Answer:
They attach due to the attraction of charges.
When the ebonite rod is rubbed with silk, the rod will aquire a negative charge and the silk will aquire a positive charge.
Thus they will attract.
Like charges repel (+)(+) (-)(-).Opposite charges attract (+)(-) (-)(+).Charged and neutral attract (+)(-+) (-)(+-)._____________________________
what is the terminal velocity of blood
Answer:
25.1
Explanation:
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Question
A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500 - kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is displaced from the origin to x = 5.00 cm and released from rest at t = 0. (a) What is the force constant of the spring? (b) What are the angular frequency ω, the frequency, and the period of the motion? (c) What is the total energy of the system? (d) What is the amplitude of the motion? (e) What are the maximum velocity and the maximum acceleration of the particle? (f) Determine the displacement x of the particle from the equilibrium position at t = 0.500 s. (g) Determine the velocity and acceleration of the particle when t = 0.500 s.
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
We observe that a moving charged particle experiences no magnetic force. From this we can definitely conclude that: (A) No magnetic field exists in that region of space. (B) The particle must be moving parallel to the magnetic field. (C) Either no magnetic field exists or the particle is moving parallel to B. (D) Either no magnetic field exists or the particle is moving vertical (perpendicular) to B.
Answer:
(C) Either no magnetic field exists or the particle is moving parallel to B.
Explanation:
The moving charged particle tends to show no magnetic forces. Due to its moving in the same direction and will be parallel to the field. As when the magnetic fed and charged particle velocity is parallel then there will be no force. As the magnetic field are made by the generation of the current with the field line concentric to the current-carrying wire.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.
Would this be C ?? Please tell me if I’m wrong ..
Answer:
yes its c
Explanation:
took it already
Can you please help me with this worksheet I will give BRAINLISET to whoever answers first?
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 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
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
what is the effect of cigarette smoke in your family
A movie stunt double is supposed to run across the top of a train (in the opposite direction that the train is moving) and just barely jump off before reaching a tunnel, but after running the entire length of the train (starting from the front). If the train is moving at 50 km/hr, is 3 km long and the tunnel is 40 km away from the end (where the stunt double is going to jump from), how fast (in km/hr) will the stunt double need to run
Answer:
3.49km/hr
Explanation:
This train is moving at 50km/hr
Distance = length of train + distance of tunnel
= 3 km + 40 km
= 43 km
Time taken by train to get to tunnel
Time = distance / velocity
Time = 43 km/50km perhour
= 0.86
How fast will stunt double have to run?
Velocity = 3km/0.86
= 3.49 km/hr
The stunt double will have to run as fast as 3.49 kilometer per hour
can someone pls tell me what a force diagram is
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
What order shows increasing frequency for gamma
rays, microwaves, visible light, and X-rays?
A. gamma rays, X-rays, visible light, microwaves
B. microwaves, visible light, X-rays, gamma rays
C. visible light, gamma rays, microwaves, X-rays
D. X-rays, microwaves, gamma rays, visible light
Answer:
The answer is B: microwaves, visible light, X-rays, gamma rays
Explanation:
I had the same question and I got option B as the correct answer.
The waves arranged in their order of increasing frequency will be
Microwaves,Visible light, X-rays and Gamma rays.What order shows the increasing frequency for gamma rays, microwaves, visible light, and X-rays?Generally, frequency is the numeral of occurrences of a repeating event per unit of duration.
Therefore, gamma rays, microwaves, visible light, and X-rays will be arranged in their order of increasing frequency
In conclusion, The order is given as folllows
microwaves, Visible light, X-rays, and Gamma rays.
Option B is correct
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2. Which process is used by plants to cycle water from the ground, through the roots, and out the leaves as water vapor?
sublimation
O perspiration
O transpiration
O runoff
Answer:
The answer is transpiration
they enjoy being alone for quit small time
At what height is an object that has a mass of 16 kg, it its gravitational potential energy is 7500 J.
Answer:
h = 46.87 m
Explanation:
Given that,
Mass of an object, m = 16 kg
Gravitational potential energy of the object, E = 7500 J
We need to find the height of the object. The formula for the gravitational potential energy is given by :
[tex]E=mgh\\\\h=\dfrac{E}{mg}\\\\h=\dfrac{7500}{16\times 10}\\\\h=46.87\ m[/tex]
So, the required height is equal to 46.87 m.
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
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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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
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
As thermal energy is added to a sample of water, the potential energy of its molecules increases, and then the kinetic energy of its molecules increases. Which sections of the heating curve illustrate this process?
Answer: D) B followed by D
Explanation:
A p e x
1) Marty (m = 8 kg) is sitting still in a small fishing boat (40 kg) when he decides to go for a swim. Marty
jumps off of the boat, which causes the boat to have a velocity of 1 m/s. What is the velocity of Marty's
jump?
Answer:
the velocity of the Marty's Jump is 5 m/s.
Explanation:
Given;
mass of Marty, m₁ = 8 kg
mass of the boat, m₂ = 40 kg
velocity of the boat, u₂ = 1 m/s
Let the velocity of the Marty's Jump = u₁
Apply the principle of conservation of momentum;
m₁u₁ = m₂u₂
u₁ = m₂u₂/m₁
u₁ = (40 x 1) / (8)
u₁ = 5 m/s.
Therefore, the velocity of the Marty's Jump is 5 m/s.
10. What patterns do you notice between the angles of incidence and the
angles of the reflected light rays?
1
EXPLORE/EXPLAIN Lesson 2 Reflection and Mir
Answer:
The law of reflection states that the angle of reflection equals the angle of incidence—θr = θi. The angles are measured relative to the perpendicular to the surface at the point where the ray strikes the surface.
miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.500 m/s. What is the bullet's speed
Answer:
12 m/s
Explanation:
From the question,
Applying the law of conservation of momentum,
total momentum before collision = Total momentum after collision
mu+Mu' = mv+Mv'........................... Equation 1
Where m = mass of the bullet, u = initial velocity of the bullet, M = combined mass of the gun and the puck, u' = initial velocity of the gun and the puck, v = final velocity of the bullet, v' = final velocity of the gun and the puck
make v the subeject of the equation
v = [(mu+Mu')-Mv']/m................. Equation 2
Given: m = 5.00 g = 0.005 kg, M = 120 g = 0.12 kg, u = u' = 0 m/s (at rest), v' = 0.5 m/s
Substitute these values into equation 2
v = [0-(0.12×0.5)]/0.005
v = -0.06/0.005
v = -12 m/s
The negative sign can be ignored since we are looking for the speed, which has only magnitude.
Hence the speed of the bullet is 12 m/s
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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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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When a 2kg metal is exposed to 2MJ of
heat it dissolves into liquid at 500K
Calculate the latent heat of fusion for
the metal. Betty went in to get you metal is exposed to to megajoules of heat
Answer:
1×10⁶ J/Kg
Explanation:
From the question given above, the following data were obtained:
Mass (m) of metal = 2 Kg
Heat (Q) = 2 MJ
Latent heat of Fusion (Hf) =?
Next, we shall convert 2 MJ to J. This can be obtained as follow:
1 MJ = 1×10⁶ J
Therefore,
2 MJ = 2 MJ × 1×10⁶ J / 1 MJ
2 MJ = 2×10⁶ J
Finally, we shall determine the latent heat of fusion. This can be obtained as follow:
Mass (m) of metal = 2 Kg
Heat (Q) = 2×10⁶ J
Latent heat of Fusion (Hf) =?
Q = m•Hf
2×10⁶ = 2 × Hf
Divide both side by 2
Hf = 2×10⁶ / 2
Hf = 1×10⁶ J/Kg
Thus, the latent heat of fusion for the metal is 1×10⁶ J/Kg
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
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