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Sphinxa [80]
3 years ago
7

A lever is being used to move a heavy stone from a garden. The stone weighs 250 newtons. The board used as a lever is 6 meters l

ong. If two of the board's meters will be used as he resistance arm how much force will be needed to move the rock?

Physics
1 answer:
DerKrebs [107]3 years ago
7 0

Answer:

Force = 125 [N]

Explanation:

In the attached image we can see a sketch of the lever system.

And if we make a sum of moments at the point O equal to zero (0).

In the equation showed in the image, we can determinate the force that we need

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Substituting the values of the variables into the equation, we have

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Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
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Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

    Stan's speed = 42.9 m/s

Explanation:

<u>Given:</u>

  • u_k = initial speed of Kathy = 0 m/s
  • u_s = initial speed of Stan = 0 m/s
  • a_k = acceleration of Kathy = 4.73\ m/s^2
  • a_s = acceleration of Stan = 3.9\ m/s^2

<u>Assumptions:</u>

  • v_k = final speed of Kathy when see catches Stan
  • v_s = final speed of Stan when Kathy catches him
  • s_k = distance traveled by Kathy to catch Stan
  • s_s = distance traveled by Stan when Kathy catches him
  • t_k = time taken by Kathy to catch Stan = t
  • t_s = time interval in which Kathy catches Stan = t+1

Part (a):

 Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10\\

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9

The speed of Stan at the instant Kathy catches him is 42.9 m/s.

7 0
3 years ago
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