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blsea [12.9K]
2 years ago
15

Is it possible to code the impact force using an accelerometer?

Physics
1 answer:
Naddika [18.5K]2 years ago
6 0

Yes, it possible to code the impact force using an accelerometer.

Accelerometers can be used to measure vibration on cars, machines, buildings, process control systems and safety installations

<h3>What is an Accelerometer ?</h3>

An accelerometer is a device that measures the vibration, or acceleration of motion of a structure.

  • The force caused by vibration or a change in motion (acceleration) causes the mass to "squeeze" the piezoelectric material which produces an electrical charge that is proportional to the force exerted upon it.

  • They can also be used to measure seismic activity, inclination, machine vibration, dynamic distance and speed with or without the influence of gravity.

Learn more about Accelerometer meter here:

brainly.com/question/15185937

#SPJ4

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DaniilM [7]
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3 0
4 years ago
10. Ahmad drives 60 km from his house to the petrol station in a car moving at a constant speed of
djyliett [7]

Answer:

Explanation:

Average speed is distance traveled over time taken to do so

home to station

d = 60 km

t = 60 km/60 km/hr = 1 hr

station to shop

d = 60 km

t = 60 km / 120 km/hr = ½ hr

speed

s = (60 + 60) / (1 + ½) = 120/1.5 = 80 km/h

7 0
3 years ago
If the energy bar was his only fuel, how far could a 68 kg person walk at 5 km/h?
goldfiish [28.3K]

An energy bar contains of 20grams of carbohydrates and 1gram is equal to 17 KJ. If the energy bar was his only fuel the total energy available is equal to (17,000 x 20) = 340,000j. Estimate KJ per hour is equal to 1539KJ, it has Time equal to (340,000/1539,000) is equal to 0.2209 hour and his Distance is equal to (5,000 x .2209) = 1.1045km. 

3 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

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3 years ago
An electric circuit consists of a variable resistor connected to a source of constant potential difference. If the resistance of
yulyashka [42]
Then nothing will happen
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3 years ago
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