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AleksandrR [38]

3 years ago

7

A spring can be compressed to a limit of -0.50 m and can be stretched a maximum of 0.75-m within the limit of a spring constant

k = 25.0 N/m. What is the range of force that can be applied to this spring.

1 answer:

kvv77 [185]3 years ago

4 0

Answer:

try google or the wikipedia

Explanation:

You might be interested in

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$

$R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2$

$R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2$

$-R_1 \sin(50^\circ) = -\dfrac{mg}2$

$R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}$

Solve for $R_2$ .

$\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0$

$\dfrac{R_2}2 = -mg\cot(110^\circ)$

$R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}$

8 0

2 years ago

hello friends,i need your help my home work now in physics,topic:motion.40 marks +brainliest if correct .

fenix001 [56]

Answer:

See below

Explanation:

Vertical position is given by

df = do + vo t - 1/2 a t^2 df = final position = 0 (on the ground)

do =original position = 2 m

vo = original <u>VERTICAL</u> velocity = 0

a = acceleration of gravity = 9.81 m/s^2

THIS BECOMES

0 = 2 + 0 * t - 1/2 ( 9.81)t^2

to show t =<u> .639 seconds to hit the ground </u>

During this .639 seconds it flies horizontally at 10 m/s for a distance of

10 m/s * .639 s =<u> 6.39 m </u>

5 0

1 year ago

A van is traveling with an initial velocity of 12 m/s. The driver takes a time of 45 seconds to speed up to a velocity of 20 m/s

Rufina [12.5K]

Initial velocity=u=12m/s
Final velocity=v=20m/s
Time=t=45s

$\\ \rm\hookrightarrow Acceleration=\dfrac{v-u}{t}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{20-12}{45}$

$\\ \rm\hookrightarrow Acceleration=\dfrac{8}{45}$

$\\ \rm\hookrightarrow Acceleration=0.1m/s^2$

Now

Distance=s

$\\ \rm\hookrightarrow v^2-u^2=2as$

$\\ \rm\hookrightarrow (20)^2-12^2=2(0.1)s$

$\\ \rm\hookrightarrow 400-144=0.2s$

$\\ \rm\hookrightarrow 256=0.2s$

$\\ \rm\hookrightarrow s=\dfrac{256}{0.2}$

$\\ \rm\hookrightarrow s=1280m$

4 0

3 years ago

The weight (W) of an object is the product of its mass (m) in kilograms and g, the acceleration due to gravity in meters/second2

Nookie1986 [14]

The weights in newtowns for the given masses are

<span> masses 22.1, 33.5, 41.3, 59.2, 78
weights 216.58N 328.3N 404.74N 580.16N 764.4N

e.g, for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>

7 0

3 years ago

An engineering firm is designing a ski lift. The wire rope needs to travel with a linear velocity of 2.0 meters per second, and

Mazyrski [523]

Answer:

The diameter of the bull-wheel is 3.82

Explanation:

Given that,

Velocity = 2.0 m/s

Angular velocity = 10 rev/m

$\omega=10\times\dfrac{2\pi}{60}$

$\omega=1.0472\ rad/s$

We need to calculate the diameter of bull-wheel

Using formula of angular velocity

$v= r\omega$

$r=\dfrac{v}{\omega}$

Put the value into the formula

$r=\dfrac{2.0}{1.0472}$

$r=1.91\ m$

The diameter of the bull-wheel

$D=2r$

$D=2\times1.91$

$D=3.82\ m$

Hence, The diameter of the bull-wheel is 3.82 m.

6 0

3 years ago