Answer:
[tex]f=1.98\times 10^5\ Hz[/tex]
Explanation:
The magnetic field of a plane-polarized electromagnetic wave moving in the z-direction is given by :
[tex]B=1.2\times 10^{-6}\sin [2\pi (\dfrac{z}{240}-\dfrac{10^7t}{8})][/tex] .....(1)
The general equation of the magnetic field wave is given by :
[tex]B=B_o\sin (kz-\omega t)[/tex] ....(2)
Equation (1) is in form of equation (2), if we compare equation (1) and (2) we find that,
[tex]\omega=\dfrac{10^7}{8}[/tex]
We need to find the frequency of the wave. It is given by :
[tex]f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{10^7}{8\times 2\pi}\\\\f=1.98\times 10^5\ Hz[/tex]
So, the frequency of the wave is [tex]1.98\times 10^5\ Hz[/tex]
A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to accelerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him? (a) 70 N (b) 370 N (c) 80 N (d) 380 N
Answer:
Choice a. [tex]70\; \rm N[/tex], assuming that the skating rink is level.
Explanation:
Net force in the horizontal directionThere are two horizontal forces acting on the boy:
The pull of his friend, andFrictions.The boy should be moving in the direction of the pull of his friend. The frictions on this boy should oppose that motion. Therefore, the frictions on the boy would be in the opposite direction of the pull of his friend.
The net force in the horizontal direction should then be the difference between the pull of the friend, and the friction on this boy.
[tex]\text{Net force, horizontal} = 75\; \rm N - 5\; \rm N = 70\; \rm N[/tex].
Net force in the vertical directionThe net force on this boy should be zero in the vertical direction. Consider Newton's Second Law of motion. The net force on an object is proportional to its acceleration. In this question, the net force on this boy in the vertical direction should be proportional to the vertical acceleration of this boy.
However, because (by assumption) the ice rink is level, the boy has no motion in the vertical direction. His vertical acceleration will be zero. As a result, the net force on him should also be zero in the vertical direction.
Net forceTherefore, the (combined) net force on this boy would be:
[tex]\sqrt{(70\; \rm N)^2 + (0\; \rm N)^2} = 70\; \rm N[/tex].
As a professional teacher who has been assigned to teach science in an elementary school class design activities to teach source of energy to your learners
Answer:
For solar energy, I would show them how a magnifying glass works when exposed to the sun.
For wind energy, I would teach them how to make a paper windmill and explain how it works.
For the hydroelectric energy, I would have them make a plastic turbine and explain to them how to use it in rivers or streams.
For electromagnetic energy, I would tell them to rub a balloon until their hairs stand on end.
And for electricity, I would teach them how the other energy sources create electricity and what electricity works for in these times.
Explanation:
To explain something so complicated to a child is not as easy as it would be with a teenager or an adult.
To make the children learn about the forms of energy, I would use the nemotechnique rule, using short and easy-to-remember sentences and explaining with many examples about how to get each type of energy and its use, in addition to adding didactic, visual and auditory content, which are the most common types of learning in children.
The maximum energy a bone can absorb without breaking is surprisingly small. Experimental data show that both leg bones, together, of a healthy human adult can absorb about 200 J before breaking. From what maximum height could a 75 kg person jump and land rigidly upright on both feet without breaking their legs
Answer:
0.21m
Explanation:
Note that the 200 J before breaking legs is Kinetic energy
So if potential energy= kinetic energy
Then 200J = mgh
h= 200/75x 9.8
= 0.21m so the maximum height will be 0.21m
An electron moves along the z-axis with vz=4.5×10^7m/s. As it passes the origin, what are the strength and direction of the magnetic field at the following (x, y, z) positions?
a. (1 cm , 0 cm, 0 cm)
b. (0 cm, 0 cm, 1 cm )
c. (0 cm, 2 cm , 1 cm )
A man walks 7 km, east in 2 hours and 2 km in 1 hour in the same direction. a) what is
the man's average speed for the whole journey? b) what is the man's average velocity
for the whole journey?
Explanation:
Average speed = distance / time
|v| = (7 km + 2 km) / (2 hr + 1 hr)
|v| = 3 km/hr
Average velocity = displacement / time
v = (7 km east + 2 km east) / (2 hr + 1 hr)
v = 3 km/hr east
A 2.0 kg block on a horizontal frictionless surface is attached to a spring whose force constant is 590 N/m. The block is pulled from its equilibrium position at x = 0 m to a displacement x = +0.080 m and is released from rest. The block then executes simple harmonic motion along the x-axis (horizontal). The velocity of the block at time t = 0.10 s is closest to:________.
Answer:
The value is [tex]v = -0.04 \ m/s[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m = 2.0 \ kg[/tex]
The force constant of the spring is [tex]k = 590 \ N/m[/tex]
The amplitude is [tex]A = + 0.080[/tex]
The time consider is [tex]t = 0.10 \ s[/tex]
Generally the angular velocity of this block is mathematically represented as
[tex]w = \sqrt{\frac{k}{m} }[/tex]
=> [tex]w = \sqrt{\frac{590}{2} }[/tex]
=> [tex]w = 17.18 \ rad/s[/tex]
Given that the block undergoes simple harmonic motion the velocity is mathematically represented as
[tex]v = -A w sin (w* t )[/tex]
=> [tex]v = -0.080 * 17.18 sin (17.18* 0.10 )[/tex]
=> [tex]v = -0.04 \ m/s[/tex]
The velocity of the block at the given time is -0.04 m/s.
Angular speed of the blockThe angular speed of the block is determined by using the following wave equation;
[tex]\omega = \sqrt{\frac{k}{m} } \\\\\omega = \sqrt{\frac{590}{2} } \\\\\omega = 17.176 \ rad/s[/tex]
Velocity of the blockThe velocity of the block at the given time is calculated as follows;'
[tex]v = - A sin(\omega t)\\\\v =- 0.08 \times sin(17.176 \times 0.1)\\\\v = -0.04 \ m/s[/tex]
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A 2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a. Determine the pressure Pb.
b. Determine the work done on the gas during the process ab.
c. Determine the change in internal energy of the gas during the process ab.
d. Determine the heat transferred to the gas during the process ab.
Answer:
a) Pb= 200 PA
b).work done= -3600 joules
c).3600joules
D).the system works under isothermal condition so no heat was transferred
Explanation:
2.0 moles of a monatomic ideal gas expands isothermally from state a to state b, Pa = 600 Pa, Va = 3.0 m3, and Vb = 9.0 m3.
a). PbVb= PaVa
Pb= (PaVa)/VB
Pb= (600*3)/9
Pb= 1800/9
Pb= 200 PA
b). work done= n(Pb-Pa)(Vb-Va)
Work done= 2*(200-600)(9-3)
Work done= -600(6)
Work done=- 3600 Pam³
work done= -3600 joules
C). Change in internal energy I the work done on the system
= 3600joules
D).the system works under isothermal condition so no heat was transferred
Determine the inductance L of a 0.65-m-long air-filled solenoid 3.2 cm in diameter containing 8400 loops.
Answer:
The inductance is [tex]L = 0.1097 \ H[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 0.65 \ m[/tex]
The diameter is [tex]d = 3.2 cm = 0.032 \ m[/tex]
The number of loops is [tex]N = 8400[/tex]
Generally the radius is evaluated as
[tex]r = \frac{ 0.032 }{2} = 0.016 \ m[/tex]
The inductance is mathematically represented as
[tex]L = \frac{ \mu_o * N^2 * A }{ l }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
A is the cross-sectional area which is mathematically evaluated as
[tex]A = \pi r^2[/tex]
=> [tex]A = 3.142 * (0.016)^2[/tex]
=> [tex]A = 0.000804 \ m^2[/tex]
=> [tex]L = \frac{ 4\pi * 10^{-7} * 8400^2 * 0.000804 }{ 0.65 }[/tex]
=> [tex]L = 0.1097 \ H[/tex]
IN A FORCE COMPRESSION GRAPH, WHAT IS THE STORED POTENTIAL ENERGY OF THE SPRING WHEN IT'S COMPRESS 0.60M ?
Answer:
La energía potencial elástica es la energía asociada con los materiales elásticos. Por ejemplo, un resorte al ser comprimido o elongado almacena energía potencial elástica y, al ser soltado, puede realizar trabajo sobre un objeto.
Para mantener el resorte comprimido o alargado una cierta longitud x, a partir de su largo natural, es necesario que, en este caso, la mano aplique una fuerza F_{M} sobre el resorte; esta fuerza es directamente proporcional a x.
Explanation:
ón conocida como ley de Hooke.
Para encontrar una expresión que describa la energía potencial asociada con la fuerza del resorte, se determina el trabajo que se requiere para comprimir el resorte desde su posición de equilibrio hasta cierta posición final arbitraria x. Debido a que la fuerza varía desde O hasta kx, se utiliza la fuerza promedio \frac{(F_{0}+F_{X})}{2}.
\[ \bar{F}=\frac{0+K X}{2}=\frac{1}{2}kx \]
fuerza-sobre-un-resorte
Fuerza sobre un resorte. La fuerza para estirar un resorte aumenta linealmente con su elongación .
El trabajo realizado por la fuerza aplicada será: W=\bar{Fx}=\frac{1}{2}kx^{2}
El trabajo realizado se almacena en el resorte comprimido en forma de energía potencial elástica como:
\[ \boxed{ Ep_{elas}=\frac{1}{2}kx^{2}} \]
Una vez que se ha comprimido o estirado el resorte respecto a su posición de equilibrio, la energía potencial elástica se puede considerar como la energía almacenada en el resorte deformado. Esta energía siempre es positiva en un objeto deformado al depender de x^{2}.
Por ejemplo, en la figura se observa que un resorte realiza trabajo sobre un bloque. El resorte que se encuentra sin deformar (a) cuando es empujado por un bloque de masa m, se comprime una distancia x (b). Cuando el bloque se suelta (c), partiendo del reposo, la energía potencial plástica almacenada en el sistema se transforma en energía cinética del bloque.
energia-potencial
Which of the following is NOT an observation? a.The apple tastes sour b.The apple weighs about 38 g c.The apple is light green in color d.Apples are the best fruit
Answer:
d apple's are the best fruit
A solenoid with a certain number of turns N and carrying a current of 2.000 A has a length of 34.00 cm. If the magnitude of the magnetic field generated at the center of the solenoid is 9.000 mT, what is the value of N?
Answer:
The number of turns of the solenoid is 1217 turns
Explanation:
Given;
current in the solenoid, I = 2 A
length of the solenoid, L = 34 cm = 0.34 m
magnitude of the magnetic field, B = 9 mT = 0.009 T
Number of turns of the solenoid = N
The magnitude of magnetic field at the center of the solenoid is given by;
B = μnI
Where;
μ is permeability of free space = 4π x 10⁻⁷ m/A
I is the current in the solenoid
n is the number of turns per length
n = B/μI
n = (0.009) / (4π x 10⁻⁷)(2)
n = 3580.52 turns/m
N = nL
N =(3580.52 turns/m) x (0.34 m)
N = 1217 turns
Therefore, the number of turns of the solenoid is 1217 turns
A cannonball is fired on flat ground at 420 m/s at a 53.0° angle. how far away does it land?
Answer:
17,300 m
Explanation:
Using kinematic equations, first find the time it takes to land.
Δy = v₀ t + ½ at²
0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²
t = 0 s or 68.5 s
The horizontal distance it moves in that time is:
Δx = v₀ t + ½ at²
Δx = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²
Δx = 17,300 m
Alternatively, you can use the range equation:
R = v₀² sin(2θ) / g
R = (420 m/s)² sin(2 × 53.0°) / (9.8 m/s²)
R = 17,300 m
The distance a cannonball will land if it is fired on flat ground at 420 m/s at a 53.0° angle is 17,300 meters.
What is the distance?The complete movement of an object, regardless of direction, is referred to as distance. The amount of ground a thing travels from its starting point to its destination is also referred to as distance.
Given:
A cannonball is fired on flat ground at 420 m/s at a 53.0° angle,
Calculate the time to land on the ground as shown below,
[tex]\Delta y = v_o t +1/2 at^2[/tex]
0 m = (420 sin 53.0° m/s) t + ½ (-9.8 m/s²) t²
t = 0 s or 68.5 s
Calculate the distance as shown below,
[tex]\Delta x = v_o t +1/2 at^2[/tex]
Δ x = (420 cos 53.0° m/s) (68.5 s) + ½ (0 m/s²) (68.5 s)²
Δ x = 17,300 m
Thus, the total distance covered by the cannonball fired with a speed of 420 m/s is 17300 meters.
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Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV. What is the maximum kinetic energy of the photoelectrons emitted from the metal
Answer:
1.41eV
Explanation:
Kinetic Energy of photoelectron(K. Emax)
E = Wo + K.Emax
E = hc/ λ
h = planck's constant = 6.63 * 10^-34
c = speed of light = 3×10^8 m/s
λ = 400nm
Work function (Wo) = 1.70eV
1 eV = 1×10^-19
E = [(6.63×10^-34) * (3×10^8)] / 400×10^-7
E = (19.89 × 10^(-26))/400×10^-7
E = 0.049725×10^-19
K.Emax = E - Wo
K.Emax = (0.049725×10^-19) - (1.7×10^-38)
0.049725×10^-19 interms of eV = (0.049725/1.6)×10^-19 =
K.Emax = 3.11eV - 1.70eV
K.Emax = 1.41eV
The homogeneous beam of mass 5 kg indicated in the figure is in equilibrium and supported at points A and B. Calculate the reactions at the supports.
Explanation:
Sum of moments about point A:
∑τ = Iα
-mg (L/2) + Rb x = 0
-(5 kg) (10 m/s²) (0.75 m) + Rb (0.70 m) = 0
Rb = 53.6 N
Sum of forces in the y direction:
∑F = ma
Ra + Rb − mg = 0
Ra = mg − Rb
Ra = (5 kg) (10 m/s²) − 53.6 N
Ra = -3.6 N
g Radiation of an unknown wavelength is used in a photoelectric effect experiment on a sodium surface. The maximum kinetic energy of the observed electrons is 0.7 eV. What is the wavelength of the light
Answer:
λ = 4.1638 10⁻⁷ m
Explanation:
The photoelectric effect was explained by Einstein assuming that the radiation acts like particles and the equation that describes the process is
K = h f -Ф
where K is the kinetic energy of the emitted electrons, hf the energy of the photons according to Planck's equation and Ф the work function of the material
In this case they give us the kinetic energy of the electrons
K = 0.7 eV
The sodium work function is tabulated Ф = 2.28 eV
Let's find the frequency of the photons
f = (K + Ф) / h
Planck's constant is
h = 6.626 10⁻³⁴ J s (1 eV / 1.6 10⁻¹⁹ J) = 4.136 10⁻¹⁵ eV s
f = (0.7 + 2.28) / 4.136 10⁻¹⁵
f = 7.2050 10¹⁴ Hz
let's find the wavelength using the relationship between speed and frequency and wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 7.205 10¹⁴
λ = 4.1638 10⁻⁷ m
A discus thrower achieves a high throw of 100m with the discus released at an angle of 30° calculate the initial speed of the discuss.
Answer:
u = 88.54 m/s
Explanation:
Given that,
A discus thrower achieves a high throw of 100 m.
Angle of projection is 30°
We need to find the initial speed of the discuss. It is a cse of projectile motion. The maximum height reached by the discus is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}[/tex]
u is the initial speed of the discus
So,
[tex]u^2=\dfrac{2\times 9.8\times 100}{\sin^2(30)}\\\\u=\sqrt{7840}\\\\u=88.54\ m/s[/tex]
So, the initial speed of the discus is 88.54 m/s.
A plate having an area of 0.6 m2 is sliding down the inclined plane at 300 to the horizontal with a velocity of 0.36 m/s. There is a cushion of fluid 1.8 mm thick between the plane and the plate. Calculate the viscosity of the fluid if the weight of the plate is 280 N.
Answer:
The viscosity of the fluid is 1.16 N-s/m²
Explanation:
Given that,
Area = 0.6 m²
Angle = 30°
Velocity = 0.36 m/s
Thickness = 1.8 mm
Weight = 280 N
We need to calculate the viscosity of the fluid
Using balance equation
[tex]w\sin\theta=\dfrac{\mu\times v}{t}\times A[/tex]
Put the value in the equation
[tex]280\sin30=\dfrac{\mu\times0.36}{1.8\times10^{-3}}\times(0.6)[/tex]
[tex]140=\mu\times120[/tex]
[tex]\mu=1.16\ N-s/m^2[/tex]
Hence. The viscosity of the fluid is 1.16 N-s/m².
A man weighing 700 NN and a woman weighing 440 NN have the same momentum. What is the ratio of the man's kinetic energy KmKmK_m to that of the woman K
Answer:
The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Explanation:
Given;
weight of the man, W = 700 N
Weight of the woman, W = 440 N
momentum is given by;
[tex]P = mv\\\\v = \frac{P}{m}[/tex]
Kinetic energy of the man;
[tex]K_m = \frac{1}{2}m_m(\frac{P_m}{m_m})^2 \\\\K_m = \frac{P_m^2}{2m_m}[/tex]
Momentum of the man is calculated as;
[tex]P_m^2 = 2m_mK_m[/tex]
The kinetic energy of the woman is given by;
[tex]K_w = \frac{P_w^2}{2m_w}[/tex]
The momentum of the woman is given;
[tex]P_w^2 = 2m_wK_w[/tex]
Since, momentum of the man = momentum of the woman
[tex]P_m^2 = P_w^2[/tex]
[tex]2m_mK_m = 2m_wK_w\\\\\frac{K_m}{K_w} = \frac{2m_w}{2m_m}\\\\\frac{K_m}{K_w} = \frac{m_w}{m_m}[/tex]
mass of the mas = 700 / 9.8 = 71.429
mass of the woman is = 440 / 9.8 = 44.898
[tex]\frac{K_m}{K_w} = \frac{44.898}{71.429}\\\\\frac{K_m}{K_w} =0.629[/tex]
Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.
Base of wall of water dam is made wider.Give reason.
In this picture the bike rider starts at point A rides to his friend's house at point B(4 miles away),rides to point C (3 miles away)
and then returns to point A (5 miles away).
Explain-- What is his displacement and why? What is his total distance and how did you calculate it? In general, what is the
difference between distance and displacement?
BONUS QUESTION-Why can displacement never be greater than distance?
Answer:
bonus questions:becz displacement work only with the help of any force or object and distance is the totl length of two points
Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the direction in which such pressure fluctuations travel
Answer:
the pressure fluctuation is LONGITUDINAL
Explanation:
Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.
The expression for the wave is
ΔP = Δo sin (kx - wt)
Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL
A metal sphere of radius 19 cm has a net charge of 2.4 × 10–8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 370 V?
Answer:
1.29*10^6 N/C
1135.6 V
9.18 cm
Explanation:
Given that
radius of the metal, r = 19 cm
charge of the metal, q = 2.4*10^-8 C
coulomb's constant, k = 8.99*10^9
to find the electric field, we use the formula E = kq/r², where
E = electric field
k = coulomb constant
q = charge on the metal and
r = radius of the metal
E = (8.99*10^9 * 2.4*10^-8) / 0.19²
E = 215.76 / 0.0361
E = 1.29*10^6 N/C
to find the electric potential, we use this relation
V = kq/r
V = (8.99*10^9 * 2.4*10^-8) / 0.19
V = 215.76 / 0.19
V = 1135.6 V
V = kq/r,
r = kq/V
r = (8.99*10^9 * 2.4*10^-8)/ (1135.6 - 370
r = 215.76 / 765.6
r = 0.281 = 28.1 cm
distance from the sphere
28.18 - 19 = 9.18 cm
a 16.0 kg cart is being pulled by a 95.4 N force to the right and a 36.0 N force to the left. What is the acceleration of the cart?
Answer:
The cart's acceleration is [tex]\approx 3.71\,\,\frac{m}{s^2}[/tex]
Explanation:
Let's start by finding the net force acting on the cart, and then find its acceleration using Newton's 2nd Law.
Net force = 95.4 N -36.0 N = 59.4 N
Now, since we know the cart's mass, we can use Newton's 2nd Law to find the cart's acceleration:
[tex]F=m\,*\,a\\a=\frac{F}{m} \\a=\frac{59.4}{16} \,\,\frac{m}{s^2} \\a\approx 3.71\,\,\frac{m}{s^2}[/tex]
Answer:
3.71 m/s²
Explanation:
Which statement is true regarding the waves shown?
A)
Doubling the frequency of the bottom waves by will cause it to match the
top waves.
B)
Cutting the frequency of the bottom waves in half will cause it to match
the top waves
C)
There is no way to match the bottom and top waves to the same frequency.
D)
Decreasing the frequency of the top waves by half will cause it to match
the bottom waves.
Answer:
The correct answer is B)
Cutting the frequency of the bottom waves in half will cause it to match top waves.
Explanation:
USATESTPREP
Cutting the frequency of the bottom waves will cause it to match the top waves. So the correct option is B.
What are waves?A propagation of disturbance, from one point to another point is called a wave. Waves are either mechanical or non-mechanical. Electromagnetic waves are non-mechanical waves.
The mechanical waves cannot travel without a medium e.g sound waves. Non-mechanical waves do not require a medium to travel, this means that they can even travel through a vacuum.
Waves are of two types. Transverse waves and longitudinal waves.
If the direction of propagation of wave is perpendicular to the direction of movement of particles of the medium, it results in a transverse wave.
If the direction of propagation of wave is parallel to the direction of movement of particles in a medium, it results in a longitudinal wave.
Five properties of waves are amplitude, frequency, wavelength, time period and speed.
Maximum displacement from the mean position is the amplitude. The number of vibrations in a fixed point in unit time is the frequency. The distance between two identical points is the wavelength. Time taken by a wave to pass through a point is the time period and the distance travelled between particular points in a unit of time is the speed.
Therefore the correct option is B.
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A scientist claims that a certain chemical will make fabric waterproof. Which
option describes a controlled experiment that will produce evidence that will
support or refute her claim?
Answer: She adds different amounts of the chemical to the material and then puts them in the water
Explanation:
Answer: One group of fabric is treated with the chemical, and the other group is not. Then each group is exposed to water.
Explanation:
Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
What was his average speed in mph over the last 400 m?
Answer: His average speed in mph over the last 400 m is 7.7 m/s.
Explanation:
Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.
We know that , speed = [tex]\dfrac{distance}{time}[/tex]
Here , distance = 400m and time = 51.9 s
Then, speed = [tex]\dfrac{400}{51.9}\approx7.7\ m/s[/tex]
Hence, his average speed in mph over the last 400 m is 7.7 m/s.
Answer: 17.2 MPH
Explanation:
Average speed = distance/time
400m÷ 51.9s= 7.7 m/s
Now convert m/s →MPH
m→km→mi and s→min→hr
[tex]\frac{7.7m}{s} x[/tex] [tex]\frac{1 km}{1000 m} x\frac{0.6214 mi}{1 km} x\frac{60 s}{1 min} x\frac{60 mins}{1 hr} = 17.2 mph[/tex]
A 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck has traveled a distance of
Answer:
250 m
Explanation:
1. First, we have to calculate the acceleration:
[tex]\boxed{\mathsf{v=v_o+a\cdot t}}[/tex]
where:
v = present velocity
vo = initial velocity
a = acceleration
t = time
2. Let us use given information and substitute in the expression above. Hence:
[tex]\mathsf{50=0+a\cdot10}\\\\\mathsf{50=10\cdot a}\\\\\mathsf{a=\dfrac{50}{10}}\\\\\therefore \boxed{\mathsf{a=5\,m/s^2}}}[/tex]
3. Now we can calculate traveled distance with Torricelli's equation:
[tex]\boxed{\mathsf{v^2=v_o^2+2\cdot a \cdot d}}[/tex]
where:
v = present velocity
vo = initial velocity
a = accelerarion
d = distance
4. So, we get:
[tex]\mathsf{50^2=0^2+2\cdot 5 \cdot d}\\\\\mathsf{2500=10\cdot d}\\\\\mathsf{d=\dfrac{2500}{10}}\\\\\therefore \boxed{\mathsf{d=250\,m}}[/tex]
Conclusion: during 10 seconds the truck has traveled a distance of 250 m.
Have a nice day! : )
If a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds. During the 10 seconds, the truck traveled a distance of 250 meters.
What are the three equations of motion?There are three equations of motion given by Newton
v = u + at
S = ut + 1/2×a×t²
v² - u² = 2×a×s
By using the first equation of motion,
v = u + at
50 = 0 + 10 a
a = 50/10
a= 5 meters/second²
As given in the problem a 1000-kilogram truck accelerates uniformly from rest, reaching a speed of 50 meters per second in 10 seconds.
The total distance traveled by truck,
S = ut + 0.5at²
S = 0 + 0.5 ×5 ×10²
S = 250 meters
Thus, the total distance traveled by truck would be 250 meters.
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For the microscope to be in focus, how far should the objective lens be from the specimen?
Answer:
p ≈ f_ objective Therefore for the object to be in focus it must be close to the focal length
Explanation:
A microscope is an optical instrument that uses two lenses, or a long focal length objective lens that forms a real image of the object and an eyepiece that forms a virtual image of the object. Therefore for the object to be in focus it must be close to the focal length
p ≈ f_ objective
p distance objetive
Rank these significant figures numbers from the least to the most
a. 357
b. 0.006
c. 9520.00
d. 9256.0
e. 700.003
f. 6010
Answer:
0.006<357<700.003<6010<9256.0<9520.00
A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%. What is her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started? Show all your work.
Her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s
What is mechanical energy?Mechanical energy is the combination of all the energy in motion represented by total kinetic energy and the total potential energy stored energy in the system which is represented by total potential energy.
As given in the problem A cyclist is standing still at the top of a hill and then begins to coast down the hill. The mass of the cyclist and bicycle is 64 kg in total. The cyclist’s gravitational potential energy is converted into kinetic energy with an efficiency of 52%
The potential energy is getting converted into kinetic energy with an efficiency of 52 %
1/2mv² = 0.52 (mgh)
v²= 1.04gh
v = √(1.04gh)
v= √(1.04×9.81×10)
v = 10.1 m/s
Thus, her speed when she reaches a point that is a vertical distance of 10 m lower than the point at which she started would be 10.1 m/s
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