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:
________________________
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.
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
How much power is required to do 10 J of work on an object in 2 seconds?
Answer:
5 watts
power = work done ÷ time taken
10 j ÷ 2 = 5w
a disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular acceleration?
b. how long did it take the disk to reach this velocity?
Answer: [tex]1.875\ rad/s^2, 8\ s[/tex]
Explanation:
Given
Radius of disc [tex]r=10\ cm[/tex]
Angle turned [tex]\theta=60\ rad[/tex]
Initial angular velocity [tex]\omega_0=0[/tex]
Final angular velocity [tex]\omega_f=15\ rad/s[/tex]
using [tex]\omega^2-\omega_o^2=2\alpha \cdot \theta[/tex]
Substitute values
[tex]\Rightarrow 15^2-0=2\times \alpha \times 60\\\Rightarrow \alpha=1.875\ rad/s^2[/tex]
using [tex]\omega=\omega_o+\alpha t[/tex]
Substitute values
[tex]\Rightarrow 15=0+1.875\times t\\\Rightarrow t=8\ s[/tex]
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
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.
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
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.
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
A car travels 60km in 45 minutes. At the same average speed, how far will it travel in 1 hour 30 minutes?
Drink water at least every
weather.
minutes while exercising in hot
Answer:
Drink water at least every 5 to 20 minutes while exercising in hot weather.
Explanation:
Hope this helps brainliest please
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
A circular loop of radius 15 cm carries a current of 11 A. A flat coil of radius 0.79 cm, having 66 turns and a current of 1.9 A, is concentric with the loop. The plane of the loop is perpendicular to the plane of the coil. Assume the loop's magnetic field is uniform across the coil. What is the magnitude of (a) the magnetic field produced by the loop at its center and (b) the torque on the coil due to the loop
Answer:
[tex]4.61\times 10^{-5}\ \text{T}[/tex]
[tex]1.05\times 10^{-6}\ \text{Nm}[/tex]
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi10^{-7}\ \text{H/m}[/tex]
[tex]r_l[/tex] = Radius of loop = 15 cm
[tex]I_l[/tex] = Current in loop = 11 A
[tex]r_c[/tex] = Radius of coil = 0.76 cm
N = Number of turns of coil = 66
[tex]I_c[/tex] = Current in coil = 1.9 A
Magnetic field is given by
[tex]B=\dfrac{\mu_0I_l}{2r_l}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}\times 11}{2\times 0.15}\\\Rightarrow B=4.61\times 10^{-5}\ \text{T}[/tex]
Magnitude of magnetic field produced by the loop at its center is [tex]4.61\times 10^{-5}\ \text{T}[/tex].
Torque is given by
[tex]\tau=BI_c\pi r_c^2N\sin90^{\circ}\\\Rightarrow \tau=4.61\times 10^{-5}\times 1.9\times \pi\times (0.76\times 10^{-2})^2\times 66\sin90^{\circ}\\\Rightarrow \tau=1.05\times 10^{-6}\ \text{Nm}[/tex]
Magnitude of torque on the coil due to the loop is [tex]1.05\times 10^{-6}\ \text{Nm}[/tex]
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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The first law of Thermodynamics is another way to describe the law of conservation of Energy. It states that:
Answer:
C. The change of internal energy of a system is the sum of work and heat spent on it.
Explanation:
The law of conservation of Energy states that energy cannot be destroyed but can only be converted or transformed from one form to another. Therefore, the sum of the initial kinetic energy and potential energy is equal to the sum of the final kinetic energy and potential energy.
Mathematically, it is given by the formula;
Ki + Ui = Kf + Uf .......equation 1
Where;
Ki and Kf are the initial and final kinetic energy respectively.
Ui and Uf are the initial and final potential energy respectively.
The law of conservation of Energy is another way to describe the law of Thermodynamics. It states that the change of internal energy of a system is the sum of work and heat spent on it.
Mathematically, it is given by the formula;
ΔU = Q − W
Where;
ΔU represents the change in internal energy of a system.
Q represents the net heat transfer in and out of the system.
W represents the sum of work (net work) done on or by the system.
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:
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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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.A ray in glass is incident onto a water-glass interface. The angle of incidence equals 0.75 times the critical angle for that interface. The index of refraction for water is 1.33 and for glass is 1.78. What is the angle that the refracted ray in the water makes with the normal?
52 degrees
42 degrees
48 degrees
63 degrees
Which is true regarding a child standing up for their own rights?
Answer:
hey mate......looks like the question is incomplete
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.
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
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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3 CO2
How many atoms of Oxygen (O) are there?
Answer:
6 oxygen
Explanation:
3 multiply O2 and 6 Oxygen
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:
Heat transfer in liquids or gases that happens due to currents of hot and cold is called
Entropy
Conduction
Convection
Radiation
Answer:
convection it's your answer plz make me brainy least
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
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
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