The electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
What is electric field?An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.
Thus, for the charged balloon and a charged piece of paper 0.04 m away, we can conclude that, the electric field is all around the balloon and paper including between them, where, depending on their charge, there would be a force of either attraction or repulsion.
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An electron with an initial speed of 700,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the electron? What was the initial kinetic energy of the electron, in electron volts?
Answer:
See below.
Explanation:
According to the question, we know that,
work done is given by, [tex]W=qV[/tex]
and change in kinetic energy is, Δ [tex]KE=W=1=1/2[mv^{2} ][/tex]
therefore equating both the equations we get,
[tex]qV=1/2[mv^{2} ][/tex] ⇒ [tex]V=\frac{mv^{2} }{2q}[/tex]
m= mass of electron = [tex]9.1*10^{-31} kg[/tex]
q= charge on an electron = [tex]1.6*10^{-19} C[/tex]
v= speed of electron= 700000m/s
substituting the values in the above equation, we get
[tex]V=\frac{9.1*10^{-31} *(700000)^{2} }{2*1.6*10^{-19} } =1.39V[/tex]
(1). the potential difference that stopped the electron is 1.39 volts.
now the kinetic energy equation is : 2 ways[tex]KE=1/2[mv^{2} ]=\frac{9.1*10^{-31} *700000^{2} }{2} =2.22*10^{-19} J\\[/tex]
or [tex]KE=\frac{2.22*10^{-19} }{1.6*10^{-19} } =1.39eV[/tex]
(2). the initial kinetic energy of the electron is 1.39eV.
A small block, with a mass of 250 g, starts from rest at the top of the apparatus shown above. It then slides without friction down the incline, around the loop, and then onto the final level section on the right. The maximum height of the incline is 80 cm, and the radius of the loop is 15 cm.
a.) Find the initial energy of the block.
b.) Find the velocity of the block at the bottom of the loop.
c.) Find the velocity of the block at the top of the loop.
(a) The initial energy of the block due to its position is 1.96 J.
(b) The velocity of the block at the bottom of the loop is 3.96 m/s.
(c) the velocity of the block at the top of the loop is 3.13 m/s.
Initial energy of the blockThe initial energy of the block due to its position is calculated as follows;
P.E = mgh
P.E = 0.25 X 9.8 X 0.8
P.E = 1.96 J
Conversation of the energyThe velocity of the block at the bottom of the loop is determined by applying the principle of conservation of energy as shown below;
P.Ei + P.Ef = K.Ei + K.Ef
1.96 + 0 = 0 + ¹/₂mvf²
vf² = 2(1.96)/m
vf² = (2 x 1.96) / (0.25)
vf² = 15.68
vf = √15.68
vf = 3.96 m/s
Velocity of the block at top of the loopThe velocity of the block at the top is calculated by applying principle of conservation of energy,
P.Ei + P.Ef = K.Ei + K.Ef
1.96 = mghf + ¹/₂mvf²
where;
hf is the position of the ball at the top of the loop = 2r = 2 x 15 cm = 30 cm = 0.31.96 = 0.25 x 9.8 x 0.3 + 0.5 x 0.25vf²
1.225 = 0.125vf²
vf² = 1.225/0.125
vf² = 9.8
vf = 3.13 m/s
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Which planet is the farthest?
how r u
________________.
Which of the following particles is similar to a He nucleus?
alpha
beta
gamma
neutrino
Alpha
I hope this helps you
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A bucket of mass m is attached to a rope that is wound around the outside of a solid sphere (I = 2/5 M^2) of radius R. When the bucket is allowed to fall from rest, it falls with an acceleration of a down. What is the mass of the sphere in terms of m, R, a, and g?
Answer:
[tex]\displaystyle \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{M^{2}\, a}}[/tex], assuming that the tension in the rope is the only tangential force on the sphere ([tex]g[/tex] denote the gravitational acceleration.)
Explanation:
The forces on the bucket are:
Weight of the bucket: [tex]m\, g[/tex] (downward.)Tension in the rope (upward.)Since the weight of the bucket and the tension from the rope are in opposite directions, the magnitude of the net force would be:
[tex]\begin{aligned} \|\text{Net Force}\| =\; & \|\text{Weight}\| - \|\text{Tension}\| \end{aligned}[/tex].
The upward tension in the rope prevents the bucket from accelerating at [tex]g[/tex] (free fall.) Rather, the bucket is accelerating at an acceleration of only [tex]a[/tex]. The net force on the bucket would be thus [tex]m\, a[/tex].
Rearrange the equation for the net force on the bucket to find the magnitude of the tension in the rope would be:
[tex]\begin{aligned} & \|\text{Tension}\| \\ =\; & \|\text{Weight}\| - \|\text{Net Force}\| \\ =\; & m\, g - m\, a \\ =\; & (g - a)\, m\end{aligned}[/tex].
At a distance of [tex]R[/tex] from the center of the sphere, the tension in the rope [tex](g - a)\, m[/tex] would exert a torque of [tex](g - a)\, m\, R[/tex] on the sphere. If this tension is the only tangential force on this sphere, the net torque on the sphere would be [tex](g - a)\, m\, R\![/tex].
Let [tex]M[/tex] denote the mass of this sphere. The moment of inertia of this filled sphere would be [tex]I = (2/5)\, M^{2}[/tex].
Therefore, the magnitude of the angular acceleration of this sphere would be:
[tex]\begin{aligned}& \|\text{Angular Acceleration}\| \\ =\; & \frac{\|\text{Net Torque}\|}{(\text{Moment of Inertia})} \\ =\; & \frac{(g - a)\, m\, R}{(2/5)\, M^{2}} \end{aligned}[/tex].
The bucket is accelerating at a magnutide of [tex]a[/tex] downwards. The rope around the sphere need to unroll at an acceleration of the same magnitude, [tex]a\![/tex]. The tangential acceleration of the sphere at the surface would also need to be [tex]\! a[/tex].
Since the surface of the sphere is at a distance of [tex]R[/tex] from the center, the angular acceleration of this sphere would be [tex](a / R)[/tex].
Hence the equation:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \|\text{Angular Acceleration}\| = \frac{a}{R} \end{aligned}[/tex].
Solve this equation for [tex]M[/tex], the mass of this sphere:
[tex]\begin{aligned}& \frac{(g - a)\, m\, R^{2}}{(2/5)\, M^{2}} = \frac{a}{R} \end{aligned}[/tex].
[tex]\begin{aligned}M^{2} &= \frac{(g - a)\, m\, R^{2}}{(2/5)\, a} \\ &= \frac{(5/2)\, (g - a)\, m\, R^{2}}{a}\end{aligned}[/tex].
[tex]\begin{aligned}M&= \sqrt{\frac{(5/2)\, (g - a)\, m\, R^{2}}{a}}\end{aligned}[/tex].
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. Which of the following statements correctly describes the relationship between m1 and m2 and provides evidence from the graphs?
Answer:
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1 describes the displacement
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2
M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
What is the graph represents?The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system. For the system consisting of the two blocks, the pulley and the Earth, the change in the total mechanical energy of the system is zero.
The two graphs shown represent the motion of two blocks with different masses, m1 and m2. The blocks are oscillating on identical springs. For the system consisting of the two blocks, the change in the kinetic energy of the system is equal to work done by gravity on the system.
Therefore, M1 would seem to be slower because of a larger mass
x1 = A1 sin ω1 t1
ω1 / ω2 = ((k1 / k2) / (m1 / m2))^1/2 = (m2 / m1)^1/2 since k's are equal
ω1 / ω2 = 1/2 from graph (frequency of 2 is greater)
(m1 / m2)^1/2 = ω2 / ω1 from above
m1 / m2 = 2^2 = 4 so m1 would have 4 times the mass of m2.
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Which of the following is an example of the characteristic of excretion?
A) We shiver when we get cold.
B) Moss on the side of the tree is active even though it looks still.
C) Human kidneys produce urine.
D) A rabbit gets nutrients from a carrot.
Answer:
C
because urine is waste product
What type of heat transfer occurs in your stomach when you eat hot soup and an ice cold beverage
The type of heat transfer occurs in your stomach when you eat hot soup and an ice cold beverage is Conduction.
What is Conduction?This is the process by which heat energy is transmitted through collisions between neighboring atoms or molecules.
This happens when they are in close contact with each other which was why Conduction was chosen as the most appropriate choice.
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A 7 kg ball of clay traveling at 12 m/s collides with a 25 kg ball of clay traveling in the
same direction at 6 m/s. What is their combined speed if the two balls stick together
when they touch?
Answer:
Given:
m1 = 7 kg
V1 = 12 m/s
m2 = 25 kg
V2 = 6 m/s
To find:
Combined speed of two balls stick together after collision V = ?
Solution:
According to law of conservation of momentum,
m1V1 + m2V2 = (m1+m2)V
7×12 + 25×6 = (7+25)V
84 + 150 = 32V
V = 234/32
V = 7.31 m/s
Combined speed of two ball is 7.31 m/s
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How much momentum, in the x-direction, was transferred to the more massive cart, in kilogram meters per second
The momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.
Momentum transfered to the more massive cartThe momentum transfered to the more massive cart is determined by applying the principle of conservation of linear momentum as shown below;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
m₁ is the mass of the smaller cartu₁ is the initial velocity of the samller cartm₂ is the mass of the bigger cart = 3m₁u₂ is the initial velocity of the bigger cartv₁ is the final velocity of the smaller cartv₂ is the final veocity of the bigger cart⁻ΔP₁ = ΔP₂
ΔP₂ = m₂v₂ - m₂u₂
ΔP₂ = m₂(v₂ - u₂)
ΔP₂ = 3m₁(v₂ - u₂)
ΔP₂ = 3 x 3.8 x (1.7 - 0)
ΔP₂ = 19.38 kgm/s
Thus, the momentum, in the x-direction, that was transferred to the more massive cart after the collision is 19.38 kgm/s.
The complete question is beblow
A cart of mass 3.8 kg is traveling to the right (which we will take to be the positive x-direction for this problem) at a speed of 6.9 m/s. It collides with a stationary cart that is three times as massive. After the collision, the more massive cart is moving at a speed of 1.7 m/s, to the right.
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What is the force of the drag for a 65 kg bicyclist, initially at rest at the top of a hill coasts down the hill, reaching a speed of 15.5 m/s at the bottom of the hill. The distance is 60M. neglect any friction impeding the motion and the rotational energy of the wheels.
Height is 19M
Intial GPE is 12350J
KE is 7808J and loss is 4542 J
Read the text below. Each sentence is about one, two or no energy at all. (5 points) Name the type (s) of energy for each sentence, or leave the space blank (if in the sentence no energy is mentioned). Artan decided to paint the house. He moved the furniture, climbed the stairs, and began work. After two hours he took a break, ate lunch and turned on the radio to listen to some music. When done, turn on a heater to allow the paint to dry as quickly as possible. At dinner everything had ended. a) ............................................................................................................................................ b) ............................................................................................................................................ c) ............................................................................................................................................ d) ............................................................................................................................................ e) ............................................................................................................................................
What are the three symbols used in Ohm's law. Explain what each symbol represents and give the units for each of the variables.
Answer:
Step by step explanation:
the radius of a ball is increasing at a rate of 2 mm per second. how fast is the volume of the ball increasing when the diameter is 40 mm
Step 1: Define an equation that relates the volume of a sphere to its radius.
V = 4/3*π*r3
Step 2: Take the derivative of each side with respect to time (we will define time as "t").
(d/dt)V = (d/dt)(4/3*π*r3)
dV/dt = 4πr2*dr/dt
Step 3: We are told in the problem statement that diameter is 100m, so therefore r = 50mm. We are also told the radius of the sphere is increasing at a rate of 2mm/s, so therefore dr/dt = 2mm/s. We are looking for how fast the volume of the sphere is increasing, or dV/dt.
dV/dt = 4π(50mm)2*(2mm/s)
dV/dt = 62,832 mm3/s
You see a car that appears very small, so you assume that it must be far from you. You are using the monocular cue of
The monocular cue of relative size
A 1980-kg car is traveling with a speed of 15.5 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 39.2 m
Answer: 6067.5 N
Explanation:
Work = Change in Energy. To start, all of the energy is kinetic energy, so find the total KE using: KE = 1/2(m)(v^2). Plug in 1980 kg for m and 15.5 m/s for v and get KE = 237847.5 J.
Now, plug this in for work: Work = Force * Distance; so, divide work by distance to get 6067.5 N.
When resting, a person generates about 412005 joules of heat from the body. The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C. If the heat from the person goes only into the water, find the water temperature.
If a person generates about 412005 joules of heat from the body, the water temperature is mathematically given as
t=21.6296C
What is the water temperature.?Question Parameter(s):
The person is submerged neck-deep into a tub containing 2124 kg of water at 20.9 °C
Generally, the equation for the Heat is mathematically given as
Heat gained =Heat loess
Thereofore
mw*cw*(t-2160)=1.5*10^5
[tex]t=21.60+\frac{1.5*10^5}{mw*Cw}\\\\t=21.60+\frac{1.5*10^5}{1.2*10^3*4186}[/tex]
t=21.6296C
In conclusion, the tempreature
t=21.6296C
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Someone please help me !!
Answer:25
Explanation: because higher means less kinetic energ
If the range of a projectile's trajectory is six times larger than the height of the trajectory, then what was the angle of launch with respect to the horizontal? (Assume a flat and horizontal landscape.)
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
a 1. You found that the MCB was tied with a thread and the thread was fixed with a nail on the wall in your friend's house. i. Is it good idea to do this? ii. What could be the possible hazard of this? iii. What should have done to keep the circuit safe?
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
What is a miniature circuit breaker?A miniature circuit breaker is a circuit breaker that is used in homes as a means of guarding against damage to appliances due to a very high current.
This miniature circuit breaker is also harzardous in the sense that it could lead to an electrical fault related fire outbreak especially when it is being blown freely by wind as you tie it with a thread. Doing this a very bad idea because of the risk of a fire hazard.
The miniature circuit breaker should rather be fastned to a wall using nails and other neccessary tools.
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explain how a deflection magnetometer can be used to find the horizontal component of the Earth's magnetic field
A source of light emits photons with a wavelength of 8.1 x 10-8 meters. What is the frequency of this light
Answer:
Explanation:
Speed of light v = 3 x 10⁸ m/s
wavelength λ = 8.1 x 10⁻⁸ m
frquency f = v/λ = 3.7 x 10¹⁵ Hz
If a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then the frequency of the light would be 3.7 × 10¹⁵ Hz, as the wavelength and the frequency of the photon are inversely proportional to each other.
What is Wavelength?It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.
C = λν
As given in the problem if a source of light emits photons with a wavelength of 8.1 x 10⁻⁸ meters, then we have to find out the frequency of the light,
The frequency of the light = 3 × 10⁸ / 8.1 x 10⁻⁸
=3.7 × 10¹⁵ Hz
Thus, the frequency of the light would be 3.7 × 10¹⁵ Hz
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How much force is required to pull a spring 3.0 cm from its equilibrium position if the spring constant is 20 N/m?
[tex]\\ \rm\rightarrowtail F=-kx[/tex]
[tex]\\ \rm\rightarrowtail F=-20(0.03)[/tex]
[tex]\\ \rm\rightarrowtail F=-0.6N[/tex]
5 waves with a length of 4m hit the shore every 2 seconds, what is the frequency?
The frequency of the 5 waves with a length of 4m hit the shore every 2 seconds is 2.5 Hz.
What is frequency?This is the number of cycles completed by a wave in one second. The s.i
unit of frequency is Hert (Hz).
From the question, to calculate the frequency of 5 waves with length of 4 m that hit the shores every 2 seconds, we use the formula below.
Formula:
F = n/t........... Equation 1Where:
n = Number of waveF = Frequencyt = timeFrom the question,
Given:
n = 5 waves t = 2 secondsSubstitute these values into equation 1
F = 5/2F = 2.5 Hz.Hence, The frequency of the wave is 2.5 Hz.
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A car of mass 1000 kg moves 3 km east in a straight line and then 4 km north. What is the total distance and displacement of the car from the initial position?
The net (resultant) force on the car is
Select one:
a) distance = 7 km and displacement = 5 km
b) distance = 5 km and displacement =7 km.
c) distance = 25 km and displacement =7 km.
d) distance = 7 km and displacement = 25 km
Answer:
a
Explanation:
Distance is simply the distance travelled which in this case would be 4km + 3km = 7km
To work out displacement, try to imagine the situation.
Draw a straight line to the east (label it 3) and then draw another line from the end of the first line upwards (label this one 4). Thus, you've created a right angles triangle. Now use pythagorean theorem to work out the displacement
4^2 + 3^2 = 25
sqrt 25 = 5 = displacement
Which statement best describes the circular flow model?
Alex (31kg) and Cassie (19Kg) sit on a 10kg metre-long see-saw at the local park. The pivot of the see-saw is in the middle of its length. If Cassie sits at one end of the see-saw, where relative to the other end must Alex sit so the net torque is balanced? (unit:metres)
Answer:
M1 g L1 = 19 kg * 9.8 m/s^2 * 5 m = counter clockwise torque - Cassie at left end
M1 g L1 = M2 g L2 for torques to balance
L2 = M1 L1 / M2 = 19 * 5 / 31 = 3.06 M
Alex should sit at 3.1 m from the fulcrum (at 5 m from each end)
a 2.99 kg sphere makes a perfectly inelastic collision with a second sphere that is intially at rest. the composite moves with a speed equal to one third the original speed of the 2.99kg. what is the mass of the second sphere?
Answer:
5.98 kg
Explanation:
To solve this problem, let use the Linear Momentum Conservation Law:
Before collision: [tex]\sum p=p_{1}+p_{2}=m_{1}v_{1}+m_{2}v_{2}=(2.99v_{1})+0=2.99v_{1}[/tex]
After collision: [tex]\sum p'=p_{1}'+p_{2}'=(2.99+m_{2})(v_{1}/3)[/tex]
So, we obtain:
[tex]\sum p =\sum p' \rightarrow 2.99v_{1}=(2.99+m_{2})(v_{1}/3) \rightarrow 8.97 = 2.99 + m_{2}[/tex]
[tex]m_{2}=8.97-2.99=5.98 kg[/tex]
A magnet gets demagnetized when it is heated.
Answer:
The delicate balance between temperature and magnetic domains is destabilized when a magnet is subjected to high temperatures. If a magnet is exposed to this temperature for an extended length of time or heated over its Curie temperature, it will lose its magnetism and become irreversibly demagnetized.
Explanation: