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2 years ago

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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti

on?
A. It's not moving.
B.
It's moving at a constant speed.
O C. It's moving at a constant velocity.
OD. It's speeding up.

1 answer:

Art [367]2 years ago

7 0

Answer:

B

Explanation:

on USAtestprep

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A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,

Art [367]

Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$ ..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{5.6}{3}$

let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

$-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}$

$\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}$

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

6 0

3 years ago

La distancia por carretera de Chitré a Parita es de 12 km; exprese en pies ésta distanciaLa distancia por carretera de Chitré a

densk [106]

Answer:

La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.

Explanation:

La regla de tres es una forma de resolver problemas de proporcionalidad entre tres valores conocidos y un valor desconocido, estableciendo una relación de proporcionalidad entre todos ellos.

Si la relación entre las magnitudes es directa, es decir, cuando una magnitud aumenta, también lo hace la otra (o cuando una magnitud disminuye, también lo hace la otra), se debe aplicar la regla directa de tres. Para resolver una regla directa de tres, se debe seguir la siguiente fórmula, siendo a, b y c los valores conocidos y x el valor a determinar:

a ⇒ b

c ⇒ x

Entonces $x=\frac{c*b}{a}$

La regla directa de tres es la regla que se aplica en este caso donde hay un cambio de unidades. Para realizar esta conversión de unidades, primero debes saber que 1 km = 3280,84 pies. Entonces, si 1 km son 3280,84 pies, ¿cuántos pies son 12 km?

1 km ⇒ 3280.84 pies

12 km ⇒ x

$x=\frac{12 km*3280.84 pies}{1 km}$

x= 39370.08 pies

<u><em>La distancia por carretera de Chitré a Parita es de 12 km o 39370.08 pies.</em></u>

7 0

2 years ago

What is the station's orbital speed? the radius of earth is 6.37×106m, its mass is 5.98×1024kg.

Paha777 [63]

Answer:

$v_o=2503.08\frac{m}{s}$

Explanation:

Orbital velocity is the speed that a body that orbits around another body must have, for its orbit to be stable. For orbits with small eccentricity and when one of the masses is almost negligible compared to the other mass, like in this case, the orbital speed is given by:

$v_o=\sqrt{\frac{GM}{r}}$

Where M is the greater mass around which this negligible body is orbiting, r is the radius of the greater mass and G is the universal gravitational constant. So:

$v_o=\sqrt{\frac{6.674*10^{-11}\frac{m^3}{kg\cdot s^2}(5.98*10^{24}kg)}{6.37*10^6m}}\\v_o=2503.08\frac{m}{s}$

3 0

3 years ago

Read 2 more answers

A turtle and a rabbit are in a 150 meter race. The rabbit decides to give the turtle a 1 minute head start. The turtle moves at

yan [13]

Answer:

a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

e) The time required for the rabbit to finish the race can be determined by the following expression:

$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

$t' = 11.667\,s$

The time required for the rabbit from rest to maximum speed is 11.667 seconds.

f) The animal with the lowest time wins the race. Now, each running time is determined:

Turtle:

$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

7 0

3 years ago

A 70 kg man's arm, including the hand, can be modeled as BIO 75-cm-long uniform rod with a mass of 3.5 kg. When the man raises b

Butoxors [25]

Answer:

Δy = 7.1 cm

Explanation:

The center of mass of a body is defined

$y_{cm}$ = 1 /M ∑ $m_{i} y_{i}$ i

Where M is the total mass of the body, m mass of each part and ‘y’ height

Let's apply this equation to our case

We locate the reference system on the shoulders

The height of the arms is at its midpoint

y = -75/2 = 37.5 cm

With arms down

$y_{cm}$ = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)

$y_{cm}$ = 1/70 (63 y)₀ - 7 37.5)

With arms up

$y_{cm}$ ’= 1/70 (63 y₀ + 3.5 y + 3.5 y)

$y_{cm}$ ’= 1/70 (63y₀ + 7 35.5)

let's subtract the two equations

$y_{cm}$ ’ - $y_{cm}$ = 1/70 2 (7 35.5)

Δy = $y_{cm}$ ’ - $y_{cm}$ = 2 7 35.5 / 70

ΔY = 7.1 cm

7 0

2 years ago