About how many times greater is the density of a neutron star compared to a white dwarf?

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

vlada-n [284]

Answer:

$v=2.02\frac{m}{s}$

Explanation:

Assuming no energy lost, according to the law of conservation of energy, the kinetic energy of the automobile becomes potential energy after the crash:

$K=U\\\frac{mv^2}{2}=\frac{kx^2}{2}$

Here m is the automobile's mass, v is the speed of the car before impact, k is the "bumper" constant and x is the compression of the bumper due to the collision. Solving for v:

$v=x\sqrt\frac{k}{m}\\v=2.63*10^{-2}m\sqrt{\frac{5.9*10^6\frac{N}{m}}{10^3kg}}\\v=2.02\frac{m}{s}$

8 0

2 years ago

Applied force is the force of support exerted by an object that holds up another object.

kenny6666 [7]

False, applied force is when a person or an object pushes on another object

6 0

3 years ago

Read 2 more answers

Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi

max2010maxim [7]

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

3 0

2 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Viktor [21]

Answer:

6.75 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 16 m/s²

g = Acceleration due to gravity = 9.81 m/s²

Let y be the distance the rocket is accelerating

960-y is the distance traveled in free fall

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 16\times y+0^2}\\\Rightarrow v^2=32y\ m/s$

In free fall

$v^2-u^2=2g(960-y)\\\Rightarrow 0-32y=2g(960-y)\\\Rightarrow -32y=2\times -9.81(960-y)\\\Rightarrow 960-y=\dfrac{-32}{2\times -9.81}y\\\Rightarrow 960-y=1.63098878695y\\\Rightarrow 960=2.63098878695y\\\Rightarrow y=\dfrac{960}{2.63098878695}\\\Rightarrow y=364.881828749\ m$

The distance the rocket will keep accelerating is 364.881828749 m

After which it will travel 960-364.881828749 = 595.118171251 m in free fall

$s=ut+\frac{1}{2}at^2\\\Rightarrow 364.881828749=0t+\frac{1}{2}\times 16\times t^2\\\Rightarrow t=\sqrt{\frac{364.881828749\times 2}{16}}\\\Rightarrow t=6.75353452598\ s$

The time the rocket is accelerating is 6.75 seconds

5 0

3 years ago

In what order does light pass through structures of the eye? lens, cornea, retina cornea, pupil, lens pupil, iris, lens pupil, c

Mrrafil [7]

Answer: Light passes through the front of the eye (cornea) to the lens. The cornea and the lens help to focus the light rays onto the back of the eye (retina). The cells in the retina absorb and convert the light to electrochemical impulses which are transferred along the optic nerve and then to the brain.

Explanation:

When light rays reflect off an object and enter the eyes through the cornea (the transparent outer covering of the eye), you can then see that object. The cornea bends, or refracts, the rays that pass through the round hole of the pupil.

4 0

3 years ago