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3 years ago

Find an equation for the plane that passes through the point [1, 1, 8] and that has [-1, -6, 3] as normal vector.

1 answer:

3 0

The normal to the plane is (Fx, Fy, Fz) = (1, 1, 8)
We have (x0, y0, z0) = (-1, - 6, 3)

Equation of the plane:
Fx(x - x0) + Fy(y - y0) + Fz(z - z0) = 0
=> (1)(x + 1) + (1)(y + 6) + (8)(z - 3) = 0
=> x + 1 + y + 6 + 8z - 24 = 0
=> x + y + 8z = 17

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During a collision with a wall, the velocity of a 0.200-kg ball changes from 20.0 m/s toward the wall to 12.0 m/s away from the

mixas84 [53]

So the problem ask to calculate the magnitude of the average force applied to the ball if its mass is 0.2kg changes its velocity from 20m/s to 12m/s and the time contact with the ball with the wall is 60 ms. In my calculation the best answer would be 107N.

8 0

3 years ago

How much force is needed to accelerate a 1,500 kg car at a rate of 7 m/s2?

Sidana [21]

Newton's 2nd law of motion:

   Force = (mass) x (acceleration)
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   (1,500 kg) x (7 m/s²) = 10,500 newtons .

4 0

4 years ago

Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something li

Troyanec [42]

Which of the following statements about this experiment is FALSE? Before each trial one should reshape the bob into something like a ball. You may assume the collision between the bob and the box is completely inelastic. The initial position for the box should be just touching the pendulum bob when it is hanging straight down. To make the box move, the pendulum bob should hit close to the bottom of the box during the collision is given below

Explanation:

So, The bob is held at angle theta from initial position, P. Find the displacement from P.

displacement= L sin Θ. I hope you get it here. The pendulum forms a triangle with length L, the initial position and horizontal displacement.

So,the point is at displacement point the kinetic energy of bob is 0 and when it is released it will have maximum kinetic energy at P. It will hit the box and all he kinetic energy of bob will get transferred into driving force(F.D) of box. Now kinetic energy = Force * displacement.

K.E= 0.5 m V^2= F.D * Lsin Θ.

Find F.D

F.D = (0.5 m V^2) / Lsin Θ.

Now for the box, F.D - Friction = m(box) a.

Friction = F.D + m(box) a.

Find the accelaration of box from an equation of motion. ( u=0, find displacement of box,s,time taken and so on)

U get the friction. Now,

coeff of friction = µ = friction/ reaction.

Note that reaction here = weight of box= m(box) g. g= acc of free fall= 9.81.

So here u go.. U get the coeff of friction.. I hope am right here.. and made no mistake.. Anyway try it with the values to confirm! ;)

7 0

3 years ago

To measure the coefficient of kinetic friction by sliding a block down an inclined plane the block must be in equilibrium.

lozanna [386]

Answer:

Explanation:

A block sliding down an inclined plane, is subject to two external forces along the slide.
One is the component of gravity (the weight) parallel to the incline.
If the inclined plane makes an angle θ with the horizontal, this component (projection of the downward gravity along the incline, can be written as follows:

$F_{gp} = m*g* sin \theta (1)$

(taking as positive the direction of the movement of the block)

The other force, is the friction force, that adopts any value needed to meet the Newton's 2nd Law.
When θ is so large, than the block moves downward along the incline, the friction force can be expressed as follows:

$F_{f} = \mu_{k} * N (2)$

The normal force, adopts the value needed to prevent any vertical movement through the surface of the incline:

$N = m*g* cos \theta (3)$

In equilibrium, both forces, as defined in (1), (2) and (3) must be equal in magnitude, as follows:

$m*g* sin \theta = \mu_{k} * m*g* cos \theta$

As the block is moving, if the net force is 0, according to Newton's 2nd Law, the block must be moving at constant speed.
In this condition, the friction coefficient is the kinetic one (μk), which can be calculated as follows:

$\mu_{k} = tg \theta$

8 0

3 years ago

Jason is taking a physical exam and has to do push ups for sixty seconds. He does 30 pushups in the sixty seconds. Each push up

Alexandra [31]

Answer: 2 seconds is the unit rate.

Explanation:

We know that Jason does 30 pushups in 60 seconds at a constant rate, and we know that each push up takes 2 seconds.

then, we can write this as 30 pushups/60 seconds = (1/2) pushups per second.

Here, two seconds represents the time needed to do one pushup, is the unit rate (this means that we need 2 seconds to have a unit "one pushup")

8 0

3 years ago