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docker41
11 days ago
8

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

ion while momentarily coming to rest.
1)If the initial speed of the box were doubled, how far x2 would the spring compress?

x2 = ?2 x1
x2 = 2 x1
x2 = 4 x1
Physics
1 answer:
serg [1.1K]11 days ago
7 0

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

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Which, if any, of the following statements concerning the work done by a conservative force is NOT true? All of these statements
Yuliya22 [1153]

Answer:

If the starting and ending points are identical, the overall work equals zero.

Explanation:

Option (D) is correct.

A force is considered conservative when the work performed by it while moving an object from point A to point B does not rely on the path taken and remains consistent across all paths. The work done is determined solely by the initial and final locations of the particle. Thus, when the initial and final positions in a conservative field coincide, the work is said to be zero.

8 0
12 days ago
Rosa studies the position-time graph of two race cars. A graph titled Position versus Time shows time in hours on the x axis, nu
Sav [1105]

Answer:

B. Truck X was ahead, not truck Y.

Explanation:

Let's analyze the information provided.

Truck X moved from the point (0,20) to (2.8,50). This indicates that it began at the 20th kilometer and reached 50 km in 2.8 hours. Thus, its speed is v1 = (s2 - s1) / t

v1 = (50 - 20) / 2.8

v1 = 10.7 km/h

Given that it started from the 20th km, it indeed had a head start. Since the line on the graph is linear, this shows its speed was constant without any change in direction.

On the other hand, Truck Y's movement went from the origin (0,0) to (5,20), meaning it took 5 hours to travel 20 km, resulting in a speed of v2 = 20 / 5

v2 = 4 km/h

Again, the straightness of its graph line signifies it maintained a constant speed in a single direction.

Thus, it is evident that Rosa erred in her assumption that Truck Y had a head start.

5 0
7 days ago
Read 2 more answers
Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
ValentinkaMS [1149]

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

V=\frac {1}{3}\times \pi\times r^2\times h

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Additionally, we differentiate the volume expression in relation to time:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given that \frac {dV}{dt} = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

The rate at which the root beer level is decreasing is 0.08603 cm/s.

3 0
11 days ago
Two planets having equal masses are in circular orbit around a star. Planet A has a smaller orbital radius than planet B. Which
serg [1198]

Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.

6 0
13 days ago
In 2014, the Rosetta space probe reached the comet Churyumov Gerasimenko. Although the comet's core is actually far from spheric
ValentinkaMS [1149]

To tackle this issue, we will utilize concepts related to gravity based on Newtonian definitions. To find this value, we'll apply linear motion kinematic equations to determine the required time. Our parameters include:

Comet mass M = 1.0*10^{13} kg

Radius r = 1.6km = 1600 m

The rock is released from a height 'h' of 1 m above the surface.

The relationship for gravity's acceleration concerning a body with mass 'm' and radius 'r' is described by:

g = \frac{GM}{R^2}

Where G represents the gravitational constant and M denotes the mass of the planet.

g = \frac{(6.67408*10^{-11})(1*10^{13})}{1600^2}

g = 2.607*10^{-4} m/s^2

Now, let’s compute the time value.

h = \frac{1}{2} gt^2

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(1)}{2.607*10^{-4}}}

t = 87.58s

Ultimately, the time for the rock to hit the surface is t = 87.58s.

8 0
7 days ago
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