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3 years ago

An elf pushes a sleigh with force of 9N over a distance of 11m. How much work did the elf. Do on the sleigh

Physics

1 answer:

Whitepunk [10]3 years ago

5 0

Answer:

Explanation:

The work done by an object can be found by using the formula

workdone = force × distance

From the question we have

workdone = 9 × 11

We have the final answer as

Hope this helps you

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A boat leaves the dock at t = 0.00 s and, starting from rest, maintains a constant acceleration of (0.461 m/s2)i relative to the

liberstina [14]

Answer:

At t=4.82 s, the boat is moving at 3.464 m/s.

At t=4.82 s, the boat is 13.112 m from the dock.

Explanation:

The speed of the boat in j'th direction remains constant for all times (vj=2.16 m/s), however, the speed in i'th direction is changing due to the constant acceleration (0.461 m/s^2)i.

In order to find the velocity of the boat a t=4.82 s, first we need to compute the velocity of the boat relative to the water in the i direction (vi_b) at t=4.82 s:

vi_b = a*t = (0.461 m/s^2)*(4.82 s) = 2.222 m/s

Now, we add this velocity to the velocity of the water in the i direction:

vi = vi_b + vi_w = 2.222 m/s + 0.486 m/s = 2.708 m/s

Therefore, the speed of the boat at t = 4.82 s is: v = (vi, vj) = (2.708, 2.16) m/s. Finally, to find its speed, we just calculate the magnitude of v and obtain that the speed is: 3.464 m/s.

For the second question, first we will find the distance that the boat moved in the i'th direction and then in the j'th direction.

The speed in the i'th direction, for all times, is given by:

(0.485 + 0.461*t) and in order to find the distance advanced in the i'th direction (di) during 4.82 s, we need to integrate this velocity:

di = 0.485*t + (0.461*t^2)/2 (evaluated from t=0 to t =4.82) = 0.485*(4.82) + (0.461*(4.82)^2)/2 = 2.337 + 5.634 = 7.971 m

The speed in j'th direction, for all times, is given by:

2.16 and in order to find the distance advanced in the j'th direction (dj) during 4.82 s, we need to integrate this velocity:

dj = 2.16*t (evaluated from t=0 to t =4.82) = (2.16)*(4.82) = 10.411 m

Using Pythagoras' Theorem, we find that the the boat is at 13.112 m from the dock at t = 4.82 s.

4 0

3 years ago

A boater is floating 1060 meters down a river. The person takes 40 minutes. At what speed is the boat moving?

Mashutka [201]

26.5, I’m not sure if it’s right but 1060/40 gets that

7 0

2 years ago

In general, an organism will be more likely to develop phobias of __________.

BARSIC [14]

Answer:

a) dangers faced during natural circumstances

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2 years ago

Two soccer players kick the same 2-kg ball at the same time in opposite directions one kicks with a force of 15 n the other kick

Illusion [34]

Because the direction of the kicks are opposite, the net force between the applied forces is their difference.
Fn = F₂ - F₁
Substituting,
Fn = 15 N - 5 N
Fn = 10 N

From Newton's second law of motion,
Fn = m x a
where m is mass and a is acceleration. Manipulating the equation so that we are able to calculate for a,
a = Fn / m

Substituting,
a = (10 N) / 2 kg
a = 5 m/s²

<em>ANSWER: 5 m/s²</em>

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3 years ago

Read 2 more answers

un bus sale de florencia hacia el municipio de san vicente del caguan con una velocidad de 30km/h. 2 horas mas tarde sale otro b

goldenfox [79]

Recuerda que distancia=velocidad x tiempo ( $d=vt$ ). vamos a llamar $d _{1}$ a la distancia que reccore el primer bus, y $d _{2}$ la distancia que recoore el segundo. Los se econtraran cuando la ditancia recorrida por cada bus sea la misma, es decir $d _{1} =d _{2}$ .
Ahora sabemos que la velocidad del primer bus es 30 km/h y su tiempo es $t$ ; en cuanto al segundo bus, su velocidad es 45 km/h y como sale 2 horas mas tarde su tiempo será $t-2$ . Ahora podemos reemplazar los valores en nuestra ecuación de de distanica para hallar el tiempo:
$30t=45(t-2)$
$30t=45t-90$
$30t-45t=-90$
$-15t=-90$
$t= \frac{-90}{-15}$
$t=6$
Los buses se encontraran en 6 horas, 4 horas después de haber salido el segundo bus.

Ahora para hallar la distancia a la que los buses estaran de <span>san vicente del caguan solo necesitaremos reemplazar la el tiempo en nuestra equación de distancia para el primer bus:
</span> $d _{1} =(30)(6)$
$d _{1} =180$
<span>En el punto de encuentro los buses estarán a 180 km de San Vicente del Caguan.
</span>

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3 years ago