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3 years ago

A bullet 2cm log is fired at 420m/s and passes straight a 10cm thick board exiting at 280m/s

2 answers:

7 0

Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

(I was only able to do A and B)

Alexeev081 [22]3 years ago

6 0

Answer:

Explanation:

(a)Solving for the acceleration of the bullet

acceleration = (vf^2 – vi^2) / 2d

acceleration = ((280 m/s)^2 – (420 m/s)^2) / (2 * 0.12 m)

acceleration = (78400 - 176400) / 0.24 m

acceleration = -98000 / 0.24

acceleration = -408333 m/s^2

(a)Solving for contact time with board

t^2 = 2d/a

t^2 = 2 * 0.12 m / 408333 m/s^2

t^2 = 0.24 m / 408333 m/s^2

t^2 = 5.8775558 x 10^-7

t = 0.0007666 s or 767 microseconds

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During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi

vladimir2022 [97]

Answer:

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 × 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

7 0

3 years ago

A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move awa

DerKrebs [107]

Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, $\lambda=53\ cm=0.53\ m$

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

$v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s$

So, the wave move with a speed of 7.95 m/s.

5 0

2 years ago

What’s the best conductor for electricity

rjkz [21]

So the best conductor for electricity is silver.
However because silver is so expensive we use copper which is the next best conductor

3 0

3 years ago

Read 2 more answers

A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?

Yanka [14]

Answer:

$\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}$

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

$\boxed{ \bold{\sf KE = \frac{1}{2} m {v}^{2} }}$

Substituting values of m & v in the equation:

$\sf \implies KE = \frac{1}{2} \times 380.2 \times {11}^{2}$

$\sf \implies KE = \frac{1}{ \cancel{2}} \times \cancel{2} \times 190.1 \times 121$

$\sf \implies KE = 190.1 \times 121$

$\sf \implies KE = 23002.1 \: J$

$\therefore$

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0

3 years ago

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

4 0

3 years ago