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3 years ago

When the car carrying the unbelted driver crashes, what brings the driver to a stop

1 answer:

4 0

When the driver is not wearing a seat belt and the car hits a brick wall,
the driver keeps going and doesn't stop, (because of the inertia of the
driver's body).

-- If the driver is lucky, the car is going slow enough, AND it has a
driver-side airbag, AND the airbag works, AND it inflates fast enough,
AND it cushions the driver well enough to avoid the driver's death.

-- If an airbag doesn't stop the driver, then one or more of these brings
the driver to a stop:

... the car's engine block crashing in through the dashboard
... the steering wheel
... the windshield
... the same brick wall that stopped the car.

Any one of these is probably fatal.

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You are assigned the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to

Sergeeva-Olga [200]

Answer:

630.93 kN of force.

Explanation:

Pressure inside the tank is 150 kPa

The acceleration due to gravity on Mars g is 3.71 m/s^2.

The depth of water h is 13.6 m.

Pressure due to air outside tank is 93 kPa

The density of water p is 1000 kg/m^3

Pressure of the water on the tank bottom will be equal to pgh

Pressure of water = pgh

= 1000 x 3.71 x 13.6 = 50456 Pa

= 50.456 kPa.

Total pressure at the bottom of the tank will be pressure within tank and pressure due to water and pressure outside tank.

Pt = (150 + 50.456 + 93) = 293.456 kPa

Force at the bottom of the tank will be pressure times area of tank bottom.

F = Pt x A

F = 293.456 x 2.15 m^2 = 630.93 kN

7 0

3 years ago

Explain what physicists mean when they say that the de Broglie wavelength relates to the probability distribution wave function.

Ksju [112]

Answer:

Because it unites particle and wave nature.

Explanation:

De Broglie wavelength can be defined as the,

$\lambda =\frac{h}{mv}$

Here, h is planks constant, m is the mass of electron, v is the velocity of electron.

Since the de Broglie wavelength can behave like the photon wavelength with respect to the momentum

It unites particles and waves nature ,so De Broglie wavelengths is probability waves associated with the wave function according to physicists.

7 0

2 years ago

Why do telescopes give modern astronomers an advantage over people in the past

pochemuha

its because it allows modern astronomers the ability to see farther out and more accuratly

6 0

3 years ago

Which two factors does the power of a machine depend on?

lianna [129]

The power of a machine depend on two factors are work and time.

Option C

<u>Explanation:</u>

In science, power defined as the amount of work done in a unit time. i.e. delivering work in a rate of time or energy supply, expressed in input of work or transmitted energy divided by the time interval (t) or W/t.

Example: Some work can be done in the long run with a low-power engine or in a short time with a motor with high performance. The equation for power can be given as

$Power\ (in\ watts) =\frac{\text { work (joules) }}{\text { time (seconds) }}$

$P=\frac{W}{t}$

7 0

2 years ago

so I know you can solve this either by using Vox or Voy. I'm getting 3.08s when using Vox and 3.14s for Voy way. For Voy I'm usi

Tatiana [17]

The person's horizontal position is given by

$x=v_0\cos40^\circ t$

and the time it takes for him to travel 56.6 m is

$56.6\,\mathrm m=\left(24.0\,\dfrac{\mathrm m}{\mathrm s}\right)\cos40^\circ t\implies t=3.08\,\mathrm s$

so your first computed time is the correct one.

The question requires a bit of careful reading, and I think there may be a mistake in the problem. The person's vertical velocity $v_y$ at time $t$ is

$v_y=v_{0y}-gt$

which tells us that he would reach the ground at about $t=3.15\,\mathrm s$ . In this time, he would have traveled

$x=v_{0x}(3.15\,\mathrm s)=57.9\,\mathrm m$

But we're told that he is caught by a net at 56.6 m, which would mean that the net cannot have been placed at the same height from which he was launched. However, it's possible that the moment at which he was launched doesn't refer to the moment the cannon went off, but rather the moment at which the person left the muzzle of the cannon a fraction of a second after the cannon was set off. After this time, the person's initial vertical velocity $v_{0y}$ would have been a bit smaller than $\left(24.0\,\frac{\mathrm m}{\mathrm s}\right)\sin40^\circ$ .

7 0

2 years ago