Answer:
P.E = 0.027 Joules.
Explanation:
Given the following data;
Spring constant, k = 15N/m
Extension, e = 6cm to meters = 6/100 = 0.06m
To find the potential energy;
P.E = ½ke²
Substituting into the equation, we have;
P.E = ½*15*0.06²
P.E = 7.5 * 0.0036
P.E = 0.027 Joules
can someone pls tell me what a force diagram is
Badll
Which of the following is an example of
the Law of Inertia?
A. Sitting in a chair and breaking it
B. Throwing a ball in outer space and it goes on forever
unless acted upon by another force
C. Eating a salad to bring chemical energy into the body
D. Driving a car on a track
Which of the following parallel plate diagrams would have the greatest electric fields between them?
The greatest electric field between them is Option B.
An electric field is the physical subject that surrounds electrically charged particles and exerts a force on all other charged debris inside the area, either attracting or repelling them. It also refers back to the bodily field of a machine of charged particles.
Reasoning:-
Electric field = Kq/r²
Since the distance is inversely proportional to the square root of the distance between them. therefore decreasing the distance electric field will be the greatest. Hence Option B.
The electrical field is defined mathematically as a vector discipline that can be related to each point in space, the force in step with unit charge exerted on a fine take a look at the price at rest at that factor. the electric field is generated by the electrical rate or through time-varying magnetic fields. electric fields provide us with the pushing force we need to set off the contemporary float.
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A beam of protons is directed in a straight line along the z direction through a region of space in which there are crossed electric and magnetic fields The electric field is 550 V m in the y direction and the protons move at a constant speed of 105 m s 1 What must be the magnitude of the magnetic field such that the beam of protons continues along its straight line trajectory Express your answer using two significant figures
Answer:
B = 5.23 T
Explanation:
Given that,
Electric field, E = 550 V/m
The speed of the proton, v = 105 m/s
We need to find the magnitude of the magnetic field such that the beam of protons continues along its straight-line trajectory.
To move in a straight line, the magnitude of the electric force from the field and the magnetic field must be equal i.e.
[tex]qE=qvB\\\\B=\dfrac{E}{v}\\\\B=\dfrac{550}{105}\\\\B=5.23\ T[/tex]
So, the magnitude of the magnetic field is equal to 5.23 T.
An airplane of mass 13300 kg is flying in a straight line at a constant altitude and with a speed of 560.0 km/hr. The force that keeps the airplane in the air is provided entirely by the aerodynamic lift generated by the wings. The direction of this force is perpendicular to the wing surface. Calculate the magnitude of the lift generated by the wings of this airplane.
Answer:
The magnitude of the lift generated by the wings of the airplane is 130,340 N.
Explanation:
Given;
mass of the airplane, m = 13,300 kg
speed of the airplane, v = 560 km/h = 155.56 m/s
The magnitude of the lift generated by the wings of the airplane is calculated as;
[tex]F_l = mg\\\\where;\\\\F_l \ is \ the \ magnitude \ of \ the \ lift \ generated\\g \ is \ acceleration \ due \ to \ gravity = 9.8 \ m/s^2\\\\F_l = 13,300 \times \ 9.8\\\\F_l = 130,340 \ N[/tex]
Therefore, the magnitude of the lift generated by the wings of the airplane is 130,340 N.
Given three different locations on Earth's surface, where will the weight of a person be greatest?
Answer:
Explanation:
In order to answer this question, we simply have to refer to the laws of the equations of gravitational mechanics.
The equation given by Newton tells us that
[tex]F = \frac{Gm_{1} m_{2} }{r^{2} }[/tex]
In the case where we compare a specific place where the Force of Gravity is greater or lesser, we focus on the term assigned to the Planet's Radius.
In the case of [tex]G, m_{1} ,m_{2}[/tex], we understand that they are constant.
We can easily notice that the more the Radius (Height seen from a viewer on the ground), the lower the force will be.
In other words, the smaller the radius in which the measurement is made with respect to the center of the earth, the greater the gravitational force.
In that order of ideas the smallest radio has South Pole, which is about 6356 km from the center of the Earth on the Equator line
A person drops a pebble of mass m1 from a height h, and it hits the floor with kinetic energy KE. The person drops another pebble of mass m2 from a height of 4h, and it hits the floor with the same kinetic energy KE. How do the masses of the pebbles compare
Hello,
QUESTION)✔ We have: KE = PE (potential energy)
PE = m x g x h
The potential energy that the pebble of mass 1 has is called PE1 and the potential energy that the pebble of mass 2 has is called PE2
PE1 = PE2 ⇔ PE1/PE2 = 1
[tex]\frac{m_1\times g\times h}{m_2\times g\times 4h} = 1 \\ \\ \frac{m_1}{m_2\times 4} = 1 \\ \\ \frac{m_1}{m_2} = 4[/tex]
The mass m1 is therefore 4 times greater than that of the stone of mass m2.
Can you please help me with this worksheet I will give BRAINLISET to whoever answers first?
In the United States, household electric power is provided at a frequency of 60 HzHz, so electromagnetic radiation at that frequency is of particular interest. On the basis of the ICNIRP guidelines, what is the maximum intensity of an electromagnetic wave at this frequency to which the general public should be exposed
Answer:
the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²
Explanation:
Given the data in the question;
To determine the maximum intensity of an electromagnetic wave, we use the formula;
[tex]I[/tex] = [tex]\frac{1}{2}[/tex]ε₀cE[tex]_{max[/tex]²
where ε₀ is permittivity of free space ( 8.85 × 10⁻¹² C²/N.m² )
c is the speed of light ( 3 × 10⁸ m/s )
E[tex]_{max[/tex] is the maximum magnitude of the electric field
first we calculate the maximum magnitude of the electric field ( E[tex]_{max[/tex] )
E[tex]_{max[/tex] = 350/f kV/m
given that frequency of 60 Hz, we substitute
E[tex]_{max[/tex] = 350/60 kV/m
E[tex]_{max[/tex] = 5.83333 kV/m
E[tex]_{max[/tex] = 5.83333 kV/m × ( [tex]\frac{1000 V/m}{1 kV/m}[/tex] )
E[tex]_{max[/tex] = 5833.33 N/C
so we substitute all our values into the formula for intensity of an electromagnetic wave;
[tex]I[/tex] = [tex]\frac{1}{2}[/tex]ε₀cE[tex]_{max[/tex]²
[tex]I[/tex] = [tex]\frac{1}{2}[/tex] × ( 8.85 × 10⁻¹² C²/N.m² ) × ( 3 × 10⁸ m/s ) × ( 5833.33 N/C )²
[tex]I[/tex] = 45 × 10³ W/m²
[tex]I[/tex] = 45 × 10³ W/m² × ( [tex]\frac{1 kW/m^2}{10^3W/m^2}[/tex] )
[tex]I[/tex] = 45 kW/m²
Therefore, the maximum intensity of an electromagnetic wave at the given frequency is 45 kW/m²
they enjoy being alone for quit small time
Ashley decides to enter her pet turtle in a race. She knows her turtle can travel at a rate of 2 meters per hour. The race track measures 5 meters from start to finish. How long will it take Ashley’s turtle to finish the race?
A) 3 hours
B) 10 hours
C) .4 hours
D) 2.5 hours
Thanks for the help! And if it’s not a bother could you please explain how you got the answer? However it’s not necessary if you don’t want to.
BRAINLEST FOR CORRECT ANSWER PLEASE
Which has more momentum: a 3 kg sledgehammer swung at 1.5 m/s OR a 4 kg sledgehammer swung at 0.9 m/s? SHOW YOUR WORK
Answer:
Sledgehammer A has more momentum
Explanation:
Given:
Mass of Sledgehammer A = 3 Kg
Swing speed = 1.5 m/s
Mass of Sledgehammer B = 4 Kg
Swing speed = 0.9 m/s
Find:
More momentum
Computation:
Momentum = mv
Momentum sledgehammer A = 3 x 1.5
Momentum sledgehammer A = 4.5 kg⋅m/s
Momentum sledgehammer B = 4 x 0.9
Momentum sledgehammer B = 3.6 kg⋅m/s
Sledgehammer A has more momentum
can someone help with me 1,2,3 please I will mark brainless .
Answer:
1) A. .33 hr
2) B. 6ft
3) A. 58mi/hr
what is the terminal velocity of blood
Answer:
25.1
Explanation:
how much work is done if a force of 20 N moves an object a distance of 6 m?
Explanation:
W=F×s
w=20N×6m=120J
W=120J
3. A record 0.30 cm in diameter rotates 33.5 times per minute.
a. What is its frequency?
b. What is its period?
c. What is the linear speed of a point on its rim?
d. What is the centripetal acceleration of a point on its rim?
Explanation:
The diameter of a record, d = 0.3 cm
Radius, r = 0.15 cm
It rotates 33.5 times per minute.
(a) Frequency, f = 33.5 rotation/minute
= (33.5/60) rotation/second
= 0.55 rotation/second
(b) Time period,
T = 1/f
So,
[tex]T=\dfrac{1}{0.55}\\\\T=1.81\ s[/tex]
(c) Linear speed of a point on its rim,
[tex]v=r\omega\\\\v=0.15\times 10^{-2}\times 2\pi \times 0.55\\\\v=5.18\times 10^{-3}\ m/s[/tex]
(d) Centripetal acceleration,
[tex]a=\dfrac{v^2}{r}\\\\a=\dfrac{(5.18\times 10^{-3})^2}{0.15\times 10^{-2}}\\\\a=0.017\ m/s^2[/tex]
Hence, this is the required solution.
The energy that a substance contains is
a. equilibrium
b. heat
c. thermal energy
d. entropy
The energy that a substance contains is known as heat. Thus, the correct option for this question is B.
What is Entropy?Entropy may be characterized as the measurement of the degree of randomness or in other words, it is the increase in the disorganization within a system. It is the measure of a system's thermal energy per unit temperature that is unavailable for doing useful work.
Equilibrium is a state of balance between opposing forces or actions that is either static or dynamic. It symbolizes the circumstance of being balanced in nature.
Thermal energy is a type of energy that is significantly contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.
Therefore, the energy that a substance contains is known as heat. Thus, the correct option for this question is B.
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PLEASE HELP ASAP!!!
A car is sitting on a hill which is 25 m higher than ground level. What is the mass of the car if it contains 360,800 J of potential energy?
Answer:
14,432 g
Explanation:
PE=mh
360,800 J = (m)(25)
[tex]\frac{360,800 J}{25 m}[/tex] = m
14,432 g = m
In which regions can the gravitational field strength due to the two planets be zero? Explain.
A. None
B. A, B and C
C. A and C
D. B
Answer:
I believe its a and c but my notes are all kinds of messed up so im sorry if its wrong
Explanation:
None of the regions can the gravitational field strength due to the two planets be zero.
What is gravitational field?
A gravitational field is a model used in physics to explain the effects that a large thing has on the area surrounding it, exerting a force on smaller, less massive bodies. Consequently, a gravitational field, which is measured in newtons per kilogram, is employed to describe gravitational processes.
Field due to given masses can not be zero in the given region they do not cancel each other.
None of the regions can the gravitational field strength due to the two planets be zero.
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50 points help please
Answer:
C?
Explanation:
Yep,It's C all right.
Answer:
Yep,It's C all right.
Explanation:
An Atwood's machine consists of two masses, m1 and m2, connected by a string that passes over a pulley. If the pulley is a disk of radius R and mass M , find the acceleration of the masses. Express your answer in terms of the variables m1, m2, R, M, and appropriate constants.
Answer:
Explanation:
Suppose m₂ is greater than m₁ and it is going down . m₁ will be going up.
Let tension in string be T₁ and T₂ . Let common acceleration of system be a
For motion of m₁
T₁-m₁g = m₁a ----- (1)
For motion of m₂
m₂g- T₂ = m₂a ------- (2)
For motion of pulley
(T₂-T₁ )R represents net torque
(T₂-T₁ )R = I x α where I is moment of inertia of disc and α is angular acceleration of disc
(T₂-T₁ )R = 1/2 M R² x a / R
(T₂-T₁ ) = M a /2
Adding (1) and (2)
(m₂-m₁)g = (m₂+m₁)a + (T₂-T₁ )
(m₂-m₁)g = (m₂+m₁)a + Ma/2
(m₂-m₁)g = (m₂+m₁+ 0.5M)a
a = (m₂-m₁)g / ( (m₂+m₁+ 0.5M)
The acceleration of the masses is [tex]\dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex].
Given to us
Masses = m₁, m₂
The radius of the pulley = R
Mass of the pulley = M
Assumption
Let the mass m₁ > m₂. therefore, the mass m₁ is going down due to its weight while m₂ is going up.Assuming the tension in the string be T₁ and T₂, respectively.Also, the common acceleration in the system is a.Tensions in stringsWe know the acceleration due to gravity is denoted by g,
Tension in string 1, T₁[tex]T_1 = m_1(a+g)[/tex]......... equation 1
Tension in string 2, T₂[tex]T_2 = m_2(a-g)[/tex]......... equation 2
Inertia and acceleration of the pulleyThe inertia of the pulley, [tex]I = \dfrac{1}{2}MR^2[/tex]
Acceleration of the pulley, [tex]a = {\alpha }\times {R}[/tex]
Torque in the pulley[tex](T_2-T_1)R = I \times \alpha \\\\[/tex]
Substitute the values we get,
[tex][m_2(a-g)-m_1(a+g)]R = \dfrac{1}{2} MR^2 \times \dfrac{a}{R}\\\\[/tex]
[tex][m_2a-m_2g-m_1a-m_1g]R= \dfrac{1}{2} MR \times a[/tex]
[tex]m_2a-m_1a-m_2g-m_1g= \dfrac{1}{2} M \times a[/tex]
[tex]a(m_2-m_1)-g(m_2+m_1)= 0.5M \times a[/tex]
[tex]a(m_2-m_1)-0.5Ma=g(m_2+m_1)\\\\a(m_2-m_1-0.5M) = g(m_2+m_1)\\\\a = \dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex]
Hence, the acceleration of the masses is [tex]\dfrac{g(m_2+m_1)}{(m_2-m_1-0.5M)}[/tex].
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ou have been called to testify as an expert witness in a trial involving a head-on collision. Car A weighs 660.0 kg and was traveling eastward. Car B weighs 490.0 kg and was traveling westward at 64.0 km/h. The cars locked bumpers and slid eastward with their wheels locked for 6.00 m before stopping. You have measured the coefficient of kinetic friction between the tires and the pavement to be 0.750
Answer:
vₐ₀ = 29.56 m / s
Explanation:
In this exercise the initial velocity of car A is asked, to solve it we must work in parts
* The first with the conservation of the moment
* the second using energy conservation
let's start with the second part
we must use the relationship between work and kinetic energy
W = ΔK (1)
for this part the mass is
M = mₐ + m_b
the final velocity is zero, the initial velocity is v
friction force work is
W = - fr x
the negative sign e because the friction forces always oppose the movement
we write Newton's second law for the y-axis
N -W = 0
N = W = Mg
friction forces have the expression
fr =μ N
fr = μ M g
we substitute in 1
-μ M g x = 0 - ½ M v²
v² = 2 μ g x
let's calculate
v² = 2 0.750 9.8 6.00
v = ra 88.5
v = 9.39 m / s
Now we can work on the conservation of the moment, for this part we define a system formed by the two cars, so that the forces during the collision are internal and therefore the tsunami is preserved.
Initial instant. Before the crash
p₀ = + mₐ vₐ₀ - m_b v_{bo}
instant fianl. Right after the crash, but the cars are still not moving
p_f = (mₐ + m_b) v
p₀ = p_f
+ mₐ vₐ₀ - m_b v_{bo} = (mₐ + m_b) v
mₐ vₐ₀ = (mₐ + m_b) v + m_b v_{bo}
let's reduce to the SI system
v_{bo} = 64.0 km / h (1000m / 1km) (1h / 3600s) = 17.778 m / s
let's calculate
660 vₐ₀ = (660 +490) 9.39 + 490 17.778
vₐ₀ = 19509.72 / 660
vₐ₀ = 29.56 m / s
we can see that car A goes much faster than vehicle B
What is the gravitational potential energy of a 3 kg ball that is 1 meter above the floor?
Answer:
PE=mgh
M= Mass (kg)
G= Gravitational field strength (N/kg)
H= Hight (m)
PE= Gravitational Potential Energy (J)
Explanation:
Gravitational Potential Energy is the energy stored in a object due to its position above the Earth's surface.
The potential energy of the ball is 29.4 joules
The parameters given are;
mass= 3 kg
height= 1 meter
g= 9.8 m/s²
P.E= mgh
= 3 × 9.8 × 1
= 29.4
Hence the potential energy is 29.4 joules
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Tom has a mass of 50,000 g and runs up a flight of stairs 4 m high in 12.5 seconds.
Calculate Tom’s power. (g = 10 m/s2)
Answer:
160 watts.
Explanation:
Remark
Power = Work / Time
Work = F * d
Note: Since he is running up stairs he is doing work against gravity.
Givens
m = 50000 g kg / 1000 grmsm = 50000 / 1000 = 50 kgh = 4 mtime = 12.5 secondsg = 10 m/s^2Formula
P = W * d/tW = m*g *d / tSolution
P = 50kg * 10 m/s^2 * 4 m / 12.5 P = 160 watts.
A 100 gram piece of ice is placed in an insulated calorimeter of negligible heat capacity containing 100 grams of
water at 373K.
a. What is the final temperature of the water once thermal equilibrium is established?
b. Find the entropy change of the universe for this process.
Answer:
a) 10.23°C
b) - 0.8795 kJ / k
Explanation:
Given data :
mass of ice = 100 gram = 0.1 kg
mass of water in calorimeter = 100 grams = 0.1kg
temperature of water in calorimeter = 373K
a) Determine the Final temperature of the water once thermal equilibrium is established
Heat lost by water = heat gained by Ice
Mwater * CPwater * Δt = Mice * Lice + Mice * CPwater * Δt
0.1 * (4.2 * 10^3 ) * ( 100 - t ) = ( 0.1 * 334 * 10^3 ) + 0.1 * (4.2 *1063 ) * ( T - 0 )
42000 = 33400 + 0.2 * 4200 * T
∴ equilibrium temp ( T ) = 10.23°C
B) Determine the entropy change of the universe of the process
Entropy change of ice = 33400/ 273 + 0.1 * 42 * 10^3 * In 283.23 / 273
= 276.84 J/K
Entropy change of water = 0.1 * 42* 10^3 * In 283.23 / 273
= -1156.34 J/K
Hence Entropy change of variance = 276.84 J/K + ( - 1156.34 J/K )
= - 0.8795 kJ / k
An electron has a charge of 1.602 X 10-19.coulomb. When two electrons are separated by 1.2 X 10-9m, what force will they exert on each other?(-19 & -9 are exponents)
6.2 X 10-6N
1.6 X 10-10N
3.4 X 10-9N
5.3 X 10-7 N
Answer:
The force they will exert on each other is 1.6*10⁻¹⁰ N
Explanation:
The electromagnetic force is the interaction that occurs between bodies that have an electric charge. When the charges are at rest, the interaction between them is called the electrostatic force. Depending on the sign of the interacting charges, the electrostatic force can be attractive or repulsive. The electrostatic interaction between charges of the same sign is repulsive, while the interaction between charges of the opposite sign is attractive.
Coulomb's law is used to calculate the electric force acting between two charges at rest. This force depends on the distance "r" between the electrons and the charge of both.
Coulomb's law is represented by:
[tex]F=k*\frac{q1*q2}{r^{2} }[/tex]
where:
F = electric force of attraction or repulsion in Newtons (N). Like charges repel and opposite charges attract. k = is the Coulomb constant or electrical constant of proportionality. q = value of the electric charges measured in Coulomb (C). r = distance that separates the charges and that is measured in meters (m).In this case:
k= 9*10⁹ [tex]\frac{N*m^{2} }{C^{2} }[/tex]q1= 1.602*10⁻¹⁹ Cq2= 1.602*10⁻¹⁹ Cr= 1.2*10⁻⁹ mReplacing:
[tex]F=9*10^{9} \frac{N*m^{2} }{C^{2} }*\frac{1.602*10^{-19} C*1.602*10^{-19} C}{(1.2*10^{-9} )^{2} }[/tex]
and solving you get:
F=1.6*10⁻¹⁰ N
The force they will exert on each other is 1.6*10⁻¹⁰ N
A forward honzontal force of 50 N is applied to a crate. A second horizontal force o
180 N is applied to the crate in the opposite direction. Determine the magnitude and
direction of the resultant force acting on the crate
Answer:
Explanation:
magnitude: 180-50=130N
Direction: in the direction same as the second horizontal force
Would this be C ?? Please tell me if I’m wrong ..
Answer:
yes its c
Explanation:
took it already
POSSIBLE POINTS: 1
After the Frisbee falls to the ground, you throw one of them back up into the tree, where it comes to rest on a branch.
Compared to when the Frisbee was on the ground, which of the following is true about the potential energy of the system?
Answer:
the effect on energy is due to the force applied to the frisbee to throw it back in the tree.
Explanation:
At the end of the previous experiment, aclumsy scientist drops the coil, while still in the magnetic field, and still oriented with its plane perpendicular to the magnetic field, denting it and changing its shape to a semi-circle. The new shape has the same perimeter, but a different area, and it takes 0.036s to deform. What isthe average induced EMF during this mishap
Answer:
hello your question has some missing parts below are the missing parts
A Circular, 10-turn coil has a radius of 10.7 cm and is oriented with its plane perpendicular to a 0.2-T magnetic field.
answer : 1 volt
Explanation:
Determine the Average induced EMF during this mishap
A' = A/2 ( for a semi circle )
where A = [tex]\frac{\pi r^2}{2}[/tex]
To determine the Average induced EMF apply the relation below
| E | = η * [tex]\frac{\beta A}{T}[/tex] ----- ( 1 )
Replace A in equation 1 with A = [tex]\frac{\pi r^2}{2}[/tex]
hence equation becomes : | E | = η * βπr^2 / 2T'
where : T' = 0.0365 , β = 0.2 , η = 10 , r = 0.107
∴| E | = 0.999 ≈ 1volts