Ask question

Ask question

All categories

2 years ago

9

In a common but dangerous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land

hard on the floor. Suppose the victim falls by 0.50 m, the mass that moves downward is 70 kg, and the collision on the floor lasts 0.082 s. What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

1 answer:

sattari [20]2 years ago

4 0

Answer:

a) Impulse |J|= 219.4 kgm/s

b) Force F = 2672 N

Explanation:

Given

Height of fall h = 0.50 m

Mass M = 70 kg

Period of collision t = 0.082 s

Solution

The final velocity of the person v is zero since the person will come to rest.

The initial velocity of the person can be calculated by using the "law of conservation of energy".

Initial Kinetic energy = Final potential energy

$\frac{1}{2} mu^2=mgh\\\\u = \sqrt{2gh} \\\\u = \sqrt{2 \times 9.81 \times 0.50} \\\\u = 3.13 m/s$

a) Impulse

J = final momentum - initial momentum

$J = mv -mu\\\\J = 0 - (70 \times 3.13)\\\\J = -219.2 kgm/s$

Magnitude of impulse

$|J| = 219.1 kgm/s$

b) Force

$F = \frac{J}{t} \\\\F = \frac{219.1}{0.082} \\\\F = 2672 N$

You might be interested in

With countercurrent flow, diffusion happened in all regions of the filter. Explain why

jeka94

Answer:

When the blood and the dialysate are flowing in the same direction, as the the dialysate and the blood move away from the region of higher concentration of the urea, to a region distant from the source, the concentration of urea in the blood stream and in the dialysis reach equilibrium and diffusion across the semipermeable membrane stops within the higher filter regions such as II, III, IV or V

However, for counter current flow, as the concentration of the urea in the blood stream becomes increasingly lesser the, it encounters increasingly unadulterated dialysate coming from the dialysate source, such that diffusion takes place in all regions of the filter

Explanation:

3 0

2 years ago

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what

Alexxx [7]

<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>

4 0

2 years ago

A hot piece of iron is thrown into the ocean and its temperature eventually stabilizes. Which of the following statements concer

777dan777 [17]

Answer:

E. The ocean gains more entropy than the iron loses.

Explanation:

When there is a spontaneous process , entropy of the system increases . Here hot iron is losing entropy and ocean is gaining entropy . Net effect will be gain of entropy . That means entropy gained by ocean is more than entropy lost by iron .

Hence option E is correct .

8 0

2 years ago

A team of astronauts is on a mission to land on and explore a large asteroid. In addition to collecting samples and performing e

Blizzard [7]

<h2>Answer: 117.626m/s</h2>

Explanation:

The escape velocity $V_{esc}$ is given by the following equation:

$V_{esc}=\sqrt{\frac{2GM}{R}}$ (1)

Where:

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M$ is the mass of the asteroid

$R$ is the radius of the asteroid

On the other hand, we know the density of the asteroid is $\rho=3.84(10)^{8}g/m^{3}$ and its volume is $V=2.17(10)^{12}m^{3}$ .

The density of a body is given by:

$\rho=\frac{M}{V}$ (2)

Finding $M$ :

$M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})$ (3)

$M=8.33(10)^{20}g=8.33(10)^{17}kg$ (4) This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

$V=\frac{4}{3}\piR^{3}$ (5)

Finding $R$ :

$R=\sqrt[3]{\frac{3V}{4\pi}}$ (6)

$R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}$ (7)

$R=8031.38m$ (8) This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

$V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}$ (9)

Finally:

$V_{esc}=117.626m/s$ This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

6 0

2 years ago

A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

$CPR=\frac{KW}{DAT}$

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6 to give CPR in mm/yr

mpy

$CPR=\frac{KW}{DAT}$

$=\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}$

=37.4mpy

mm/yr

$CPR=\frac{KW}{DAT}$

$=\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}$

=0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0

2 years ago