The forces that make a passenger speed up, slow down, or
turn a curve are the same forces that have the same effect
on the driver and anybody else in the car.
-- Speeding up . . .
the back of the seat
friction between the car seat and the seat of your pants
-- Slowing down . . .
the seat belt
friction between the car seat and the seat of your pants
-- Turning away from a straight line . . .
the seat belt
friction between the car seat and the seat of your pants
the door, or whatever or whomever you're leaning against
False. An Electromagnetic induction is the production of an electromotive force across an electrical conductor in a changing magnetic field.
Answer:
The ballon will brust at
<em>Pmax = 518 Torr ≈ 0.687 Atm </em>
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Explanation:
Hello!
To solve this problem we are going to use the ideal gass law
PV = nRT
Where n (number of moles) and R are constants (in the present case)
Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.
--- (*)
Our initial state is:
P1 = 754 torr
V1 = 3.1 L
T1 = 294 K
If we consider the final state at which the ballon will explode, then:
P2 = Pmax
V2 = Vmax
T2 = 273 K
We also know that the maximum surface area is: 1257 cm^2
If we consider a spherical ballon, we can obtain the maximum radius:

Rmax = 10.001 cm
Therefore, the max volume will be:

Vmax = 4 190.05 cm^3 = 4.19 L
Now, from (*)

Therefore:
Pmax= P1 * (0.687)
That is:
Pmax = 518 Torr
<span>To begin, the formula for finding frequency when wavelength is known is "f = c / w" when c is the constant velocity (3 * 10^8 m/s). To convert the wavelength into a common form (m/s), it will have to be multiplied by 10^-2. This leaves the equation as "f = 3.0 * 10^8 / (2.4 * 10^-5 * 10^-2), or 2.4 * 10^-7. This gives 1.25 * 10^15 m/s as the frequency.</span>
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²