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1 year ago

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

1 answer:

3 0

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

<h3>What is displacement ?</h3>

A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.

The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector

To learn more about displacement from the given link:

brainly.com/question/321442

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Are There some sounds for which we are all deaf

lana66690 [7]

Answer: Infrasound is the span of low-frequency sounds below 20 Hz that fall below the hearing range of humans. While these sounds escape our ears, scientific instruments can detect them—and tell us some interesting things about the planet.

Explanation:

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3 years ago

An object is 1.0 cm tall and its inverted image is 4.0 cm tall. what is the exact magnification

gtnhenbr [62]

-4.0 (negative since the image is inverted)

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3 years ago

Margaret wants to go for a swim, and decides to jump in using the diving board that measures 3-m long.

otez555 [7]

Answer:

The horizontal component of her velocity is approximately 1.389 m/s

The vertical component of her velocity is approximately 7.878 m/s

Explanation:

The given question parameters are;

The initial velocity with which Margaret leaps, v = 8.0 m/s

The angle to the horizontal with which she jumps, θ = 80° to the horizontal

The horizontal component of her velocity, vₓ = v × cos(θ)

∴ vₓ = 8.0 × cos(80°) ≈ 1.389

The horizontal component of her velocity, vₓ ≈ 1.389 m/s

The vertical component of her velocity, $v_y$ = v × sin(θ)

∴ $v_y$ = 8.0 × sin(80°) ≈ 7.878

The vertical component of her velocity, $v_y$ ≈ 7.878 m/s.

6 0

3 years ago

A circular curve of radius 150 m is banked at an angle of 15 degrees. A 750-kg car negotiates the curve at 85.0 km/h without ski

Crazy boy [7]

Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.

Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:

Pcos 15°-N=0

Psin15°-f= m*ac

from the first we obtain N, the normal force

N=750Kg*9.8* cos (15°)= 7.1 *10^3 N

Then to calculate the frictional force (f) we can use the second equation

f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r

f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N

6 0

3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago