Answer:
Force on front axle = 6392.85 N
Force on rear axle = 8616.45 N
Explanation:
As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels
Now we know that



now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle
so we can write torque balance about its center of mass


now from above equation

now we have

now the other force is given as

Answer:
V=20cm/s
Explanation:
The average speed is the distance total divided the time total:

First stage:
T1=5s

But,
(decelerates to rest)
then: 
on the other hand:

X1=75cm
Second stage:
T2=5s

X2=125cm
Finally:
X=X1+X2=200cm
T=T1+T2=10s
V=X/T=20cm/s
A roller coaster accelerates from an initial velocity of 6.0 m/s to a final velocity of 70 m/s over 4 seconds. whats the acceleration
The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
<u>Explanation:</u>
H₂ + Br₂ ⇒ 2HBr
PH₂ = 0.782atm
PBr₂ = 0.493atm
Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹
At equilibrium:
Let 2x = pressure of HBr
PH₂ = 0.782 -x
PBr₂ = 0.493 - x
Kp = (2x)^2 / (0.782-x)(0.493-x)
Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.
Then,
Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)
x = 1.2X10⁻¹¹
PHBr = 2x = 2.4 X 10⁻¹¹ atm
Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
Explanation:
It is given that,
Magnetic field, B = 0.5 T
Speed of the proton, v = 60 km/s = 60000 m/s
The helical path followed by the proton shown has a pitch of 5.0 mm, p = 0.005 m
We need to find the angle between the magnetic field and the velocity of the proton. The pitch of the helix is the product of parallel component of velocity and time period. Mathematically, it is given by :





So, the angle between the magnetic field and the velocity of the proton is 50.58 degrees. Hence, this is the required solution.