Question 3 of 10

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

sage has a walking speed of 300 feet per minute on the way to gate 14c at the airport sage has the option of using a moving side

nalin [4]

Answer:

2 min 40 s.

Explanation:

Distance = 800 ft

Speed (walking speed) = 300 ft/min

Speed = distance/time

Time, t = 800/300

= 8/3

= 2 min 40 s.

4 0

3 years ago

Podrian ayudarme? Llevo dias intentando pero no se como 1. el profesor no explico

Paraphin [41]

You have to upload this in the area of mathematicians..!

8 0

3 years ago

Which step is part of the information of a batholith?

Kruka [31]

Many plutons clump together as they grow through the crust.

6 0

4 years ago

Read 2 more answers

The mass of the Sun is 2 × 1030 kg, and the distance between Neptune and the Sun is 30 AU. What is the orbital period of Neptune

Veronika [31]

Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
$\frac{r^3}{T^2}= \frac{GM}{4 \pi^2}$ (1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun

Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
$1 AU = 1.5 \cdot 10^{11} m$
so the radius of the orbit is
$r=30 AU = 30 \cdot 1.5 \cdot 10^{11} m=4.5 \cdot 10^{12} m$

And if we re-arrange the equation (1), we can find the orbital period of Neptune:
$T=\sqrt{ \frac{4 \pi^2}{GM} r^3} = \sqrt{ \frac{4 \pi^2}{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )(2\cdot 10^{30} kg)}(4.5 \cdot 10^{12} m)^3 }= 5.2 \cdot 10^9 s$

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
$T=5.2 \cdot 10^9 s /(60 \cdot 60 \cdot 24 \cdot 365)=165 years$

3 0

3 years ago