Answer:
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
Explanation:
La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.
Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?
[tex]distancia=\frac{1 segundo*90 km}{3600 segundos}[/tex]
distancia= 0.025 km
Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?
[tex]distancia=\frac{0.025 km*1,000 metros}{1 km}[/tex]
distancia= 25 metros
La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h
La cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.
Para obtener la velocidad de la motocicleta en un segundo, necesitaremos convertir 90 km / h en metros por segundo
Usando la tasa de conversión;
1000 m = 1 km
1 hora = 3600 segundos
[tex]\frac{90km}{hr} = \frac{90km \times 1000 m}{1km \times 3600s} \\\\\frac{90km}{hr} = \frac{90,000 m}{3600s} =25m/s \\[/tex]
Esto muestra que la cantidad de metros que recorre una motocicleta en un segundo si viaja a una velocidad de 90 km / h es de 25 m / s.
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If the outer conductor of a coaxial cable has radius 2.6 mm , what should be the radius of the inner conductor so that the inductance per unit length does not exceed 50 nH per meter? Express your answer using two significant figures.
Answer:
Inner radius = 2 mm
Explanation:
In a coaxial cable, series inductance per unit length is given by the formula;
L' = (µ/(2π))•ln(R/r)
Where R is outer radius and r is inner radius.
We are given;
L' = 50 nH/m = 50 × 10^(-9) H/m
R = 2.6mm = 2.6 × 10^(-3) m
Meanwhile µ is magnetic constant and has a value of µ = µ_o = 4π × 10^(−7) H/m
Plugging in the relevant values, we have;
50 × 10^(-9) = (4π × 10^(−7))/(2π)) × ln(2.6 × 10^(-3)/r)
Rearranging, we have;
(50 × 10^(-9))/(2 × 10^(−7)) = ln((2.6 × 10^(-3))/r)
0.25 = ln((2.6 × 10^(-3))/r)
So,
e^(0.25) = (2.6 × 10^(-3))/r)
1.284 = (2.6 × 10^(-3))/r)
Cross multiply to give;
r = (2.6 × 10^(-3))/1.284)
r = 0.002 m or 2 mm
During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial speed v0 = 38 m/s at an angle θ = 35° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.
Part (a) Express the magnitude of the ball's initial horizontal velocity Or in terms of vo and 20%
Part (b) Express the magnitude of the ball's initial vertical velocity vOy in terms of vo and 0. 20%
Part (c) Find the ball's maximum vertical height Amat in meters above the ground.
Part (d) Create an expression in terms of vo-e, and g for the time-ur İt takes te ball to travel to its maximum vertical height.
Part (e) Calculate the horizontal distance in meters the ball has traveled when it returns to ground level.
Answer:
a) v₀ₓ = v₀ cos θ , b) v_{oy} = v₀ sin θ , c) y = v_{oy}² / 2g, y = 24.25 m
e) R = 138.46 m
Explanation:
This is a projectile launch exercise
a) let's use trigonometry to find the components of the initial velocity
cos θ = v₀ₓ / v₀
v₀ₓ = v₀ cos θ
v₀ₓ = 38 cos 35
v₀ₓ = 31.13 m / s
b) sin θ = [tex]v_{oy}[/tex] / v₀
v_{oy} = v₀ sin θ
v_{oy} = 38 sint 35
v_{oy} = 21 80 m / s
c, d) to find the maximum height, the vertical speed is zero
v_{y}² = v_{oy}² - 2 g y
0 = [tex]v_{oy}[/tex]² - 2 gy
y = v_{oy}² / 2g
let's calculate
y = 21.80 2 / (2 9.8)
y = 24.25 m
e) They ask to find the horizontal distance
for this we can use the expression of reaches
R = v₀² sin 2θ / g
let's calculate
R = 38² sin (2 35) / 9.8
R = 138.46 m
5.Which of the following does not affect rate of evaporation?
O Wind speed
O Surface area
O Temperature
O Insoluble heavy impurities
Answer:
D
Explanation:
Insoluble impurities would not change the constituent of the substance. Soluble would for example salt water takes longer time for the water to become vapour when subjected to the same temperature that normal water.
Wind would affect, the more windy the tendency for particles of the liquid to be moved into the atmosphere.
With an increase in surface area, the evaporation rate increase . Take a clue from water placed on the ground and exposed to the atmosphere and that same quantity of water is placed in a cup. That on the floor would evaporate faster.
Similarly the higher the temperature a substance is subjected to the easier is it's rate of evaporation. Take for instance water in a cup placed in the sun and that same placed in a room with mild temperatures than that of the sun.With time that in the sun decreases in volume faster than that in the room.
The only force acting on a 3.2 kg canister that is moving in an xy plane has a magnitude of 6.7 N. The canister initially has a velocity of 3.3 m/s in the positive x direction, and some time later has a velocity of 6.9 m/s in the positive y direction. How much work is done on the canister by the 6.7 N force during this time
Answer:
The work done by the force is 5.76 J
Explanation:
Given;
mass of canister , m = 3.2 kg
magnitude of force, f = 6.7 N
initial velocity of the canister on x-axis, [tex]v_i[/tex]= 3.3i m/s
final velocity of the canister on y- axis, [tex]v_f[/tex] = 6.9j m/s
The work done on the canister = change in the kinetic energy of the canister
[tex]W = K.E_f - K.E_i[/tex]
where;
K.Ei is the initial kinetic energy
K.Ef is the final kinetic energy
The initial kinetic energy:
[tex]K.E_i = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_i = \frac{1}{2} *3.2\sqrt{3.3^2 +0^2+0^2}\\\\K.E_i = 5.28 \ J[/tex]
The final kinetic energy:
[tex]K.E_f = \frac{1}{2} *m\sqrt{i^2 +j^2+z^2}\\\\K.E_f = \frac{1}{2} *3.2\sqrt{0^2 +6.9^2+0^2}\\\\K.E_f = 11.04 \ J\\[/tex]
W = 11.04 - 5.28
W = 5.76 J
Therefore, work done on the canister by the 6.7 N force during this time is 5.76 J
the distance between 2 station is 5400 m find the time taken by a train to cover this distance, if the train travels with speed 60m/s
Answer:
I dont know bro
Explanation:
Ask an expert
Answer:
Time=90s
Explanation:
Speed=distance /time
[tex]60 = \frac{5400}{t} where \: t \: is \: time \\60t = 5400 \\ t = \frac{5400}{60} \\ t =90 \\ hope \: this \: helps..good \: luck [/tex]
The self-referencing effect refers to ________.