Answer:
A. Her total angular momentum has decreased
Explanation:
Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.
The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
I hope it helps you!
If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
In the photoelectric effect, if the intensity of light shone on a metal increases, what will happen?
Answer:
C) There will be more electrons ejected
Explanation:
The number of electrons ejected whenever a photoelectric effect is identified it is proportional to the intensity of the incident light
Nevertheless, the photoelectrons' maximal kinetic energy is independent of their light intensity
Therefore, the maximum speed of the electron ejected doesn't really depend on the light intensity
So, if the intensity rises, only the number of electrons ejected will rised
Therefore the option c is correct
Answer:
C) There will be more electrons ejected
Explanation:
In the photoelectric effect, photons with an energy of E are shone upon a piece of metal, and if the energy of the photons overcome the work function ϕ of the metal, then electrons with will be ejected from the metal with a kinetic energy KE.
E_photon = Φ + KE
Each photon is capable of ejecting one electron from the metal. Therefore, increasing the intensity of the light (the number of photons shone on the metal) will increase the number of electrons ejected from the metal.
A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.
Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
An 100 V/m electric field is directed along the x axis. If the potential at the origin is 300 V, what is potential at the point ( -2m, 0) point
Answer:
200volts
Explanation:
Pls see attached file
Answer:
100 V
Explanation:
Electric field E = 100 V/m
Potential at the origin = 300 V
Potential at point (-2m, 0) i.e 2 m behind the origin = ?
From the equation ΔV = EΔd,
ΔV = [tex]V_{0} - V_{x}[/tex]
where [tex]V_{0}[/tex] is the potential at origin,
and [tex]V_{x}[/tex] is the potential at point (-2, 0)
E = electric field
Δd = 0 - (-2) = 2 m
[tex]V_{0} - V_{x}[/tex] = 300 - [tex]x[/tex]
equating, we have
300 - [tex]x[/tex] = 100 x 2
300 - [tex]x[/tex] = 200
[tex]x[/tex] = 100 V
Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What is the mass of the string (in kg)?
Answer:
m = 0.0125 kg
Explanation:
Let us apply the formula for the speed of a wave on a string that is under tension:
[tex]v = \sqrt{\frac{F}{\mu} }[/tex]
where F = tension force
μ = mass per unit length
Mass per unit length is given as:
μ = m / l
where m = mass of the string
l = length of the string
This implies that:
[tex]v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }[/tex]
Let us make mass, m, the subject of the formula:
[tex]v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}[/tex]
From the question:
F = 20 N
l = 4.50 m
v = 85 m/s
Therefore:
[tex]m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg[/tex]
symbol of science hhshsjsiwtwwisjzhJava
Answer:
is this a company name.? java is a computer software right..
A friend throws a heavy ball toward you while you are standing on smooth ice. You can either catch the ball or deflect it back toward your friend. What should you do in order to maximize your speed right after your interaction with the ball?
A. You should catch the ball.
B. You should let the ball go past you without touching it.
C. You should deflect the ball back toward your friend.
D. More information is required to determine how to maximize your speed.
E. It doesn't matter. Your speed is the same regardless of what you do.
Answer:
C You should deflect the ball back toward your friend.
Explanation:
This is because it would result in a completely inelastic collision, and the final velocity of me would be found using,
with m= mass, V=velocity, i=initial, f=final:
mV(me,i) +mV(ball,i) = [m(me)+m(b)]V(f)
So V(f) would be just the momentum of the ball divided by just MV mass of the ball and it will be higher resulting in inelastic collision
Answer:
A. You should catch the ball.
Explanation:
Catching the ball maximizes your speed by converting most of the momentum of the flying ball into the momentum of you and the ball. Since the ice is smooth, the friction between your feet and the ice is almost negligible, meaning less energy is needed to set your body in motion. Catching the ball means that you and the ball undergoes an inelastic collision, and part of the kinetic energy of the ball is transferred to you, setting you in motion. Deflecting the ball will only give you a relatively small speed compared to catching the ball.
Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing through the aorta (in grams of blood per second)?
Answer:
94.248 g/sec
Explanation:
For solving the total current of the blood passing first we have to solve the cross sectional area which is given below:
[tex]A_1 = \pi R^2\\\\A_1 = \pi (1)^2\\\\A_1 = 3.1416 cm^2[/tex]
And, the velocity of blood pumping is 30 cm^2
Now apply the following formula to solve the total current
[tex]Q = \rho A_1V_1\\\\Q = (1)(3.1416)(30)\\\\[/tex]
Q = 94.248 g/sec
Basically we applied the above formula So, that the total current could come
Which of the following is not considered a behavior?
A. eating
B. anxiety
C. sleeping
D. crying
━━━━━━━☆☆━━━━━━━
▹ Answer
B. Anxiety
▹ Step-by-Step Explanation
Anxiety isn't a behavior since it's a feeling. Behavior and feeling are different things therefore, anxiety is the correct answer.
Hope this helps!
- CloutAnswers ❁
Brainliest is greatly appreciated!
━━━━━━━☆☆━━━━━━━
Eating, sleeping, and crying all are considered as behaviors. However, anxiety cannot be considered as a behavior because it is a feeling. Thus, the correct option is B.
What is Anxiety?Anxiety is an intense feeling of excessive, and persistent worry and the fear about everyday situations. This includes fast heart rate, rapid breathing, sweating, and feeling tired constantly may occur.
Behavior is the range of actions and mannerisms which are made by individuals, organisms, systems or the artificial entities in some environment. These systems can include other systems or organisms as well as the inanimate physical environment. Behaviors include eating, sleeping, and crying. Anxiety is not a behavior, it is a feeling.
Therefore, the correct option is B.
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In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
The angular speed is [tex]w = 5.89 \ rad/s[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 1.6 s[/tex]
The number of somersaults is n = 1.5
The total angular displacement during the somersault is mathematically represented as
[tex]\theta = n * 2 * \pi[/tex]
substituting values
[tex]\theta = 1.5 * 2 * 3.142[/tex]
[tex]\theta = 9.426 \ rad[/tex]
The angular speed is mathematically represented as
[tex]w = \frac{\theta }{t}[/tex]
substituting values
[tex]w = \frac{9.426}{1.6}[/tex]
[tex]w = 5.89 \ rad/s[/tex]
Two long, parallel wires carry currents in the same direction. If I1 = 10 A, and I2 = 20 A, and they are d = 1.0 m apart, what is the magnetic field at a point P midway between them?
Answer:
The magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T
Explanation:
Given;
current in the first wire, I₁ = 10 A
current in the second wire, I₂ = 20 A
distance between the two wires, d = 1.0 m
Magnetic field at mid point between two parallel wires is calculated as;
[tex]B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r}(I_1 +I_2)[/tex]
where;
r is the midpoint between the wires, = 0.5 m
μ₀ is the permeability of free space, = 4π x 10⁻⁷
[tex]B = \frac{\mu_o }{2\pi r}(I_1 +I_2)\\\\B = \frac{4\pi*10^{-7} }{2\pi *0.5}(10 +20)\\\\B = \frac{4\pi*10^{-7} *30}{2\pi *0.5}\\\\B = 1.2 *10^{-5} \ T[/tex]
Therefore, the magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T
Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The longer one has length 8L, diameter 4D, and resistance R2. How do the resistances of these two resistors compare
Answer:
the resistance of the longer one is twice as big as the resistance of the shorter one.
Explanation:
Given that :
For the shorter cylindrical resistor
Length = L
Diameter = D
Resistance = R1
For the longer cylindrical resistor
Length = 8L
Diameter = 4D
Resistance = R2
So;
We all know that the resistance of a given material can be determined by using the formula :
[tex]R = \dfrac{\rho L }{A}[/tex]
where;
A = πr²
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
For the shorter cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{ 4 \rho L }{\pi \ D ^2}[/tex]
For the longer cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{32\rho L }{\pi \ (4 D) ^2}[/tex]
[tex]R = \dfrac{2\rho L }{\pi \ (D) ^2}[/tex]
Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :
[tex]\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}[/tex]
[tex]\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}[/tex]
[tex]\dfrac{R_s}{R_L} =2[/tex]
[tex]{R_s}=2{R_L}[/tex]
Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.
Refer the attached photo
Answer:
A
Explanation:
since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet
You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answer:
D
Explanation:
The power equation is P= V^2/R
Please let me know if this helped! Please rate it the brainlist if possible!
As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3
Answer:
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
Explanation:
Density is the mass per unit volume
Density = mass/volume = m/V
Volume of a cylinder V = πr^2 h
Given;
Height h = 48.5mm = 0.0485 m
Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m
Substituting the values;
Volume V = π×(0.0245^2)×0.0485
V = 0.000091458438030 m^3
V = 0.000091458 m^3
The mass is given as;
Mass = 1 kg
So, the density can be calculated as;
Density = 1/0.000091458
Density = 10933.92825785 kg/m^3
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
A long straight metal rod has a radius of 2.0 mm and a surface charge of density 0.40 nC/m2. Determine the magnitude of the electric field 3.0 mm from the axis.
Answer:
Explanation:
Gauss Theorem
E2πrL=o2πRL/εo
then
E=oR/(rεo)
E=(0.4*10^-9*2*10^-3) / (3*10^-3*8.85*10^-12)
= 30.13 N/C
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
The complete question is;
In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.
Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?
Answer:
F_top = 385.36 N
Explanation:
We are given;
mass;m = 52 kg
Time;t = 4.3 s
Diameter;d = 16m
So,Radius;r = 16/2 = 8m
The formula for the centrifugal force is given as;
F_c = mω²R
Where;
R = radius
Angular velocity;ω = 2πf
f = frequency = 1/t = 1/4.3 Hz
F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.
The force at top would be;
F_top = F_c - mg
F_top = 905.29 - (9.81 × 53) N
F_top = 385.36 N
The force at the top of ride will be "385.36 N".
Force and mass:According to the question,
Rider's mass, m = 52 kg
Time, t = 4.3 s
Diameter, d = 16 m
Radius, r = [tex]\frac{16}{2}[/tex] = 8 m
Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz
We know the formula,
Centrifugal force, [tex]F_c[/tex] = mω²R
or,
Angular velocity, ω = 2πf
By substituting the values in the above formula,
[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]
[tex]= 905.29[/tex] N
hence,
The top force will be:
→ [tex]F_{top} = F_c[/tex] - mg
By substituting the values,
[tex]= 905.29-(9.81\times 53)[/tex]
[tex]= 385.36[/tex] N
Thus the above response is correct.
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(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it
Answer:
The minimum speed required is 2.62m/s
Explanation:
The value of gravitational acceleration = g = 9.81 m/s^2
Radius of the vertical circle = R = 0.7 m
Given the mass of the pail of water = m
The speed at the highest point of the circle = V
The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.
Below is the calculation:
[tex]\frac{mV^{2}}{R} = mg[/tex]
[tex]V = \sqrt{gR}[/tex]
[tex]V = \sqrt{9.81 \times 0.7}[/tex]
[tex]V = 2.62 m/s[/tex]
How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere
Answer:
1.007 × 10^(10) electron
Explanation:
We are given;
Electric Field;E = 1450 N/C
Diameter;d = 20 cm = 0.2 m
So, Radius: r = 0.2/2 = 0.1 m
Formula for Electric field just outside the sphere is given by the formula;
E = kq/r²
Where;
E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge
k is a constant with a value of 9 x 10^(9) N.m²/C²
Making q the subject, we have;
q = Er²/k
Thus,
q = 1450 × 0.1²/(9 × 10^(9))
q = 1.61 × 10^(-9) C
Now, total charge q is also given by the formula;
q = Ne
Where;
e is charge on electron which is 1.6 × 10^(-19)
N is number of excess electrons
Making N the formula, we have;
N = q/e
N = (1.61 × 10^(-9))/(1.6 × 10^(-19))
N = 1.007 × 10^(10) electron
A woman is standing at the rim of a nonuniform cylindrical horizontal platform initially at rest. The platform is free to rotate about frictionless orthogonal axle that goes through its center and has 4 m in diameter and moment of inertia of 500 kgm2. The woman then starts walking along the rim in clockwise direction at a constant speed of 1.50 m/s relative to the Earth. If a woman has 60 kg, how much work does she do to set herself and the platform into motion?
Answer:
e
Explanation:
i took it myself and got it right
supose at 20 degree celsius the resistance of Tungsten thermometer is 154.9. WHen placed in a particular solution , the resistance is 207.4 What is the temerature (in degree Celsius of this solution
Answer:
T₂ = 95.56°C
Explanation:
The final resistance of a material after being heated is given by the relation:
R' = R(1 + αΔT)
where,
R' = Final Resistance = 207.4 Ω
R = Initial Resistance = 154.9 Ω
α = Temperature Coefficient of Resistance of Tungsten = 0.0045 °C⁻¹
ΔT = Change in Temperature = ?
Therefore,
207.4 Ω = 154.9 Ω[1 + (0.0045°C⁻¹)ΔT]
207.4 Ω/154.9 Ω = 1 + (0.0045°C⁻¹)ΔT
1.34 - 1 = (0.0045°C⁻¹)ΔT
ΔT = 0.34/0.0045°C⁻¹
ΔT = 75.56°C
but,
ΔT = Final Temperature - Initial Temperature
ΔT = T₂ - T₁ = T₂ - 20°C
T₂ - 20°C = 75.56°C
T₂ = 75.56°C + 20°C
T₂ = 95.56°C
Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answer:
C is the correct answer
Explanation:
A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? Group of answer choices
Answer:
it attracts
Explanation:
since in a magnetic body there are two poles
(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole
A rod has length 0.900 mm and mass 0.500 kgkg and is pivoted at one end. The rod is not uniform; the center of mass of the rod is not at its center but is 0.500 mm from the pivot. The period of the rod's motion as a pendulum is 1.49 ss. What is the moment of inertia of the rod around the pivot
Answer:
The moment of inertia is [tex]I =0.14 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The length of the rod is [tex]l = 0.900 \ m[/tex]
The mass of the rod is [tex]m = 0.500 \ kg[/tex]
The distance of the center of mass from the pivot is [tex]d = 0.500 \ m[/tex]
The period of the rod's motion is [tex]T = 1.49 \ s[/tex]
Generally the period of the motion is mathematically represented as
[tex]T = 2 \pi * \sqrt{\frac{I}{m* g * d} }[/tex]
Where [tex]I[/tex] is the moment of inertia about the pivot so making [tex]I[/tex] the subject of formula
[tex]I = [\frac{T}{2\pi } ]^2 * m * g * d[/tex]
substituting values
[tex]I = [\frac{1.49}{2* 3.142 } ]^2 * 0.5 * 9.8 * 0.5[/tex]
[tex]I =0.14 \ kg \cdot m^2[/tex]
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)
Answer:
The temperature of silver at this given resistivity is 2971.1 ⁰C
Explanation:
The resistivity of silver is calculated as follows;
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]
where;
Rt is the resistivity of silver at the given temperature
Ro is the resistivity of silver at room temperature
α is the temperature coefficient of resistance
To is the room temperature
T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]
Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m
When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;
[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]
Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C
Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The objectstarts from rest and in 1.20 s experiences a displacement of (4.20 i - 3.30 j) m, where the direction of jis the upward vertical direction. Determine the other force.
Answer:
the other force= (16.3i + 14.6j)N
EXPLANATION:
Given:
Mass=2.80-kg
t= 1.2s
Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.
We will need to calculate the magnitude of the displacement using the below formula;
d = (1/2)at2 + v0t + d0
But note that
d0 = 0,( initial position)
v0 = 0( initial position)
a is the net acceleration
d = √[4.202 + (-3.30)2] m = 5.34 m
Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)
d = (1/2)at2
a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2
the net acceleration is 7.42 m/s2
Acceleration in terms of the vector can be calculated as
a=2(ri - r0)/t^2
Where t =1.2s which is the time
a= 2(4.2i - 3.30j)/ 1.2^2
a=( 5.83i - 4.58j)m/s
now the net force can now be calculated since we have known the value of acceleration, using the formula below;
F(x) = ma - mg
Where a = 5.83i - 4.58j)m/s and g= 9.8m/s
2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2
Therefore, the other force= (16.3i + 14.6j)N
Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?
Answer:
17.94 kN/C is the electric field intensity at the origin due to the charges.
Explanation:
From the question, we are told that
The distance of 1 μC from origin = 1 m
The distance of 2 μC from origin = 2 m
The distance of 3 μC from origin = 3 m
The distance of 4 μC from origin = 5 m
Therefore, for us to find the electric field intensity, we'll solve below:
The formula for Electric field intensity = ( k * q ) / ( r * r )
where , r is distance ,
k = 9 * 10^9 ,
and , q is charge .
now ,
electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]
=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C
=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C
=> electric field intensity at the origin = 17.94 kN/C
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It strikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± [tex]\frac{1}{2} at^2[/tex] ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± [tex]\frac{1}{2} at^2[/tex] ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
From the question;
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt - [tex]\frac{1}{2} at^2[/tex]
h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m
I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation
Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
where;
T is the period oscillation
m is mass of the spring
k is the spring constant
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]
Now, determine the amplitude of oscillation, A;
[tex]E = \frac{1}{2} kA^2[/tex]
where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation
[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]
Therefore, the amplitude of the oscillation is 2.82 cm