B. Chemical energy
O C. Thermal energy
D. Sound energy
Answer:
B. Chemical energy
Explanation:
chemical energy is the energy stored in the bonds between atoms
Una masa de aire de 20g absorbe 780 cal. Teniendo en cuenta que su temperatura inicial es de 30°C. Calcular la
temperatura que alcanzo al absorber el calor. Ce:0,24 cal/g °C
Answer:
sorry I don't know the answer
A race car rounding a corner at a constant speed of 200 miles per hour.
Answer:
Yes since it is changing direction
Explanation:
A race car rounding a corner at a constant speed of 200 miles per hour
A closed, uninsulated system fitted with movable piston, so no matter is exchanged with the surroundings, was assembled. Introduction of 430 J of heat caused the system to expand, doing 238 J of work against a constant pressure of 101 kPa. What is the value of for this process
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
The gas state of water
Answer:
The gas state of water is water vapour. It is formed by boiling liquid water or from the sublimation of ice (sublimation is the process of converting a solid to a liquid). Hope this helps!
In 1962 measurements of the magnetic field of a large tornado were made at the Geophysical Observatory in Tulsa, Oklahoma. If the magnitude of the tornado's field was B = 17.50 nT pointing north when the tornado was 9.10 km east of the observatory, what current was carried up or down the funnel of the tornado? Model the vortex as a long, straight wire carrying a current. A (conventional current) flowing ---Direction--- the tornado.
Answer:
796.25 A
Explanation:
B= (μI)/(2πr)
I= (B*2πr)/(μ)
μ= 4π*10^-7
I= ((17.50*10^-9)(2π)(9.10*10^3)) / (4π*10^-7)
= 796.25 A
A (conventional current) flowing "DOWN" the tornado.
tính chỉ số ampe kế thay ampe kế bằng vô kế có điện trở rất lớn thì vô kế chỉ bao nhiêu
Answer:
I don't understand it.....
how can I become a good science student ?
Answer:
Study hard , focus on your studies and alyways ask questions .
Study, revise, write notes, listen in class, don't let yourself be distracted by others, and do the work in class...maybe join stem or science club if you wanna
Hope this helped you- have a good day bro cya)
I need help ASAP please.....
I NEED A 100% ACCURATE ANSWER FOR THIS QUESTION ASAP NO LINKS !!!
What describes the pressure in a contained fluid?
1: unknown as long as the fluid is contained
2: the same throughout the fluid
3: higher in some places and lower in other places
Answer:
3) higher in some places nd lower In other places
Explanation:
The pressure in a fluid is given by the law
p = p0 + pgh
where
p0 is the atmospheric pressure
p is the fluid's density
g is the acceleration of gravity
h is the depth at which the pressure is calculated
As we see, the pressure depends on the value of h (depth): therefore, points which are located at more depth experience a larger pressure than points located near the fluid's surface.
What is an involuntary and a voluntary muscle action?
Answer:
Involuntary muscle action are muscles that move WITHOUT conscious control, and voluntary muscle action are muscles that move WITH conscious control.
Explanation:
A example of a involuntary muscle action would be your heart. Your heart is a muscle beats without you thinking about it and is very vital to your body. A example of a voluntary muscle would be the muscles that are attached to your bones and allow you to move your body, and it's voluntary because your body won't move on it's own.
Hope this helps.
A student is asked to calculate the centripetal acceleration of a hummingbird that makes a complete circle in 0.43 seconds. If the circle has a radius of 0.25 m, what is
the correct setup for determining the centripetal acceleration of the hummingbird?
(3.65m/s)2/0.25m
(3.65m/s)2-0.43
O (0.435)2 / 25m
O (0.432.25m
The centripetal acceleration of the hummingbird is [tex]\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2[/tex]
The given parameters;
time of motion of the hummingbird, t = 0.43 sradius of the circle, r = 0.25 mnumber of revolution of the hummingbird = 1 rev per 0.43 sThe angular speed of the hummingbird is calculated as follows;
[tex]\omega = \frac{1 \ rev}{0.43 \ s} \times \frac{2\pi \ rad}{1 \ rev} =14.61 \ rad/s[/tex]
The linear speed of the hummingbird is calculated as follows;
v = ωr
v = 14.61 x 0.25
v = 3.65 m/s
The centripetal acceleration of the hummingbird is calculated as follows;
[tex]a_c = \frac{v^2}{r} \\\\a_c = \frac{(3.65)^2}{0.25} \\\\a_c = 53.29\ m/s^2[/tex]
Thus, the centripetal acceleration of the hummingbird is [tex]\frac{(3.65)^2}{0.25} = 53.29 \ m/s^2[/tex]
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Am I correct? Plz answer ASAP! I will give brainilest:D (Science)
Answer:
I would agree with your selection.
Explanation:
A massive launcher sends a projectile vertically upwards from the surface of a planet of mass M and radius R. You may assume that this planet has no atmosphere. Part A If the projectile is launched at the escape speed, how do the magnitudes of its initial KE and gravitational potential energy (GPE) compare
Answer:
It can be shown that the potential energy of an object at the surface of the planet would be -G M / R if the potential at infinity is chosen to be zero.
Kinetic energy of G M / R would be required for the escape speed of such an object. The total energy in all such cases is zero.
This can easily be seen by considering the speed of an object falling from infinity towards the planet - the total energy will remain zero if it was zero when the object started to fall.
The potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.
What is escape speed?Escape speed is the minimum speed required for a free, non-propelled object to escape from the gravitational pull of the main body and reach an infinite distance from it in celestial physics.
It is proven that if the potential at infinity is set to zero, the potential energy of an item on the planet's surface is [tex]\rm \frac{-GM}{R}[/tex].
The escape speed of such an item would necessitate kinetic energy will be [tex]\rm \frac{GM}{R}[/tex]. In all of these circumstances, the total energy is zero.
Consider the speed of an item falling from infinity towards the planet: if the total energy was zero before the thing began to descend, the total energy will stay zero.
Hence the potential at infinity is set to zero while the kinetic energy will be [tex]\rm \frac{GM}{R}[/tex] and total energy will be zero.
To learn more about the escape speed refer to the link;
https://brainly.com/question/14178880
Car A uses tires for which the coefficient of static friction is 0.384 on a particular unbanked curve. The maximum speed at which the car can negotiate this curve is 16.7 m/s. Car B uses tires for which the coefficient of static friction is 0.809 on the same curve. What is the maximum speed at which car B can negotiate the curve
A trouble-making youth is standing on a bridge, and wants to drop a water balloon on an unsuspecting passerby. A man is jogging on a path below the bridge with a constant speed of 4.2 m/s. The bridge is 11.6 m above the ground. If the balloon is to land right at the jogger's feet, at what horizontal distance x from the bridge should he be when the youth drops the balloon?
Answer:
Explanation:
Time needed for a balloon to drop from vertical rest a distance of 11.6 m
t = √(2h/g) = √(2(11.6)/9.8) = 1.538618
d = vt = 4.2(1.538618) = 6.462197...
d = 6.5 m
Horizontal distance x = 6.5 m from the bridge he should be when the youth drops the balloon with velocity 4.2 m/s.
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.
Uniform motion an object is said to have uniform motion when object cover equal distance in equal interval of time within exact fixed direction. For a body in uniform motion, the magnitude of its velocity remains constant over time.
Given in the question time needed for a balloon to drop from vertical rest a distance of 11.6 m
t = √(2h/g) = √(2(11.6)/9.8) = 1.538618
d = vt = 4.2(1.538618) = 6.462197...
d = 6.5 m
Horizontal distance x = 6.5 m from the bridge he should be when the youth drops the balloon with velocity 4.2 m/s.
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There are 6 foundation of sports and which one you think is the most important?
I just need the points
Explanation:
But just pick any and say something like "it stans out to me most/more" or "it sounds/looks more intristing to me"
hola buen día por favor alguien me puede ayudar por favor1. Dos cargas puntuales q1=+4μC y q2=+ 6μC están separadas por 10 cm. Una carga
puntual
q3=+2μC se coloca a medio camino entre q1 y q2. Encuentra la magnitud y la dirección de la
fuerza resultante sobre q3. Asuma que q1 está a la izquierda de q3 y q2 a la derecha de q3.
Revisa el ejemplo de la clase del tema 6. Además, recuerda que las fuerzas no siempre actúan
en la misma dirección.
2. Se coloca una carga puntual de 4 μC en un punto P (x = 0.2 m, y = 0.4 m). ¿Cuál es el campo
eléctrico E debido a esta carga en el origen del plano cartesiano? Hint: asume un plano
cartesiano y dibuja el campo eléctrico que se busca. Además, recuerda el teorema de
Pitágoras.
3. Una carga puntual q = 25μC está ubicada en el centro de una esfera de radio R = 25 cm, el
campo eléctrico generado a esta distancia debido a esa carga es E=3.6 x 10^6 N/C. Se quita de
la superficie una sección circular con radio r = 5 cm. Encuentre el flujo eléctrico que pasa por
esta sección. Hint: solo quieres saber el flujo eléctrico sobre el área de la superficie que se
aisló, por lo que es importante recordar cómo se obtiene el área de un círculo.
4. En cierta región al norte del planeta tierra, existe un campo eléctrico uniforme, la dirección
del campo eléctrico es hacía el centro del planeta y según se sabe, tiene un valor de 1000 N/C.
Encuentra el cambio de la energía potencia eléctrica de una partícula que tiene una carga de
-1.6 x10 -19 C. Considera que la partícula se suelta en un punto A y llega hasta un punto B que
está a 50 m del punto A.
5. Dos cargas q1 = −2 μC y q2 = + 2 μC se fijan en sus posiciones y se separan por una distancia d
= 10 cm. ¿Cuál es el potencial eléctrico en el origen debido a estas dos cargas?
Answer:
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Explanation:
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