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allochka39001 [22]
3 years ago
8

A homeowner is trying to move a stubborn rock from his yard. By using a a metal rod as a lever arm and a fulcrum (or pivot point

) the homeowner will have a better chance of moving the rock. The homeowner places the fulcrum a distance ????=0.266 md=0.266 m from the rock, which has a mass of 325 kg,325 kg, and fits one end of the rod under the rock's center of weight. If the homeowner can apply a maximum force of 663 N at the other end of the rod, what is the minimum total length L of the rod required to move the rock? Assume that the rod is massless and nearly horizontal so that the weight of the rock and homeowner's force are both essentially vertical.
Physics
1 answer:
pogonyaev3 years ago
6 0

Answer:

L = 1.545 m

Explanation:

Let the total length of the rod is L

now the torque must applied on the other end of the rod so that it will balance the torque due to weight of rock on other side of fulcrum

so we will have

mg \times d = F(L - d)

so we have

325\times 9.8 \times 0.266 = F(L - 0.266)

F = 663 N

848 = 663(L - 0.266)

L = 1.545 m

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Explanation:

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What is the change of momentum on a 45,000 kg airplane when it changes its velocity from 242 m/s to 258 m/s?
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720,000

Explanation:

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2 years ago
A 2.72 A current flows through a wire for
agasfer [191]
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2 years ago
Vectors a and b have scalar product â6.00, and their vector product has magnitude +9.00. what is the angle between these two vec
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Answer:

Value of angle between vector a and b is 56.30^{\circ}.

Explanation:

Vectors a and b have scalar product 6.00

Let \theta be the angle between a and b.

\vec{a}.\vec{b} = 6

ab cos \theta = 6 ...(1)

Vectors a and b have magnitude of vector product 9.00

\vec{a} \times\vec{b} = 9

ab sin \theta = 9 ...(2)

Dividing equation (2) by (1) we get

\frac{ab sin \theta}{ab cos \theta}  = \frac{9}{6}

tan \theta = 1.5

\theta = tan ^{-1} (1.5)

\theta = 56.30^{\circ}

Thus, value of angle between vector a and b is 56.30^{\circ}.

3 0
3 years ago
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown
nadya68 [22]

(a) 296.6 m

The motion of the stone is the motion of a projectile, thrown with a horizontal speed of

v_x = 15.0 m/s

and with an initial vertical velocity of

v_{y0} = -20.0 m/s

where we have put a negative sign to indicate that the direction is downward.

The vertical position of the stone at time t is given by

y(t) = h + v_{0y} t + \frac{1}{2}gt^2 (1)

where

h is the initial height

g = -9.81 m/s^2 is the acceleration due to gravity

The stone hits the ground after a time t = 6.00 s, so at this time the vertical position is zero:

y(6.00 s) = 0

Substituting into eq.(1), we can solve to find the initial height of the stone, h:

0 = h + v_{0y} y + \frac{1}{2}gt^2\\h = -v_{0y} y - \frac{1}{2}gt^2=-(-20.0 m/s)(6.00 s) - \frac{1}{2}(9.81 m/s^2)(6.00 s)^2=296.6 m

(b) 176.6 m

The balloon is moving downward with a constant vertical speed of

v_y = -20 m/s

So the vertical position of the balloon after a time t is

y(t) = h + v_y t

and substituting t = 6.0 s and h = 296.6 m, we find the height of the balloon when the rock hits the ground:

y(t) = 296.6 m + (-20.0 m)(6.00 s)=176.6 m

(c) 198.2 m

In order to find how far is the rock from the balloon when it hits the ground, we need to find the horizontal distance covered by the rock during the time of the fall.

The horizontal speed of the rock is

v_x = 15.0 m/s

So the horizontal distance travelled in t = 6.00 s is

d_x = v_x t = (15.0 m/s)(6.00 s)=90 m

Considering also that the vertical height of the balloon after t=6.00 s is

d_y = 176.6 m

The distance between the balloon and the rock can be found by using Pythagorean theorem:

d=\sqrt{(90 m)^2+(176.6 m)^2}=198.2 m

(di) 15.0 m/s, -58.8 m/s

For an observer at rest in the basket, the rock is moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer at rest in the basket is

v_y (t) = gt

Substituting time t=6.00 s, we find

v_y = (-9.8 m/s)(6.00 s)=-58.8 m/s

(dii) 15.0 m/s, -78.8 m/s

For an observer at rest on the ground, the rock is still moving horizontally with a velocity of

v_x = 15.0 m/s

Instead, the vertical velocity of the rock for an observer on the ground is now given by

v_y (t) = v_{0y} + gt

Substituting time t=6.00 s, we find

v_y = (-20.0 m/s)+(-9.8 m/s)(6.00 s)=-78.8 m/s

6 0
2 years ago
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