Answer:
the speed of the electron at the given position is 106.2 m/s
Explanation:
Given;
initial position of the electron, r = 9 cm = 0.09 m
final position of the electron, r₂ = 3 cm = 0.03 m
let the speed of the electron at the given position = v
The initial potential energy of the electron is calculated as;
[tex]U_i = Fr = \frac{kq^2}{r^2} \times r = \frac{kq^2}{r} \\\\U_i = \frac{(9\times 10^9)(1.602\times 10^{-19})^2}{0.09} \\\\U_i = 2.566 \times 10^{-27} \ J[/tex]
When the electron is 3 cm from the proton, the final potential energy of the electron is calculated as;
[tex]U_f = \frac{kq^2}{r_2} \\\\U_f = [\frac{(9\times 10^9)\times (1.602 \times 10^{-19})^2}{0.03} ]\\\\U_f = 7.669 \times 10^{-27} \ J \\\\\Delta U = U_f -U_i\\\\\Delta U = (7.699\times 10^{-27} \ J ) - (2.566 \times 10^{-27} \ J)\\\\\Delta U = 5.133 \times 10^{-27} \ J[/tex]
Apply the principle of conservation of energy;
ΔK.E = ΔU
[tex]K.E_f -K.E_i = \Delta U\\\\initial \ velocity \ of \ the \ electron = 0\\\\K.E_f - 0 = \Delta U\\\\K.E_f = \Delta U\\\\\frac{1}{2} mv^2 = \Delta U\\\\where;\\\\m \ is \ the \ mass \ of\ the \ electron = 9.1 1 \times 10^{-31} \ kg\\\\v^2 = \frac{ 2 \Delta U}{m} \\\\v = \sqrt{\frac{ 2 \Delta U}{m}} \\\\v = \sqrt{\frac{ 2 (5.133\times 10^{-27})}{9.11\times 10^{-31}}}\\\\v = \sqrt{11268.935} \\\\v = 106.2 \ m/s[/tex]
Therefore, the speed of the electron at the given position is 106.2 m/s
What is the largest known star?
Answer:
UY Scuti is slightly larger than VY Canis Majoris
Explanation:
These stars are millions of miles away and cannot be seen by the naked eye.
Beetlejuice is another large star that can be seen by the eye.
What type of potential energy is a 9 volt battery an example of?
Gravitational potential energy
Elastic potential energy
Electrical potential energy
chemical potential energy
Answer:
chemical potential energy
Explanation:
A 9v battery comes in different formats, such that the most common one is the carbon-zinc and alkaline chemistry, so these are alkaline batteries (there are also rechargeable or lithium batteries, these also depend on chemical interactions).
These batteries "draw" the energy from chemical interactions of the materials inside of it, so the type of potential energy that is stored in a battery is actually chemical (regardless of the fact that the energy can be transformed into electrical energy later) the "potential" refers to how the energy is stored.
Then the correct option is chemical potential energy
Answer:
Chemical Potential Energy
Explanation:
Hope this helps!!
Have a blessed day/night!! <33
1. Una pelota rueda hacia la derecha siguiendo una trayectoria en línea recta de modo que recorre una distancia de 10m en 5 s , después cambia su trayectoria cuando es lanzada hacia arriba 25m durante 7 s. Calcular la velocidad y la rapidez al punto final (altura maxima) al que llegó la pelota.
2. Una mariposa vuela en línea recta hacia el sur recorriendo una distancia de 15 m durante 28 s, después cambia de dirección hacia el Oeste recorriendo una distancia de 50 m en un tempo de 80 s ¿cuál es la velocidad y rapidez de la mariposa?
3.- Una persona camina durante 21 minutos hacia el este de su casa una distancia de 1500 m y después cambia su dirección hacia el Norte recorriendo una distancia de 3350 m en un tiempo 32 minutos llegando al supermercado. ¿Calcula la velocidad y rapidez de la persona?
4.- Un automóvil se mueve al Oeste recorriendo una distancia de 80 km en 1.2 horas, posteriormente cambia su trayectoria hacia el Sur, recorriendo una distancia de 120 km en un tiempo 1.6 hora. ¿Calcula la velocidad y rapidez del automóvil?
Answer:
https://youtu.be/ymHHdoCGJOU
During normal beating, the heart creates a maximum 4.10-mV potential across 0.350 m of a person's chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? V/m (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? T (c) What is the wavelength of the electromagnetic wave?
Explanation:
Given that,
Maximum potential, V = 4. mV
Distance, d = 0.350 m
Frequency of the wave, f = 100 Hz
(a) The maximum electric field strength created is given by:
[tex]E=\dfrac{V}{d}\\\\E=\dfrac{4.1\times 10^{-3}}{0.350 }\\\\E=0.0117\ V/m[/tex]
(b) The corresponding maximum magnetic field strength in the electromagnetic wave is given by :
[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.0117}{3\times 10^8}\\\\B=3.9\times 10^{-11}\ T[/tex]
(c) The wavelength of the electromagnetic wave can be calculated as :
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{100}\\\\=3\times 10^6\ m[/tex]
So, the wavelength of the electromagnetic wave is [tex]3\times 10^6\ m[/tex].
calculate the pressure of water having density 1000 kilo per metre square at a depth of 20 m inside the water
Answer:
the pressure of the water at the given depth is 196,200 N/m².
Explanation:
Given;
density of the water, ρ = 1000 kg/m³
depth of the water, h = 20 m
acceleration due to gravity, g = 9.81 m/s²
The pressure at the given depth of the water is calculated as;
P = ρgh
P = 1000 x 9.81 x 20
P = 196,200 N/m²
Therefore, the pressure of the water at the given depth is 196,200 N/m².
.. Solve: 91
Find the half angular width of the central bright maximum in the Fraunhofer diffraction pattern of
a slit of width 12x10^-5 cm when the slit illuminated by monochromatic light of wave length
6000 A
[KUET’10-11)
(a) 30°
(b) 60°
(c) 15°
(d) None of these
Solution
Explanation:
bro I have no idea fam......
A(n) _______________ absorbs energy and then emits electromagnetic radiation based on its _______________. Classical physics predicted that at a high enough temperature, _______________ light would be emitted. Instead, white light was emitted, resulting in the ultraviolet _______________. The photoelectric effect occurs when light shining on a metal creates a(n) _______________. However, only light of a certain minimum _______________ causes electrons to flow. Gas atoms excited by an electric current emit bands of colors of light in a(n) _______________. Each narrow band of light is associated with _______________ of a specific energy.
Answer:
Blackbody radiator, temperature, ultraviolet, catastrophe, electric current, frequency, spectrum, photons
Explanation:
# a p e x
1 and 2 ) A blackbody radiator is an object that absorbs energy, then emits electromagnetic radiation based on the temperature of the object. This comes directly from the definition in the passages.
3 and 4 ) Ultraviolet catastrophe describes when old physicists assumed as frequency increased the waves would go from visible to ultraviolet because that is what comes next on the spectrum. Instead of this happening, the light became white light and it was an apparent 'catastrophe'
Appropriate words for blank position shown below,
Blackbody radiatortemperature ultravioletcatastropheelectric currentfrequency spectrum photonsA blackbody radiator is defined as an object that absorbs all electromagnetic radiation that falls on it at all frequencies over all angles of incidence.
Ultraviolet light is a type of electromagnetic radiation that makes black-light posters glow
A movement of positive or negative electric particles produce current.
Frequency is defined as the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
Photons are particles which transmit light.
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A pendulum is constructed from a 6 kg mass attached to a strong cord of length 1.7 m also attached to a ceiling. Originally hanging vertically, the mass is pulled aside a small distance of 7.6 cm and released from rest. While the mass is swinging the cord exerts an almost-constant force on it. For this problem, assume the force is constant as the mass swings. How much work in J does the cord do to the mass as the mass swings a distance of 8.0 cm
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;
[tex]W = \int\limits^{\theta_2} _{\theta_1} {r} \, d\theta[/tex]
[tex]W = \int\limits^{\theta_2} _{\theta_1} {-mgl sin\theta } \, d\theta[/tex]
[tex]W = -mgl \int\limits^{0.047 } _{0.045 } {sin\theta } \, d\theta[/tex]
W = [tex]-mgl[[/tex] -cosθ [tex]]^{0.047}_{0.045 }[/tex]
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J
Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 21.0 cm and carries a clockwise current of 16.0 A , as viewed from above, and the outer wire has a diameter of 32.0 cm.
Required:
a. What must be the direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
b. What must be the magnitude of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common center of the wires?
Solution :
a). B at the center :
[tex]$=\frac{u\times I}{2R}$[/tex]
Here, one of the current is in the clockwise direction and therefore, the other current must be in the clockwise direction in order to cancel out the effect of the magnetic field that is produced by the other.
Therefore, the answer is ANTICLOCKWISE or COUNTERCLOCKWISE
b). Also, the sum of the fields must be zero.
Therefore,
[tex]$\left(\frac{u\times I_1}{2R_1}\right) + \left(\frac{u\times I_2}{2R_2}\right) = 0$[/tex]
So,
[tex]$\frac{I_1}{d_1}= \frac{I_2}{d_2}$[/tex]
[tex]$=\frac{16}{21}=\frac{I_2}{32}$[/tex]
[tex]$I_2=24.38 $[/tex] A
Therefore, the current in the outer wire is 24.38 ampere.
Answer:
(a) counter clockwise
(b) 24.38 A
Explanation:
inner diameter, d = 21 cm
inner radius, r = 10.5 cm
Current in inner loop, I = 16 A clock wise
Outer diameter, D = 32 cm
Outer radius, R = 16 cm
(a) The magnetic filed due to the inner wire is inwards to the plane of paper. According to the Maxwell's right hand thumb rule, the direction of magnetic field in outer wire should be outwards so that the net magnetic field is zero at the center.
So, the direction of current in outer wire is counter clock wise in direction.
(b) Let the current in outer wire is I'.
The magnetic field due to the inner wire is balanced by the magnetic field due to the outer wire.
[tex]\frac{ \mu 0}{4\pi}\times \frac{2 I}{r}=\frac{\mu 0}{4\pi}\times \frac{2 I'}{R}\\\frac{16}{10.5}=\frac{I'}{16}\\\\I' = 24.38 A[/tex]
Light from two lasers is incident on an opaque barrier with a single slit of width 4.0 x 10^-4 m. One laser emits light of wavelength 480 nm and the other is 640 nm. A screen to view the light intensity pattern is 2.0 m behind the barrier. The distance from the center of the pattern to the nearest completely dark spot (dark for both colors) is ____ cm. (include 2 digits after the decimal point)
Answer:
a) y = 2.4 x 10⁻³ m = 0.24 cm
b) y = 3.2 x 10⁻³ m = 0.32 cm
Explanation:
The formula of Young's Double Slit experiment will be used here:
[tex]y = \frac{\lambda L}{d}\\\\[/tex]
where,
y = distance between dark spots = ?
λ = wavelength
L = distance of screen = 2 m
d = slit width = 4 x 10⁻⁴ m
a) FOR λ = 480 nm = 4.8 x 10⁻⁷ m:
[tex]y = \frac{(4.8\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 2.4 x 10⁻³ m = 0.24 cm
a) FOR λ = 640 nm = 6.4 x 10⁻⁷ m:
[tex]y = \frac{(6.4\ x\ 10^{-7}\ m)(2\ m)}{4\ x\ 10^{-4}\ m}[/tex]
y = 3.2 x 10⁻³ m = 0.32 cm
17- How much work is needed for a climber in order to climb 45 m height, where his weight is 70 kg. also, calculate the power required to climb the height in 30 minutes ? g= 9,8 m.sec
Answer:
Work Done= 3150J
Power= 1.75W
Explanation:
Work Done= Force x the distance travelled in the direction of the force (W= f x d)
Weight is a force, i think the qn. stated it wrongly, it should be 70N not 70kg.
Work Done= 70 x 45
=3150J
Power= Work Done/Time
=3150/(30x60)
*convert minutes to seconds since the S.I. unit of Power is joules/seconds(J/s) or watts(W)
=1.75W
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.010-3 kg/m/s. The diffusion coefficient of the candy solute in water is
The question is incomplete. The complete question is :
A spherical piece of candy is suspended in flowing water. The candy has a density of 1950 kg/m3 and has a 1.0 cm diameter. The water velocity is 1.0 m/s, the water density is assumed to be 1000.0 kg/m3, and the water viscosity is 1.0x10-3 kg/m/s. The diffusion coefficient of the candy solute in water is 2.0x10-9 m2/s, and the solubility of the candy solute in water is 2.0 kg/m3. Calculate the mass transfer coefficient (m/s) and the dissolution rate (kg/s).
Solution :
From flow over sphere, the mass transfer equation can be written as :
[tex]$Sh = 2 + 0.6 Re^{1/2} Sc^{1/3}$[/tex]
where, Sherood number, [tex]$Sh = \frac{K_L d}{D_{eff}}$[/tex]
Reynolds number, [tex]$Re=\frac{Vd\rho}{\mu}$[/tex]
Schmid number, [tex]$Sc= \frac{\mu}{\rho D_{eff}}$[/tex]
So,
[tex]$\frac{K_L d}{D_{eff}}=2+0.6 \left( \frac{V d \rho}{\mu} \right)^{1/2} \ \left( \frac{\mu}{\rho D_{eff}} \right)^{1/3}$[/tex]
Diameter, d = 1 cm = [tex]$1 \times 10^{-2}$[/tex] m
V = 1 m/s
[tex]$\rho = 1000 \ kg/m^3$[/tex]
[tex]$\mu = 10^{-3} \ kg/m/s$[/tex]
[tex]$D_{eff} = 2 \times 10^{-9} \ m^2/s$[/tex]
[tex]$\frac{K_L \times 10^{-2}}{2 \times 10^{-9}}=2+0.6 \left( \frac{1 \times 10^{-2} \times 10^3}{10^{-3}} \right)^{1/2} \ \left( \frac{10^{-3}}{10^3 \times 2 \times 10^{-9}} \right)^{1/3}$[/tex]
[tex]$K_L \times 5 \times 10^6=478.22$[/tex]
[tex]$K_L=9.5644 \times 10^{-5}$[/tex] m/s
So the mass transfer coefficient is 9.5644 [tex]$\times 10^{-5}$[/tex] m/s. It is given solubility,
[tex]$\Delta C = 2 \ kg/m^3$[/tex]
[tex]$N = Md^2 \times \Delta C \times K_L$[/tex]
[tex]$N= M \times (10^{-2})^2 \times 2 \times 9.5644 \times 10^{-5}$[/tex]
[tex]$N= 6 \times 10^{-8}$[/tex] kg/s (dissolution rate)
State the law of conservation of momentum
Explanation:
Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant
what are three effects of gravity
Answer:
effect on motation.effect on direction
A physics student likes to study while listening to loud music. If electricity costs 12.00$/kWh (kilowatt-hour), how much would it cost the student to run a 220 W stereo system 8.0 hours per day for 10 days of studying?
Answer:
the cost of running the stereo is $211.2
Explanation:
Given;
cost of electricity, = 12.00$/kWh
power consumed by the stereo system, P = 220 W
duration of the power consumption, t = 8 hours
number of days, = 10 days
total time of the power consumption = 8 hours x 10 = 80 hours
power consumed in kW = 220 W / 1000 = 0.22 kW
Energy consumed = 0.22 kW x 80 h = 17.6 kWh
The cost of using 17.6 kWh
= 17.6 x $12
= $211.2
Therefore, the cost of running the stereo is $211.2
A car starting at rest accelerates at 3m/s² How far has the car travelled after 4s?
Answer:
24m
Explanation:
you can use the formula
s=ut+1/2at²
s=0+1/2(3)(4)²
=1/2(3)(8)
=24m
I hope this helps
Tres personas, A, B y C jalan una caja con ayuda de cuerdas cuya masa es despreciable. Si la persona A aplica −3 en dirección horizontal y la persona B aplica a su vez 5 en dirección horizontal, ¿Cuál es el valor de la fuerza que debe ejercer la persona C, para que la caja esté en equilibrio físico?
Answer:
Un objeto se encuentra en equilibrio físico si la fuerza neta que se le aplica es igual a 0.
En este caso solo se aplican fuerzas en el eje horizontal, por lo que las podremos sumar directamente.
La persona A aplica una fuerza:
Fa = -3N
La persona B aplica una fuerza:
Fb = 5N
La persona C aplica una fuerza Fc, la cual aún no conocemos.
Pero sabemos que la caja está en equilibrio físico, por lo que:
Fa + Fb + Fc = 0N
reemplazando los valores que conocemos, obtenemos:
-3N + 5N + Fc = 0N
Ahora podemos resolver esto para Fc, la fuerza que aplica la persona C.
Fc = 0N + 3N - 5N
Fc = -2N
Podemos concluir que la persona C aplica una fuerza horizontal de -2N
what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...
(C)
Explanation:
[tex]E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})[/tex]
[tex]= 5.3×10^{-18}\:\text{J}[/tex]
Answer:
It's D
Explanation:
It's from alvs
A spherical balloon has a radius of 7.15 m and is filled with helium. The density of helium is 0.179 kg/m^3, and the density of air is 1.29 kg/m^3. The skin and structure of the balloon has a mass of 910 kg. Neglect the buoyant force on the cargo volume itself.
Determine the largest mass of cargo the balloon can lift.
The largest mass of cargo the balloon can lift is 791.06 kg
First, we need to calculate the mass of helium.
Since the radius of the spherical balloon is r = 7.15 m, its volume is V = 4πr³/3.
The volume of the balloon also equals the volume of helium present.
Now, the mass of helium m = density of helium, ρ × volume of helium, V
m = ρV
Since ρ = 0.179 kg/m³
m = ρV
m = ρ4πr³/3.
m = 0.179 kg/m³ × 4π(7.15 m)³/3
m = 0.179 kg/m³ × 4π(365.525875 m³)/3
m = 0.179 kg/m³ × 1462.1035π m³/3
m = 261.7165265π/3 kg
m = 822.207/3 kg
m = 274.07 kg
Since the mass of the skin and structure of the balloon is 910 kg, the total mass, M of the balloon = mass of skin and structure + mass of helium gas is 910 kg + 274.07 kg = 1184.07 kg.
The weight of this mass W = Mg where g = acceleration due to gravity.
The buoyant force on the balloon due to the air is the weight of air displaced, W' = mass of air, m' × acceleration due to gravity, g.
W' = m'g
Now, the mass of air m' = density of air, ρ' × volume of air displaced, V'
We know that the volume of air displaced, V' = volume of balloon, V
So, V' = V = 4πr³/3.
Since the density of air, ρ' = 1.29 kg/m³,
m' = ρ'V
m = 1.29 kg/m³ × 4π(7.15 m)³/3
m = 1.29 kg/m³ × 4π(365.525875 m³)/3
m = 1.29 kg/m³ × 1462.1035π m³/3
m = 1886.113515π/3 kg
m = 5925.4/3 kg
m = 1975.13 kg
So, the net weight W" that the balloon can lift is W" = W' - W = m'g - Mg = (m' - M )g = (1975.13 kg - 1184.07 kg)g = 791.06g.
So, the net mass m" = W"/g = 791.06g/g = 791.06 kg
This net mass is the largest mass of cargo that the balloon can lift.
Thus, the largest mass of cargo the balloon can lift is 791.06 kg
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Assume that the far point of a myopic (nearsighted) eye is 5.04 m in front of the eye. A lens is used to correct the vision, such that it can focus sharply an object at infinity. What is the power of the lens (in diopters; answer sign and magnitude)?
Answer:
[tex]P=-0.2D[/tex]
Explanation:
From the question we are told that:
Far-point [tex]v=-5.04m[/tex]
Where
u=-\alpha
Generally the equation for Lens is mathematically given by
[tex]\frac{1}{f}=\frac{1}{v}-\frac{1}{u}[/tex]
[tex]P=\frac{1}{-5.04}-\frac{1}{\alpha}[/tex]
[tex]P=-0.2D[/tex]
An object is acted upon by two and only two forces that are equal magnitude and oppositely directed. Is the objected necessarily in static equilibrium? Explain. You can draw a picture if that helps explain.
Answer:
the body is subjected to a continuous rotation and the body is not in rotational equilibrium
Explanation:
For an object to have a static equilibrium, it must meet two relationships
∑ F = 0
∑ τ =0
force acting on a body fulfills the relation of
sum F = F - F = 0
when two forces do not move from position.
To find the torque we assume that the counterclockwise rotations are positive
Σ τ = - F r - F r
Στ = -2 Fr <> 0
consequently the body is subjected to a continuous rotation and the body is not in rotational equilibrium
How can I solve this?
You have three capacitors of values 40 F, 10 F and 50 F. What would their equivalent capacitance (in F) be if they were connected in parallel with each other? Enter your answer as a number only, to one decimal place.
Explanation:
The equivalent capacitance of capacitors in parallel can be determined as
[tex]C_{eq} = C_1 + C_2 + C_3[/tex]
[tex]\:\:\:\:\:= 40\:\text{F} + 10\:\text{F} + 50\:\text{F} = 100\:\text{F}[/tex]
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 3.10 V/mV/m , what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
B = 1.03 10⁻⁸ T
Explanation:
For an electromagnetic wave, the electric and magnetic fields must oscillate in phase so that they remain between them at all times, otherwise the wave will extinguish
This relational is expressed by the relation
E /B = c
B = E / c
let's calculate
B = 3.10 / 3 10⁸
B = 1.03 10⁻⁸ T
A toy car of mass 600g moves through 6m in 2 seconds.The average kinetic energy og the toy car is?
Explanation:
kE =1/2mv²
1/2(0.6×(3m/s)²
1/2(0.6×9m/s)
2.7J I think this is the answer
The average kinetic energy of the toy car is 2.7 J.
What is kinetic energy?
The energy of the body by the virtue of its motion is known as the kinetic energy of the body. It is defined as the product of half of mass and square of the velocity.
Given data;
Mass of car is,m= 600 g = 0.6 kg
d is the distance travelled = 6 m
T is the time travelled = 2 sec
The velocity of the car is found as;
v = d /t
V = 6m / 2 sec
V = 3 m/sec
KE =1/2mv²
KE = 1/2 × 0.6 kg ×( 3 m/sec )²
KE = 2.7 J
Hence, the average kinetic energy of the toy car is 2.7 J.
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A scientist who studies insects, spiders, snails, and other bugs of an environment .
Botanist
Chemist
Ecologist
Entomologist
Question:- A scientist who studies insects, spiders, snails, and other bugs of an environment
Answer:- EntomologistExplanation:-
Entomologist word comes from two words Entomon and biologist
Entomon which means insectbiologist which means the person who study living forms15- A racehorse coming out of the gate accelerates from rest to a velocity f 15.0 m/s due west in 1.80 s. What is its average acceleration?
Answer: (15 - 0)/1.8 = 8. 33m/s^2
Explanation:
The acceleration of the racehorse is 8.33 m/s²
The given parameters;
initial velocity of the racehorse, u = 0
final velocity of the racehorse, v = 15 m/s
time of motion of the horse, t = 1.8 s
The acceleration of the racehorse is calculated from change in velocity per change in time of motion as shown below;
[tex]a = \frac{\Delta v}{\Delta t} = \frac{v-u}{t} \\\\a = \frac{15 - 0}{1.8} \\\\a = 8.33 \ m/s^2[/tex]
Thus, the acceleration of the racehorse is 8.33 m/s²
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state any 3 properties of an ideal gas as assumed by the kinetic theory.
Answer:
The simplest kinetic model is based on the assumptions that: (1) the gas is composed of a large number of identical molecules moving in random directions, separated by distances that are large compared with their size; (2) the molecules undergo perfectly elastic collisions (no energy loss) with each other and with the walls of the container, but otherwise do not interact; and (3) the transfer of kinetic energy between molecules is heat.
1. A sequence of potential differences v is applied accross a wire (diameter =0.32 mm length = 11 cm and the resulting current I are measured as follows: V 0.1 0.2 0.3 0.4 0.5 I (MA) 72 144 216 288 360 2) a) plot a graph of v against I.
b) determine the wire's resistence , R.
c) State ohm's law and try to relate it . your results.
Answer:
a. Find the graph in the attachment
b. 720 kΩ
c. The ratio V/I gives us our resistance which is 720 kΩ
Explanation:
a) plot a graph of V against I.
To plot the graph of V against I, we plot the corresponding points against each other. With the voltage V measured in volts and the current I measured in mA, the plotted graph is in the attachment.
b) Determine the wire's resistance , R.
The resistance of the wire is determined as the gradient of the graph.
R = ΔV/ΔI = (V₂ - V₁)/(I₂ - I₁)
Taking the first two corresponding measurements. V₁ = 72 V, I₁ = 0.1 mA, V₂ = 144 V and I₂ = 0.2 mA
R = (144 V - 72 V)/(0.2 - 0.1) mA
R = 72 V/0.1 mA
R = 72 V/(0.1 × 10⁻³ A)
R = 720 × 10³ V/A
R = 720 kΩ
c) State ohm's law and try to relate it your results.
Ohm's law states that the current flowing through a conductor is directly proportional to the voltage across it provided the temperature and all other physical conditions remain constant.
Mathematically, V ∝ I
V = kI
V/I = k = R
Since the ratio V/I = constant, from our results, the ratio of V/I for each reading gives us the resistance. Since we have a linear relationship between V and I, the gradient of the graph is constant and for each value of V and I, the ratio V/I is constant. So, the ratio V/I gives us our resistance which is 720 kΩ.
Since V/I is constant, we thus verify Ohm's law.
