A 6000 kg lorry is reversing into a parking space at a speed of 0.5 m/s but collides with a car. The crumple zone of the car sto

To tackle this problem, it's essential to employ concepts associated with force as per Hooke's law, alongside the forces described by Newton's second law and the concept of potential elastic energy. Since the forces are in equilibrium, the spring force matches the gravitational force. To find the spring constant k, we recognize the compression is 40cm at launch, hence applying the potential elastic energy formula results in determining the energy stored in the spring as 63.72 Joules.

Answer:

The force will rise in direct relation to the mass of the objects

Explanation:

The gravitational acceleration remains constant. It is measured in meters per second squared or m/s². The average value is 9.81 m/s², calculated from observations made on varying surfaces. In reality, the acceleration can vary based on the geographical shape of the Earth relative to the earth's magnetic field and gravitational force.

For instance, if a single washer weighs 20 kg, with the gravity at 9.81 m/s², the weight would be:

F = ma

  =

If there are three washers, the total weight calculates as:

F = 3 * 20 * 9.81

  = 588.6 N

Initially, we need to calculate the acceleration required for the car to halt from its initial speed based on the distance traveled. This can be done using the formula,
                                             2ad = Vf² - Vi²
where a represents acceleration, d is distance, and Vf and Vi are the final and initial speeds respectively. Plugging in the known quantities,
           2(a)(35 m) = (0 m/s)² - ((65 km/h) x (1000 m/ 1 km) x (1 hr / 3600 s))²
The resulting acceleration is −4.66 m/s².
To calculate the force required to stop the car, we multiply the mass by the acceleration. This calculation yields -4,660 N, and we take the absolute value, which is 4,660 N. 

A. 

The absence of a defined volume in gases causes them to disperse throughout the air. 

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v  = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = = = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s