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A 242-g block is pressed against a spring of force constant 1.62 kN/m until the block compresses the spring 10.0 cm. The spring

hammer [34]

Answer:

a) = 3.94 m

b) = 3.15 m

Explanation:

Given

Mass of the block, m = 242 g

Force constant, k = 1.62 kN/m

Compression of the spring, x = 10 cm

Angle of inclination = 60°

a) if we equate the energy at the bottom of the ramp to the energy at a distance d up the ramp, we have

1/2kx² = mgh where, h = dsinΦ

1/2kx² = mgdsinΦ

1/2 * 1.62*10^3 * 0.1² = 0.242 * 9.8 * dsin 60

1/2 * 16.2 = 2.3716 * d sin 60

d sin 60 = 8.1 / 2.3716

0.866 d = 3.415

d = 3.415 / 0.866

d = 3.94 m

b) net force on the block = mgd sin 60 + µ mgd cos 60

8.1 = d[mg sin 60 + µ mg cos 60]

8.1 = d [0.242 * 9.8 * 0.866 + 0.44 * 0.242 * 9.8 * 0.5]

8.1 = d (2.05 + 0.52)

8.1 = 2.57 d

d = 8.1 / 2.57

d = 3.15 m

3 0

3 years ago

Read 2 more answers

A typical adult human has a mass of about 70 kg.

Misha Larkins [42]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
a. <span>FM GmMmr2
</span>= 6.67 x 10-11N.m2kg27 .35 x 1022 kg 70 kg 3.78 x 108 m2
<span>= 2.40 x 10-3 N

b. </span><span>FE GmEmr2
= 6.67 x 10-11 N.m2kg 25 .97 x 1034 kg (70kg) 6.38 x 106 m2
=685 N
FMFE 2.40 x 10-3N685 N= 0.0004%</span>

3 0

3 years ago

A 1.65 kg mass stretches a vertical spring 0.260 m If the spring is stretched an additional 0.130 m and released, how long does

Irina-Kira [14]

Answer:

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

Explanation:

We notice that block-spring system depicts a Simple Harmonic Motion, whose equation of motion is:

$y(t) = A\cdot \cos \left(\sqrt{\frac{k}{m} }\cdot t +\phi\right)$ (1)

Where:

$y(t)$ - Position of the mass as a function of time, measured in meters.

$A$ - Amplitude, measured in meters.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the block, measured in kilograms.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The spring is now calculated by Hooke's Law, that is:

$k = \frac{m\cdot g}{\Delta y}$ (2)

Where:

$g$ - Gravitational acceleration, measured in meters per square second.

$\Delta y$ - Deformation of the spring due to gravity, measured in meters.

If we know that $m=1.65\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $\Delta y = 0.260\,m$ , then the spring constant is:

$k = \frac{(1.65\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{0.260\,m}$

$k = 62.237\,\frac{N}{m}$

If we know that $A = 0.130\,m$ , $k = 62.237\,\frac{N}{m}$ , $m=1.65\,kg$ , $x(t) = 0\,m$ and $\phi = 0\,rad$ , then (1) is reduced into this form:

$0.130\cdot \cos (6.142\cdot t)=0$ (1)

And now we solve for $t$ . Given that cosine is a periodic function, we are only interested in the least value of $t$ such that mass reaches equilibrium position. Then:

$\cos (6.142\cdot t) = 0$

$6.142\cdot t = \cos^{-1} 0$

$t = \frac{1}{6.142}\cdot \left(\frac{\pi}{2} \right)\,s$

$t \approx 0.255\,s$

The system will take approximately 0.255 seconds to reach the (new) equilibrium position.

4 0

3 years ago

Which is longer, 10 cm or .01 m?

motikmotik

Answer:

They´re the same.

Explanation:

Someone deleted my answer. And my brainly is gone..

Have a good night ma´am/sir.

Be safe!

6 0

3 years ago

A playground ride requires 21 seconds to make one complete revolution

luda_lava [24]

The angular speed of the playground ride is determined as 0.3 rad/s.

<h3>What is angular speed?</h3>

Angular speed is the rate at which an object changes it angles which we measure in radians in a given time.

<h3>Angular speed of the ride</h3>

The angular speed of the ride if the ride makes one complete revolution is calculated as follows;

ω = θ/t

ω = 2π/t

where;

ω is angular speed of the ride
t is time of motion of the ride

one complete revolution = 2π radians

ω = 2π/21

ω = 0.3 rad/s

Thus, the angular speed of the playground ride is determined as 0.3 rad/s.

Learn more about angular speed here: brainly.com/question/24158647

#SPJ1

The complete question is below;

A playground ride requires 21 seconds to make one complete revolution, what is angular speed of the ride in radian per second.

3 0

2 years ago