Answer:
6m/s
Explanation:
Using the law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the momentum after collision.
Using the expression
m1u1 + m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and u2 are the initial velocities
v is the final velocity after collision
Substitute the given values in the formula
0.35(10)+0.35(2) = (0.35+0.35)v
3.5+0.7 = 0.7v
4.2 = 0.7v
v = 4.2/0.7
v = 6m/s
Hence the magnitude of the velocity for Ball B after the collision is 6m/s
A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a ramp of 64.8 degrees at a speed of 25.4 m/s. What would be the largest number of buses he can clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 10.0 m long
Answer: he can only make it over 5 buses
Explanation:
Given the data in the question;
we know that range is expressed as;
R = (V₀²sin2∅₀)/g
V₀ is the initial velocity( 25.4 m/s), ∅₀ is the angle of projection( 64.8°), g is acceleration due to gravity( 9.8 m/s²),
so we substitute
R = ((25.4)²sin2(64.8))/9.8
R = 50.7 m
now, them number of buses will be;
n = R / bus length
given that bus length is 10.0 m
we substitute
n = 50.7 m / 10.0
n = 5.07 ≈ 5
Therefore, he can only make it over 5 buses
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a proton and an electron are situated 865 nm from each other and you study the forces that the particles exert on each other. As expected, the predictions of Coulomb's law are well confirmed. You find that the forces are attractive and the magnitude of each force is:______
Answer:
force F = 1.66 × [tex]10^{-13}[/tex] N
Explanation:
given data
proton and an electron = 865 nm
solution
we get here force that is express as
force F = k q1 q2 ÷ r² ......................1
put here value and we get
force F = 9 × [tex]10^{9}[/tex] × [tex]\frac{1.6\times (10^{-19})^{2}}{865 \times (10^{-9})^{2}}[/tex]
force F = 1.66 × [tex]10^{-13}[/tex] N
Required
Momentum
The magnitude of the momentum of an object is 64 kg*m/s. If the velocity of the
object is doubled, what will be the magnitude of the momentum of the object? *
32 kg*m/s
64 kg*m/s
128 kg*m/s
256 kg*m/s
Answer:
C) 128 kg*m/s
Explanation:
When you double something you multiply it by 2 most of the time. 64*2=128 or you can add it 64+64=128. Hope this helps.
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
Why does it rain more in West Ferris than in East Ferris? Explain your answer.
Answer:
This idea helps students explain why more rain forms over West Ferris than East Ferris. ... Therefore, when students explain that water vapor condenses higher in the atmosphere, they are actually explaining that water vapor condenses high in the troposphere, which is relatively low in the atmosphere.
Explanation:
Plz mark me brainliest thank u> have a good day
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 92.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 5.97 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 1.14 s. Determine the force constant of the springs in N/m.
Answer:
k = 1400.4 N / m
Explanation:
When the springs are oscillating a simple harmonic motion is created where the angular velocity is
w² = k / m
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
where angular velocity, frequency and period are related
w = 2π f = 2π / T
we substitute
2π / T = \sqrt{ \frac{k}{m} }
T² = 4π² [tex]\frac{m}{k}[/tex]
k = π² [tex]\frac{m}{T^{2} }[/tex]
in this case the period is T = 1.14s, the combined mass of the children is
m = 92.2 kg and the constant of the two springs is
k = 4π² 92.2 / 1.14²
k = 2800.8 N / m
to find the constant of each spring let's use the equilibrium condition
F₁ + F₂ - W = 0
k x + k x = W
indicate that the compression of the two springs is the same, so we could replace these subtraction by another with an equivalent cosecant
(k + k) x = W
2k x = W
k_eq = 2k
k = k_eq / 2
k = 2800.8 / 2
k = 1400.4 N / m
What is average acceleration due to gravity on Earth for a 2000 kg boulder, in proper SI units?
Answer:
9.8m/s²
Explanation:
The average acceleration due to gravity on Earth for a 2000kg boulder is 9.8m/s².
Every object on earth is accelerated towards the center by a rate of change of velocity with time value of 9.8m/s².
The acceleration due to gravity on earth is a constant value from places to places.
For other planetary bodies, the value varies and it differs.
But on earth every object is accelerated at 9.8m/s².
Surface currents are on the
of the Earth's oceans
Select the Moon and use the Info view to determine which of the following statements is correct. The Last Quarter Moon.... rises near noon and sets near midnight. rises at about 6am and sets at about 6pm. rises near midnight and sets near midday. rises and sets at the same time as the Sun. Submit Your Answer
Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).
Explanation:
Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.
There are 8 main types of the moon phases these includes:
--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.
--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.
--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.
--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.
--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.
--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker
--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.
--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.
Write the properties of Non Metals and the families containig non Metals.
Non-Malleable and Ductile: Non-metals are very brittle, and cannot be rolled into wires or pounded into sheets. Conduction: They are poor conductors of heat and electricity. Luster: These have no metallic luster and do not reflect light.
Group 15, the nitrogen family, contains two nonmetals: nitrogen and phosphorus. These non-metals usually gain or share three electrons when reacting with atoms of other elements. Group 16, the oxygen family, contains three nonmetals: oxygen, sulfur, and selenium.
Elements: Nitrogen; Oxygen; Phosphorus; Selenium...
True or False. The larger a waves wavelength, the more energy it carries. (1 Point) True O False O Maybe
Explanation:
False, it is oppisite the shorter the wavelength the more energy it carries.
* Psychology
Match the types of psychoactive drugs to their functions,
depressants
stimulants
amphetamines
hallucinogens
to excite neural activity and temporarily
elevate awareness
to increase dopamine activity and produce
schizophrenic-like paranoid symptoms
>
to inhibit the function of the central nervous
system and neural activity
to distort perceptions and effects on thinking
Answer:
See explanation below
Explanation:
Psychoactive drugs are drugs that affect the central nervous system. They alter cognitive function by changing mood and consciousness.
Examples;
Depressants: Inhibit the function of the central nervous system and neural activity.
Stimulants: Excite neural activity and temporarily elevate awareness.
Amphetamines: Increase dopamine activity and produce schizophrenic-like symptoms.
Hallucinogens: Distort perceptions and effects on thinking.
A drug is any substance that alters how the body functions.
What is a drug?A drug is any substance that alters how the body functions. There are different types of drugs that affect different parts of the body.
We shall now explain the following classifications of drugs;
depressants - to inhibit the function of the central nervousstimulants - elevate awarenesshallucinogens - to distort perceptions and effects on thinkingamphetamines - schizophrenic-like paranoid symptomsLearn more about drugs: https://brainly.com/question/6022349
Fluids
A = 2804 cm3 B = 2862 cm2 C = 2916 cm3
Three separate fluids, A, B, and C have been selected at random and each initially fills a 3000 cm3 volume at atmospheric pressure. A gage pressure of 6 x 107 N/m2 is then applied to each fluid. The final volume is given below. Determine which fluids were selected from the given list.
Acetone E = 0.92 GPa Glycerin E = 4.35 GP
Water E = 2.15 GPa Mercury E = 28.5 GPa
Benzene E = 1.05 GPa Sulfuric Acid E = 3.0 GPa
Ethyl Alcohol E = 1.06 GPa Gasoline E = 1.3 GPa
Petrol E = 1.45 GPa Seawater E = 2.34 GPa
Answer:
Explanation:
Fluid A :
Δ V = Change in volume = (3000 - 2804) x 10⁻⁶ m³ = 196 x 10⁻⁶ m³
volume strain = Δ V / V = 196 x 10⁻⁶ / 3000 x 10⁻⁶
= .06533
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .06533 = 91.84 x 10⁷ Pa = .92 GPa .
It is Acetone .
Fluid B :
Δ V = Change in volume = (3000 - 2862) x 10⁻⁶ m³ = 138 x 10⁻⁶ m³
volume strain = Δ V / V = 138 x 10⁻⁶ / 3000 x 10⁻⁶
= .046
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .046 = 130.43 x 10⁷ Pa = 1.3 GPa .
It is Gasoline .
Fluid C :
Δ V = Change in volume = (3000 - 2916) x 10⁻⁶ m³ = 84 x 10⁻⁶ m³
volume strain = Δ V / V = 84 x 10⁻⁶ / 3000 x 10⁻⁶
= .028
Δ P = increase in pressure = 6 x 10⁷ Pa
E = Δ P / volume strain = 6 x 10⁷ / .028 = 214.28 x 10⁷ Pa = 2.14 GPa .
It is Water .
What do you think about the attached scenario?
during a typical afternoon thunderstorm in the summer, an area of 66.0 km2 receives 9.57 108 gal of rain in 18 min. how many inches of rain fell during this 18 min period
Answer:
2.16 inch
Explanation:
area under water = 66 km²
= 66 x ( 3280.84 x 12 )² inch²
= 1.023 x 10¹¹ sq inch
volume of rain = 9.57 x 10⁸ gallon = 9.57 x 10⁸ x 231 inch³
= 2.21 x 10¹¹ inch³
If depth of rainfall be t
volume of rain = surface area x depth
= 1.023 x 10¹¹ x t
So ,
1.023 x 10¹¹ x t = 2.21 x 10¹¹
t = 2.16 inch
How do you calculate area when pressure and force are given to you
Answer:
This is my answer
Explanation:
First convert 150 kPa to Pa:
150 × 1,000 = 150,000.
Next substitute the values into the equation:
force normal to a surface area = pressure × area of that surface.
force = 150,000 × 180.
force = 27,000,000 N.
1.First convert 150kPato Pa:
2.150 x 1,000 + 150,000
3.next substitute the values into the equations:
4.force normal to a surface area =pressure x area of that surface.
5.force=150,000 x 180.
6.force = 27,000,000N.
can i have brainliest please
What energy store is in the torch
BEFORE it gets switched on?
Answer:
Chemical energy
Explanation:
The energy in the torch is stored as chemical energy before the torch gets switch on.
The chemical energy energy in the battery of cell will power the cell and allows it to produce light.
Chemical energy is a form of potential energy. The electrolytes within the battery are capable of producing electric current. So the chemical energy is transformed into electrical energy which is used to produce the light of the torch.A torsional pendulum is formed by attaching a wire to the center of a meter stick with a mass of 5.00 kg. If the resulting period is 4.00 min, what is the torsion constant for the wire
Answer:
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
Explanation:
The angular frequency of the torsional pendulum ([tex]\omega[/tex]), measured in radians per second, is defined by the following expression:
[tex]\omega = \sqrt{\frac{\kappa}{I} }[/tex] (1)
Where:
[tex]\kappa[/tex] - Torsional constant, measured in newton-meters.
[tex]I[/tex] - Moment of inertia, measured in kilogram-square meters.
The angular frequency and the moment of inertia are represented by the following formulas:
[tex]\omega = \frac{2\pi}{T}[/tex] (2)
[tex]I = \frac{m\cdot L^{2}}{12}[/tex] (3)
Where:
[tex]T[/tex] - Period, measured in seconds.
[tex]m[/tex] - Mass of the stick, measured in kilograms.
[tex]L[/tex] - Length of the stick, measured in meters.
By (2) and (3), (1) is now expanded:
[tex]\frac{2\pi}{T} = \sqrt{\frac{12\cdot \kappa}{m\cdot L^{2}} }[/tex]
[tex]\frac{2\pi}{T} = \frac{2}{L}\cdot \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi\cdot L}{T} = \sqrt{\frac{3\cdot \kappa}{m} }[/tex]
[tex]\frac{\pi^{2}\cdot L^{2}}{T^{2}} = \frac{3\cdot \kappa}{m}[/tex]
[tex]\kappa = \frac{\pi^{2}\cdot m\cdot L^{2}}{3\cdot T^{2}}[/tex]
If we know that [tex]m = 5\,kg[/tex], [tex]L = 1\,m[/tex] and [tex]T = 240\,s[/tex], then the torsion constant for the wire is:
[tex]\kappa = \frac{\pi^{2}\cdot (5\,kg)\cdot (1\,m)^{2}}{3\cdot (240\,s)^{2}}[/tex]
[tex]\kappa = 2.856\times 10^{-4}\,N\cdot m[/tex]
The torsion constant for the wire is [tex]2.856\times 10^{-4}\,N\cdot m[/tex].
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a driving speed of 50 mi/h. When the driver is sober, a stop can be made just in time to avoid hitting an object that is first visible 385 ft ahead. After a few drinks under exactly the same conditions, the driver fails to stop in time and strikes the object at a speed of 30 mi/h. Determine the driver's perception/reaction time before and after drinking. (Assume practical stopping distance.)
Answer:
a. 10.5 s b. 6.6 s
Explanation:
a. The driver's perception/reaction time before drinking.
To find the driver's perception time before drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)
a = - 499.52 m²/s²/234.7 m
a = -2.13 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (0 m/s - 22.35 m/s)/-2.13 m/s²
t = - 22.35 m/s/-2.13 m/s²
t = 10.5 s
b. The driver's perception/reaction time after drinking.
To find the driver's perception time after drinking, we first find his deceleration from
v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m
So, a = v² - u²/2s
substituting the values of the variables into the equation, we have
a = v² - u²/2s
a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)
a = 179.83 m²/s² - 499.52 m²/s²/234.7 m
a = -319.69 m²/s² ÷ 234.7 m
a = -1.36 m/s²
Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time
So, t = (v - u)/a
Substituting the values of the variables into the equation, we have
t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²
t = - 8.94 m/s/-1.36 m/s²
t = 6.6 s
Two 90.0-kilogram people are separated by 3.00 meters. What is the magnitude of the gravitational force that one person exerts on the other?
Answer:
the magnitude of gravitational force is 6 x 10⁻⁸ N.
Explanation:
Given;
mass of the two people, m₁ and m₂ = 90 kg
distance between them, r = 3.0 m
The magnitude of gravitational force exerted by one person on another is calculated as;
[tex]F = \frac{Gm_1m_2}{r^2} \\\\[/tex]
where;
G is gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²
[tex]F = \frac{Gm_1m_2}{r^2} \\\\F = \frac{6.67\times 10^{-11} \times \ 90 \ \times \ 90}{3^2} \\\\F = 6\times 10^{-8} \ N[/tex]
Therefore, the magnitude of gravitational force is 6 x 10⁻⁸ N.
Does changing the height of point C affect the speed of the coaster car at point D?
Without friction, NO.
The speed at D depends only on the difference in height between A and D. Whatever happens between them doesn't matter.
The speed of the coaster car at point D will be affected if the height of point C is changed.
Potencial Energy:
It is the enrgy in a body due to the position of differnt part of the object or system.
As we increase the the hight of the car the potetial enrgy increase, the gravitational acceleration on car will be more due to the high of the point C.
Therefore, the speed of the coaster car at point D will be affected if the height of point C is changed.
To know more about speed of the coaster car,
https://brainly.com/question/9178285
Under what conditions will a moving 0.030 kg marble and a moving 2.43 kg rock have the same kinetic energy
Answer:
To have the same kinetic energy the speed of the marble must be 9 times the speed of rock.
Explanation:
The general formula of kinetic energy is given as follows:
[tex]K.E = \frac{1}{2}mv^{2}[/tex]
where,
K.E = Kinetic Energy
m = mass of the object
v = speed of the object
So, for the marble and rock to have same kinetic energy, we can write:
[tex]K.E_{marble} = K.E_{rock}\\\\\frac{1}{2}m_{marble}v_{marble}^{2} = \frac{1}{2}m_{rock}v_{rock}^{2}\\\\(0.03\ kg)v_{marble}^{2} = (2.43\ kg)v_{rock}^{2}\\\\taking\ square\ root\ on\ both\ sides:\\v_{marble} = \sqrt{\frac{2.43\ kg}{0.03\ kg}}v_{rock}\\\\v_{marble} = 9\ v_{rock}[/tex]
Hence, to have the same kinetic energy the speed of the marble must be 9 times the speed of rock.
Please help. I'm stuck!
What is the mass of a catamaran moving at 7.65 m/s that has a momentum of 530145 kg x m/s?
Scientists believe that the boundary stratum between the Cretaceous and Paleogene was caused by an asteroid. What evidence is most consistent with this theory?
Answer:
Because there was fewer fossils
Explanation:
Answer:
Actually the answer is "The stratum contains iridium.".
Explanation:
Your family is moving, and you are asked
to help move some boxes. One box is so
heavy that you must push it across the
room rather than lift it. What are some
ways you could reduce friction to make
moving the box easier?
Answer:
Explanation:
I would use a hand truck dollies
What is displacement?
a. The distance an object travels.
b. The distance between the starting point and the ending point of an object's
journey.
C. The amount of time it takes an object to travel to a destination.
d. The path in which an object travels.
Answer:
displacement is the distance between the starting point and the ending point of an object's journey
12. A bag weighing 20 N CARRIED horizontally a distance of 35 m, How much
work is done on the bag in Joules? (Do not put units with your answer.) W=Fd *
Your answer
13. A child performs 10J of work in lifting a box 1 m in 2 seconds. How much
power did the child apply to the box? (Do not include units with your answer.)
P=W/t *
Your answer
Answer:
Explanation:
Well they told you the exact formula to use. Work is the force multiplied by the distance through which its applied.
W = (20N)(35m)
= 700 Joules
13.) Power is the amount of work done over the time through which the work is being done.
P = W/t
= 10J/2s
= 5J/s
Sam heaves a 16lb shot straight upward, giving it a constant upward acceleration from rest of 35 m/s^2 for 64.0 cm. He releases it 2.20m above the ground. You may ignore air resistance.
(a) What is the speed of the shot when Sam releases it?
(b) How high above the ground does it go?
(c ) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?
Answer:
6.69 m/s
4.483 m
1.42s
Explanation:
Given that:
Initial Velocity, u = 0
Final velocity, v =?
Acceleration, a = 35m/s²
1.) using the relation :
v² = u² + 2as
v² = 0 + 2(35) * 64*10^-2m
v² = 70 * 0.64
v = sqrt(44.8)
v = 6.693
v = 6.69 m/s
B.) height from the ground, h0 = 2.2
How high ball went , h:
Using :
v² = u² + 2as
Upward motion, g = - ve
0 = 6.69² + 2(-9.8)*(h - 2.2)
0= 6.69² - 19.6(h - 2.2)
44.7561 + 43.12 - 19.6h = 0
19.6h = 44.7561 - 43.12
h = 87.8761 / 19.6
h = 4.483 m
C.)
vt - 0.5gt² = h - h0
6.69t - 0.5(9.8)t²
6.69t - 4.9t² = 1.83 - 2.2
-4.9t² + 6.69t + 0.37 = 0
Using the quadratic equation solver :
Taking the positive root:
1.4185 = 1.42s
A circus tightrope walker weighing 800 N is standing in the middle of a 15 meter long cable stretched between two posts. The cable was originally horizontal. The lowest point of the cable is now at his feet and is 30 cm below the horizontal. Assume the cable is massless. What is the tension in the cable
Answer:
T = 10010 N
Explanation:
To solve this problem we must use the translational equilibrium relation, let's set a reference frame
X axis
Fₓ-Fₓ = 0
Fₓ = Fₓ
whereby the horizontal components of the tension in the cable cancel
Y Axis
[tex]F_{y} + F_{y} - W =0[/tex]
2[tex]F_{y}[/tex] = W
let's use trigonometry to find the angles
tan θ = y / x
θ = tan⁻¹ (0.30 / 0.50 L)
θ = tan⁻¹ (0.30 / 0.50 15)
θ = 2.29º
the components of stress are
F_{y} = T sin θ
we substitute
2 T sin θ = W
T = W / 2sin θ
T = [tex]\frac{ 800}{ 2sin 2.29}[/tex]
T = 10010 N
What must the charge (sign and magnitude) of a particle of mass 1.40 gg be for it to remain stationary when placed in a downward-directed electric field of magnitude 640 N/CN/C
Answer:
the charge of the particle is -2.144 x 10⁻⁵ C.
Explanation:
The force acting on the particle is calculated as;
F = EQ = mg
[tex]Q = \frac{mg}{E}[/tex]
where;
Q is magnitude of the charge of the particle
[tex]Q = \frac{(1.4\times 10^{-3})(9.8)}{640} \\\\Q = 2.144 \ \times \ 10^{-5} \ C[/tex]
since the magnetic field is acting downward, the force must be acting upward in opposite direction.
Thus, the charge of the particle will be -2.144 x 10⁻⁵ C.
