Ask question

Ask question

All categories

2 years ago

6

In a car crash, large accelerations of the head can lead to severe injuries or even death. A driver can probably survive an acce

leration of 50g that lasts for less than 30 ms, but in a crash with a 50gacceleration lasting longer than 30 ms, a driver is unlikely to survive. Imagine a collision in which a driver's head experienced a 50g accelerationWhat is the highest speed that the car could have had such that the driver survived? Express your answer with the appropriate units.

1 answer:

noname [10]2 years ago

6 0

Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

You might be interested in

A 10kg rocket is traveling at 80 m/s when the booster engine applies a constant forward force of 60 N for 3.0 seconds. What impu

Lina20 [59]

Answer:

Impulse = 90

Resulting Velocity = 89

Explanation:

Use F * change in time = m * change in velocity.

For the first part of the question, the left side of the equation is the impulse. Plug it in.

60 * (3.0 - 0) = 90.

For the second half. we use all parts of the equation. I'm gonna use vf for the final velocity.

60 * (3.0 - 0) = 10 * (vf - 80). Simplify.

90 = 10vf - 800. Simplify again.

890 = 10vf. Divide to simplify and get the answer.

The resulting velocity is 89.

4 0

1 year ago

A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both

svetlana [45]

Answer:

7 deg

Explanation:

$m$ = mass of the rod = $1.4 kg$

$W$ = weight of the rod = $mg = (1.4) (9.8) = 13.72 N$

$k_{L}$ = spring constant for left spring = $59 Nm^{-1}$

$k_{R}$ = spring constant for right spring = $33 Nm^{-1}$

$x_{L}$ = stretch in the left spring

$x_{R}$ = stretch in the right spring

$L$ = length of the rod = 0.75 m

$\theta$ = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

$k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m$

Using equilibrium of force in vertical direction for right spring

$k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m$

Angle made with the horizontal is given as

$\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg$

3 0

2 years ago

What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

1 year ago

Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b

kolbaska11 [484]

To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV

= 1.240x10^-10 m

= 1.240x10^-1 nm

3 0

2 years ago

Read 2 more answers

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to

german

Answer: v= 1.23 m/s θ = 75.3º

Explanation:

First of all, we define the direction in which both fruits are tossed as the x axis, so all initial momenta have horizontal components only.

Now, if no external forces act during collision (due to the infinitesimal time during which collision takes place) momentum must be conserved.

As momentum is a vector, both components must be conserved, so we can write the following equations:

p₁ₓ = p₂ₓ ⇒ -m₁ . vi₁ +m₂. vi₂ = m₁ . vf₁ . cos θ (1)

p₁y = p₂y ⇒ 0 =m₂ . vf₂ - m₁. vf₁. sin θ (2)

Replacing by the values of m1, m₂, vi₁, vi₂, and vf₂, we can calculate the value of the angle θ, that the apple forms with the horizontal, as follows:

(1) -0.13 Kg. 1.05 m/s + 0.15 Kg. 1.18 m/s = 0.13. vf . cos θ

(2) 0.15 Kg. 1.03 m/s = 0.13 vf. sin θ

sin θ / cos θ = 3.82 ⇒ tg θ = 3.82 ⇒ θ = arc tg (3.82) = 75.3º

Replacing this value of θ in (2), we get:

0.15 kg. 1.03 m/s = 0.13 vf . sin 75.3º = 0.13 . vf . 0.967

Solving for vf, we get:

vf = 0.15 kg. 1.03 m/s / 0.13. 0.967 = 1.23 m/s

5 0

1 year ago