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kifflom [539]
2 years ago
12

The 3kg object in figure is released from rest at height of on a curved frictionless ramp. at the foot of the ramp is a spring o

f force constant 400N/m. the object slides down the ramp and into the spring,compressing it a distance x before coming momentarily to rest.
a) find x
b) describe the motion of the object (if any) after the block momentarily comes to rest?
Physics
1 answer:
mote1985 [20]2 years ago
5 0

(a) The compression of the spring is 7.4 cm.

(b) When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

<h3>Compression of the spring [ x ]</h3>

The compression of the spring is calculated by applying Hooke's law as follows;

F = kx

x = F/k

x = (mg)/k

x = (3 x 9.8)/400

x = 0.074 m

x = 7.4 cm

When the object comes to rest its potential energy from the given height will be converted into elastic potential energy of the spring.

Learn more about potential energy here: brainly.com/question/14427111

#SPJ1

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What is the correct answer?
dexar [7]

Answer:

length=2

Explanation:

You simply find the zeros of the quartic polynomial

The factored form is (x+1)(x-5)(x-2)=0

The zeros become -1,5 and 2

Dimension 5 is already given and the question says length>width and 2>-1

Therefore length=2

5 0
3 years ago
A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi
Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

\Delta x_1=0.50m-0.35m=0.15m

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

k\Delta x_1=m_1g

Solving for the spring constant k, we get:

k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m

Finally, we sum this to the unstretched length to obtain the length of the spring:

x_2=0.35m+0.088m=0.44m

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0
4 years ago
Use the fact that the speed of light in a vacuum is about 3.00 × 108 m/s to determine how many kilometers a pulse from a laser b
Morgarella [4.7K]
The first thing that needs to be done is to find everything in the same units.  12 hours becomes 43200 seconds.  Then find the distance traveled by light in that amount of time.  Using the formula v=d/s, manipulate it so it looks like d=v*s.  Then plug in the values: d=(3x10^8)*43200,  d=1.3x10^13m.  But you need to find this in kilometers.  To do this, simply divide your answer by one thousand.  Thus, a laser beam would travel 1.3x10^10 kilometers in 12 hours.
5 0
4 years ago
A particular cylindrical bucket has a height of 36.0 cm, and the radius of its circular cross-section is 15 cm. The bucket is em
Andrew [12]

Answer:

12 cm

Explanation:

P_1 = Initial pressure = P_a=1\times 10^5\ Pa

P_2 = Final pressure = P_a+\rho_w gh

h = Depth of cylinder = 36 cm

g = Acceleration due to gravity = 10 m/s²

\rho_w = Density of water = 1000 kg/m³

h_1 = Depth of lake = 20 m

From the ideal gas relation we have

P_1V_1=P_2V_2\\\Rightarrow P_a(\pi r^2h)=(P_a+\rho_w gh_1)\pi r^2h'\\\Rightarrow 1\times 10^5\times 36=(1\times 10^5+1000\times 10\times 20)h'\\\Rightarrow h'=\dfrac{1\times 10^5\times 36}{1\times 10^5+1000\times 10\times 20}\\\Rightarrow h'=12\ cm

The height of the cylinder of air in the bucket when the bucket is at the given depth is 12 cm

6 0
3 years ago
A fidget spinner experiences a constant torque of 1.4 N.m. If the spinner is initially at rest, what is its angular momentum 2.0
motikmotik

Answer:

2.8N-msec

Explanation:

We have given torque \tau =1.4N-m

Initial time t_1=0\ sec

Final time t_2=2\ sec

There is relation between angular momentum and torque

that is \frac{dL}{dt}=\tau

dL=\tau dt

\int dL=\tau \oint_{t_1}^{t_2}dt

L=\tau (t_1-t_2)=1.4\times (2-0)=2.8N-msec

4 0
4 years ago
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