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solmaris [256]
3 years ago
13

A spring with a spring constant value of 2500 StartFraction N over m EndFraction is compressed 32 cm. A 1.5-kg rock is placed on

top of it, then the spring is released. Approximately how high will the rock rise? 9 m 17 m 27 m 85 m
Physics
2 answers:
PilotLPTM [1.2K]3 years ago
7 0

Answer:

9m

Explanation:

edge2o2o

Sedaia [141]3 years ago
5 0

Answer:

it's 9m

Explanation:

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An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring with cons
Nady [450]

To solve this problem it is necessary to apply the concepts related to the conservation of energy, specifically the potential elastic energy against the kinetic energy of the body.

By definition this could be described as

PE = KE

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

Where

k = Spring constant

x = Displacement

m = mass

v = Velocity

This point is basically telling us that all the energy in charge of compressing the spring is transformed into the energy that allows the 'impulse' seen in terms of body speed.

If we rearrange the equation to find v we have

v = \sqrt{\frac{kx^2}{m}}

Our values are given as

m = 1000kg

k = 5.75*10^6N/m

x = 3.12*10^{-2}m

Replacing at our equation we have then,

v = \sqrt{\frac{kx^2}{m}}

v = \sqrt{\frac{(5.75*10^6)(3.12*10^{-2})^2}{1000}}

v = 2.3658m/s

Therefore he speed of the car before impact, assuming no energy is lost in the collision with the wall is 2.37m/s

4 0
3 years ago
Physical Properties include...
3241004551 [841]

and C. taste, odor, and texture

6 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
Modern telescopes are capable of seeing bright galaxies up to about
marishachu [46]

Answer:

Modern telescopes are capable of seeing bright galaxies up to about 10000 millions light years away

Explanation:

A telescope is a tool that astronomers use to see faraway objects. Most telescopes work by using curved mirrors to gather and focus light from the night sky. The bigger the mirrors or lenses, the more light the telescope can gather.

Modern telescopes gather information from the electromagnetic spectrum far beyond the range of visible light.

The farthest bright galaxies, that the modern telescope is capable of seeing is 10000 millions light years away.

5 0
4 years ago
Ask Your Teacher Two long, straight wires are parallel and 11 cm apart. One carries a current of 2.9 A, the other a current of 5
dsp73

Answer:

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is away from one another

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is towards each other.

Explanation:

\mu_0 = Vacuum permeability = 4\pi\times 10^{-7}\ N/A^2

i_1 = Current in first wire = 2.9 A

i_2 = Current in second wire = 5.3 A

r = Gap between the wires = 11 cm

Force per unit length

F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is away from one another

F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N

The magnitude of force per unit length of one wire on the other is 2.7945\times 10^{-5}\ N and the direction is towards each other.

7 0
3 years ago
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