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Svet_ta [14]
3 years ago
11

If you are the driver or owner of a vehicle which is in a crash that is your fault, and you are not insured in compliance with t

he Financial Responsibility Law
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0
If a driver is at fault in an accident and is not insured by the Financial Responsibility Law then he or she may be required to pay for the damages caused to the other person's vehicle or property due to the accident before the driver at fault is given the privilege to drive again.
The Financial Responsibility Law is a law that requires drivers to prove that they have enough money saved to pay for the damages that may occur in the case of an accident.
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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.330. At what minimum rate wo
Helga [31]

Answer:3.23 m/s^2

Explanation:

Given

\mu_s =0.330

Frictional Force is balanced by force due to car acceleration

Frictional force F_s

F_s=ma_{min}

\mu_sN=ma_{min}

\mu_s\cdot mg=ma_{min}

a_{min}=\mu_s \cdot g=0.330\times 9.8=3.23 m/s^2

6 0
2 years ago
A plane has a mass of 360,000 kg takes-off at a speed of 300 km/hr. i) What should be the minimum acceleration to take off if th
melomori [17]

Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

90000 = 4 × a

a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

5 0
3 years ago
Consider the video tutorial you just watched. Suppose that we duplicate this experimental setup in an elevator. What will the sp
zloy xaker [14]

Explanation:

if the elevator is moving upward with the constant speed the spring scale will read 18 N which is the mass of each of the two blocks attached by separate springs to the scale at opposite ends.

4 0
3 years ago
You are holding a finishing sander with your right hand. THe sander has a flywheel which spins counterclockwise as seen from beh
ryzh [129]

Answer:

c. turn downward

Explanation:

From the information given:

To find the tendency of the sander;

We need to apply the right-hand rule torque; whereby we consider the direction of the flywheel, the direction at which the torque is acting, and the movement of the sander toward the right.

Since the flywheel of the sander is in counterclockwise movement, hence the torque direction will be outward placing on the wall. However, provided that the movement of the sander is toward the right, then there exists an opposite force that turns downward which showcases the tendency in the sander is downward.

3 0
2 years ago
A 0.325 g wire is stretched between two points 57.7 cm apart. The tension in the wire is 650 N. Find the frequency of first harm
Galina-37 [17]

Answer:

f = 931.1 Hz

Explanation:

Given,

Mass of the wire, m = 0.325 g

Length of the stretch, L = 57.7 cm = 0.577 m

Tension in the wire, T = 650 N

Frequency for the first harmonic = ?

we know,

v =\sqrt{\dfrac{T}{\mu}}

μ is the mass per unit length

μ = 0.325 x 10⁻³/ 0.577

μ = 0.563 x 10⁻³ Kg/m

now,

v =\sqrt{\dfrac{650}{0.563\times 10^{-3}}}

   v = 1074.49 m/s

The wire is fixed at both ends. Nodes occur at fixed ends.

For First harmonic when there is a node at each end and the longest possible wavelength will have condition

          λ=2 L

          λ=2 x 0.577 = 1.154 m

we now,

       v = f λ

      f = \dfrac{v}{\lambda}

      f = \dfrac{1074.49}{1.154}

             f = 931.1 Hz

The frequency for first harmonic is equal to f = 931.1 Hz

7 0
3 years ago
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