Is there such a thing as cubic milliliters?

LenKa [72]

Four

Sometimes I think the creators of problems out to drawn and quartered. 60 g does not mean 60 grams. It means 60 * the acceleration due to gravity.

So the question really reads. The acceleration delivered by the air bag is 60 times that of a normal gravitational. This acceleration is delivered to the person where his mass is putting up a whole lot of resistance because he and his 75 kg are moving forward with the impact of the car. The 36 msec. has nothing to do with the problem.

The Force of the Air Bag is mass * a

F_airbag = mass * acceleration = 75 kg * 60 * 9.81 mass * acceleration = 44145 newtons

The answer is 4.41 * 10^4

Answer C

Five

This problem is governed by one formula that you sort of have to get out of your hat -- a piece of magic if you will.

Fg - Bf = m * a

Fg = the Force of gravity

Bf = the braking force

The mass of the rocket is derived from its weight

The acceleration is derived from one of your big 4 equations.

m of the rocket = 75600 / 9.81 = 7706 kg

The acceleration =

vi = 1 km/s = 1000 m/s

vf = 0

t = 2 minute * 60 sec/ min = 120 seconds

a = (vf - vi)/t = (0 - 1000 m/s) / 120 sec

a = - 8.333 m/s^2 The minus sign makes perfect sense. Remember the rocket is slowing down

The net downward force = mass * acceleration = - 7706 kg * - 8.333 m/s^2

The net force = - 64217 N

So going back to the problem's equation we have

Gravitational force - Braking Force = Net Force

Gravitational Force = 75600

Net Force = - 64217

Bracking force = ?

75600 - Bracking force = - 64217 Subtract 75600 from both sides

- Bracking force = - 64217 - 75600

- Braking force = - 139817

Braking force = 139817 N = 1.398 * 10^5 N

Braking Force = 1.4 * 10^5

Answer: Last One.

Six

The first thing you should do is derive a general formula for this problem.

The force pulling both masses down is M*g where g is the acceleration due to gravity.

The formula for this problem is

Mg = (m + M) * a

Now you need to solve for a

a = [M/(M + m) ] * g

Look what is happening. is a smaller or larger than g? This is a question you should really pay attention to. If it was larger, everyone would have this system in their basement because you'd get more energy output than you put in. Something for nothing is always appealing.

So what's the answer? (I get to ask it. No one posing the question ever should).

A

A is incorrect. M never goes away. The acceleration may get very tiny, but there always is some acceleration.

B must be true. It is just what I finished saying about A

C Who said anything about velocity? It's a red herring. If the velocity became 0 the acceleration would have to turn minus. This answer sounds good, but sounds good doesn't make it right. C is wrong.

D The acceleration does not remain constant no matter what. The answer to A still applies. So D is wrong.

4 0

3 years ago

Prove that vf = vi+at.

Gnoma [55]

To make your formula neater:

$V_{f} = V_{i} + a*t$

Acceleration is equal to

$\frac{velocity}{time}$ ,

so when you multiply it by time, you will get a velocity. In fact, you would get the change in velocity, as a*t signifies the acceleration of the object over the time, aka the change in velocity.

The initial velocity plus the change in velocity will equal the final velocity, irregardless if the change in velocity is positive or negative.

Therefore,

$V_{f} = V_{i} + a*t$

6 0

3 years ago

A spring with a spring constant kk of 100 pounds per foot is loaded with 1-pound weight and brought to equilibrium. it is then s

tino4ka555 [31]

Given:

k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs

Solution:

F = -kx
mx" = -kx
x" + (k/m)x = 0

characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)

x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)

ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)

x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0

1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft

x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0

So B would be the amplitude. Therefore, the equation of motion would be x = 0.08333*cos[(2π/0.1107)t]

5 0

4 years ago

The Achilles tendon connects the muscles in your calf to the back of your foot. When you are sprinting, your Achilles tendon alt

Andrews [41]

Explanation:

It is given that,

Mass of the runner, m = 70 kg

Length of the tendon, l = 15 cm = 0.15 m

Area of cross section, $A=110\ mm^2=0.00011\ m^2$

Part A,

Let the runner's Achilles tendon stretch if the force on it is 8.0 times his weight, F = 8 mg

Young's modulus for tendon is, $Y=0.15\times 10^{10}\ N/m^2$

The formula of the Young modulus is given by :

$Y=\dfrac{F/A}{\dfrac{\Delta L}{L}}$

$0.15\times 10^{10}=\dfrac{8\times 70\times 9.8/0.00011}{\dfrac{\Delta L}{0.15}}$

$\Delta L=0.0049\ m$

Part B,

The fraction of the tendon's length does this correspond is given by :

$\dfrac{\Delta L}{L}=\dfrac{0.0049}{0.15}$

$\dfrac{\Delta L}{L}=0.0326$

Hence, this is the required solution.

6 0

3 years ago

An old iron beam was coated with an unknown metal. there is a crack on the coating and it is observed that the iron is rusting a

IgorLugansk [536]

Based on the given description above about the metal that is being used for fractures, the old metal coating that was used before was made of TIN, and now, the metal that is used to repair fractures in order to avoid further corrosion is ALUMINUM. Hope this answer helps.

3 0

4 years ago