Answer:
yes I think it is possible
True or False:
If you look at the night sky and see a very bright star, you can tell that that star is closer to Earth than all the other stars.
Answer:
hi
Explanation:
false because the light of some stars can be very bright, but it can also be very far from the earth.
have a nice day
At the end of the passage, Sarah says to Emma, “You’re not the only one with tricks up your sleeves.” Explain what Sarah means by this. Use information from the passage to support your answer.
Answer:
Sarah plans to use trickery and cunning to get back at Emma. Sarah thinks she is smarter than Emma.
Explanation:
i think this is right i dont know tho
What is the work done by the 200.-N tension shown if it is used to drag the 150-N crate 25 m across the floor at a constant speed?
Answer:
0 J
Explanation:
Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.
In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump? Assume no air resistance and that ay = −g = −9.81 m/s2 . (Ans. 2.6m/s) PLS SHOW WORK
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The minimum horizontal velocity required is "2.6 m/s".
First, we will analyze the vertical motion of the stunt person. We will use the second equation of motion in the vertical direction to find the time interval for the motion.
[tex]h=v_it+\frac{1}{2}gt^2[/tex]
where,
h = height = 3 m
vi = initial vertical speed = 0 m/s
t = time interval = ?
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]3\ m = (0\ m/s)(t) + \frac{1}{2}(9.81\ m/s^2)t^2\\\\t^2 = \frac{(3\ m)(2)}{9.81\ m/s^2}\\\\t = \sqrt{0.611\ s^2}[/tex]
t = 0.78 s
Now, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here:
[tex]s = vt\\\\v = \frac{s}{t}[/tex]
where,
s = horizontal distance = 2 m
t =0.78 s
v = minimum horizontal velocity = ?
Therefore,
[tex]v = \frac{2\ m}{0.78\ s}[/tex]
v = 2.6 m/s
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The attached picture shows the equations of motion in the horizontal and vertical directions.
A phase change is when a substance changes from one state of mind to nother because of the adding or removal of thermal energy
True
False
Answer:
true it all changes
Explanation:
________________________
Which one of the statements below is true about mechanical waves?
They must travel in empty space.
They can travel in a vacuum.
Both sound and light are examples of mechanical waves.
They require a medium to travel through.
Answer:D
Explanation:
Explain in your
own words how
you would find
the density of a
regular-shaped
object.
Answer:
Use a ruler to measure the length (l), width (w) and height (h) of the object.Place the object on the top pan balance and measure its mass.Calculate the volume of the cube using (l*w*h).Use the measurements to calculate the density of the object.True or false? Pls help
False.
Tripling the height will triple the potential energy.
Speed has nothing to do with potential energy.
3 CO2
How many atoms of Oxygen (O) are there?
Answer:
6 oxygen
Explanation:
3 multiply O2 and 6 Oxygen
Sparks occur when the electric field in air exceeds 3 x 106 N/C. This is because free electrons normally present in air are accelerated to such high speeds that their kinetic energy will overcome the potential energy holding other electrons to atoms. When those electrons rearrange themselves after such a collision, a flash of light is emitted. Let us suppose that the work done on an electron must give it an energy of 3 x 10-19 J to cause this ionization. How far does an electron involved in making in a spark travel through the air before it collides with an atom
Answer:
h = 5.38 10¹⁶ m
Explanation:
Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy
starting point. Just before ionization
Em₀ = U = qE
final point. Right after ionization
Em_f = K = ½ m v²
Energy is conserved
Em₀ = Em_f
q E = ½ m v²
v² = 2qE / m
Now we can use the relationship between net work and kinetic energy
W_net = ΔK
net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J
W - Ф = K_f - K₀
we assume that the electron converts all its initial initial kinetic energy to be zero
W -Ф = ½ m v² - 0
W = ½ m v² +Ф
we substitute
W = 1/2 m 2qE/m + E
W = qE +Ф
W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ J
When the electron is in air, its kinetic energy can be transformed into gravitational potential energy
As the electron is in the air, all work is transformed into scientific energy
W = K
starting point Em₀ = K = W
end point Em_F = U = m g h
energy conservation Em₀ = Em_f
W = m g h
h = [tex]\frac{W}{mg}[/tex]
let's calculate
h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]
h = 5.38 10¹⁶ m
Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.
What is an electric field?An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.
Energy before ionization;
[tex]\rm Em_0 = U = qE[/tex]
Energy after ionization;
[tex]Em_f = K = \frac{1}{2} mv^2[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]
The relationship between net work and kinetic energy;
[tex]W_{net} = \triangle K[/tex]
The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.
[tex]W - \phi = K_f - K_0[/tex]
[tex]W = K_f+ \phi[/tex]
[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]
EMF at starting point;
[tex]\rm Em_0 = K = W[/tex]
EMF at the endpoint;
[tex]\rm Em_F = U = m g h[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]
Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.
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Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizontal force of 100N is exerted on box A. What is the magnitude of the force that box A exerts on box B?
Answer:
2.75 N
Explanation:
Given that,
Box A has a mass 21.0 kg and box B has a mass 8.0 kg.
A horizontal force of 100N is exerted on box A.
Let a be the acceleration of the system. Using second law of motion,
[tex]F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2[/tex]
Now applying Newton's second law to box B. So,
[tex]F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N[/tex]
So, 2.75 N is the force that box A exerts on box B.
The key concept here is that when two objects interact, the forces they exert on each other are equal in magnitude but opposite in direction. The magnitude of the force that box A exerts on box B is also 100N.
Newton's third law of motion states that for every action, there is an equal and opposite reaction. It is one of the three fundamental principles described by Sir Isaac Newton in his Laws of Motion.
According to Newton's third law, when an object exerts a force on another object, the second object simultaneously exerts a force of equal magnitude but in the opposite direction on the first object. In simpler terms, if object A applies a force on object B, then object B applies an equal force in the opposite direction on object A.
Since boxes A and B are in contact on a frictionless surface, the force exerted on box A will be transmitted to box B. According to Newton's third law of motion, the magnitude of the force that box A exerts on box B will be equal in magnitude but opposite in direction to the force exerted on box A.
Therefore, the magnitude of the force that box A exerts on box B is also 100N.
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Which is true regarding a child standing up for their own rights?
Answer:
hey mate......looks like the question is incomplete
Pushing open door .action force and reaction
Answer: Newtons third law
Explanation: According to Newton's third law of motion, forces always act in equal but opposite pairs. Another way of saying this is for every action, there is an equal but opposite reaction. This means that when you push on a wall, the wall pushes back on you with a force equal in strength to the force you exerted.
Newton's third law we find that the reaction to the force is an out applied by the putting door towards to the person.
given parameters
Push the door open
to find
The reaction
Newton's third law states that the forces between bodies always occur in pairs, that is, the two bodies interacting with each other, therefore these forces are of equal magnitude, but opposite direction. Consequently each force acts on one of the bodies, these forces are called action and reaction.
By applying Newton's third law to the open door; we exert a force on the door therefore the door exerts a force of equal magnitude, but opposite direction on us.
Using Newton's third law we find that the reaction to the force is an out applied to the person.
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a stationary object experiences two forces as shown in the diagram below
Answer: the answer is B
Explanation: 80 is not the same as 150 so it will go the way 150 units of force is pulling.
The force of Earths gravity keeps earth in orbit true or false
Answer:
True
Explanation:
The force of gravity keeps all of the planets in orbit around the sun
True. The force of gravity keeps all of the planets in orbit around the sun.
What is Gravity?
The force that pulls items toward the center of a planet or other entity is called gravity. All of the planets are kept in orbit around the sun by gravity.
Gravity applies to everything that has mass. Gravity is stronger for objects with higher mass. Along with distance, gravity weakens as well. Therefore, the gravitational pull of two things becomes stronger the closer they are to one another.
The mass of the Earth is what creates gravity. The combined gravitational force of all of its mass acts on the mass in your body.
Therefore, True. The force of gravity keeps all of the planets in orbit around the sun.
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49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n times as great as that measured at corner B. What is the ratio L1/L2 of the lengths of the sides of the rectangle when n
Answer:
[tex]\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}[/tex]
Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
[tex]a_A = n \ a_B[/tex]
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A = [tex]\sqrt{L_1^2 + L_2^2}[/tex]
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
A dump truck contains a load of soil. Which action will leave the dump truck's
inertia unchanged?
A. Dump out some of the soil.
B. Add gas to its fuel tank.
C. Add more soil.
D. Increase the force applied by the engine.
Answer:
D
Explanation:
This will not change the weight and therefore not change the inertia
A dump truck contains a load of soil. The action that will leave the dump truck's inertia unchanged is that increase the force applied by the engine. Hence, option D is correct.
What is inertia?A body's ability to resist being propelled into motion or, if already moving, to modify the direction or magnitude of its velocity is known as inertia.
An object's lethargy is a passive quality that prevents it from doing anything other than obstructing active forces and torques. The only reason a moving body continues to move is the lack of a force that might slow it down, alter its trajectory, or accelerate it. This is not due to inertia.
A body's inertia moment about a certain axis and its mass, which determine how resistant it is to the application of forces to that axis, respectively, are two statistical measures of inertia.
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Which cloud types would most likely indicate that a thunderstorm is on the way?
cirrus clouds
dull, gray, stratus clouds
cumulus clouds that are small and round
cumulus clouds that are tall with a flat top
Answer:
The aswer is D
Explanation:
Answer:
d
Explanation:
cumulus clouds that are tall with a flat top
What do radio waves and gamma rays have in common?
They are both electromagnetic waves.
They are both low frequency waves.
They can only travel in a vacuum.
They both are part of the visible light spectrum.
Answer:
Both Magnetic
Explanation:
does latitude has an effect on weight?
Answer:
I think so but i could be wrong..
Explanation:
Answer: yes it does
Explanation: Yes, you weigh less on the equator than at the North or South Pole, but the difference is small. Note that your body itself does not change. Rather it is the force of gravity and other forces that change as you approach the poles. These forces change right back when you return to your original latitude.
The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will have a length of exactly 3 meters. Testing indicates that the aircraft wing will remain stable only if the component never expands to a length larger than 3.017 meters. If the component is made from the metal alloy in question, will it meet this requirement. An unknown metal alloy is being tested to discover its thermal properties to see if it is suitable for use as a component in an aircraft wing. The alloy is formed into a bar measuring 1 meter in length, and is then heated from its starting temp. of 30°C to a final temperature of 100°C. The length of the heated bar is measured to be exactly 1.002 meters in length.
Required:
What is the coefficient of thermal expansion of the alloy?
Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α = [tex]\frac{\Delta L}{L_o \ \Delta T}[/tex]
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹
What do we call the small changes that
could result in large future changes?
A. the "butterfly effect"
B. the "snowflake effect"
C. the "ripple effect"
D. the "trickle-down effect"
Answer:
The "butterfly Effect"
Explanation:
The "butterfly effect" will probably have big changes in the future.
To say that electric charge is conserved means that no case has ever been found where :_________
a. the total charge on an object has changed.
b. the total quantity of charge on an object has increased.
c. net charge has been created or destroyed.
d. quantity of negative charge on an object exactly balances positive charge.
e. none of the above
Answer:
B
Explanation:
To say that electric charge is conserved means that no case has ever been found where :_________
the total quantity of charge on an object has increased
According to conservation of energy, energy is neither created nor destroyed. But can only be converted into another form.
But in the case of charges, the number of charges can not be increased nor decreased it's only the rate of flow that can change due to the pressure supplied.
So therefore, the correct answer is B which say that:
To say that electric charge is conserved means that no case has ever been found where the total quantity of charge on an object has increased
an arrow is shot horizontally from the top of a tower at a speed of 15m/s and hits the ground with a speed of 25m/s. calculate the height of the tower
The height of the tower is 20.41 m.
To determine the height of the tower, we need to understand the concept of the energy conservation principle since the speed and acceleration due to gravity are involved in the system.
What is the energy conservation principle?The principle of energy conservation lets us know that in an isolated system, energy can neither be created nor destroyed.
It can be expressed using the formula:[tex]\mathbf{mgh = \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2}[/tex]
From the parameters given:The initial speed [tex]v_1[/tex] = 15 m/sThe final speed [tex]v_2[/tex] = 25 m/sBy applying the energy conservation principle, we have:
[tex]\mathbf{gh +\dfrac{1}{2}v_1^2 = \dfrac{1}{2}v_2^2}[/tex]
[tex]\mathbf{h = \dfrac{v_2^2- v_1^2 }{2 \times g}}[/tex]
[tex]\mathbf{h = \dfrac{25^2-15^2 }{2 \times 9.8}}[/tex]
h = 20.41 m
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what must be the mass of a rock if a boy applies a 64N force and causes it to accelerate at 4.51m/s2
first of all the formula of force is F=ma,so we are searching for m,so we can divide a on both sides F/a=m, after this substitute the values given above 64N/4.51=14.2°Kg
Time Vs Position of Battery Operated Car what type of relationship is shown in the graph?
PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
1.62 m/s²
Explanation:
There is a bell at the top of a tower that is 45 m high. The bell weighs 190 kg. The bell has
energy. Calculate it
Answer: The bell has 8550 J energy.
given, There is a bell at the top of a tower that is 45 m high. The bell weighs 190 N
i.e., bell is located at the top of tower, h = 45m
weight of the bell, F = 190 N
workdone by the gravitational force = F.hcos180°
[ gravitational force (i.e., weight ) acting downward while body is located 45m above the ground. so, angle between force and h = 180° ]
workdone by the gravitational force = 190 × 45 × (-1)
= -8550 J
we know, potential energy = negative of workdone
= -(-8550 J) = 8550 J
Answer:
200,000J
Explanation:
KE= 0.5 x m x v^2
KE= 0.5x1000x20^2
KE= 0.5 x 1000 x 400
KE= 500 x 400
KE= 200,000
which plate is made up almost entirely of oceanic crust
A. south American plate
B. Eurasian plate
C. indo-Australia plate
D. Pacific plate
Answer:
D. Pacific Plate
Explanation:
Answer:
d
Explanation:
i got it right
Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 17 cm that is placed in a spatially uniform magnetic field of magnitude 0.45 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)
Answer:
the induced electric field is 9.95 V/m
Explanation:
Given the data in the question;
Number of turns N = 52
Diameter of coil D = 17 cm = 0.17 m
Radius r = D/2 = 0.17/2 = 0.085 m
Now,
cross-section area A of the coil = πr²
A = π × ( 0.085 m )²
A = 0.0227 m²
Also given that;
Initial magnetic field B₁ = 0.45 T
Final magnetic field B₂ = 0
∴ change in magnetic field ΔB = B₁ - B₂ = 0.45 T - 0 = 0.45 T
Time taken dT = 0.10 seconds
Now, we know that;
Induced emf ∈ = N[tex](\frac{d\eta }{dt} )[/tex]
where η = BAcosθ
We know that, magnetic field is cylindrically symmetric, coil is also perpendicular to magnetic field.
Hence, the angle between B & A is 0°
∴ θ = 0°
Induced emf ε = N[tex](\frac{d }{dt} )BAcos\theta[/tex]
we substitute
ε = N[tex](\frac{d }{dt} )[/tex] (BAcos0°)
A is constant and cos0° = 1
so
ε = NA[tex](\frac{dB }{dt} )[/tex]
We now substitute in our values;
ε = 52 × 0.0227 m² × [tex](\frac{0.45T }{0.10s} )[/tex]
ε = 5.3118 V
we know that, from the relation between electric and emf
ε = ∫∈.dl or ε = ∈∫dl { for coil; ∫dl = πD }
so we have;
ε = ∈πD
solve for ∈
∈ = ε/πD
we substitute
∈ = 5.3118 V / ( π × 0.17 m )
∈ = 9.95 V/m
Therefore, the induced electric field is 9.95 V/m