Answer:
The driveway measures 4.98 m
Explanation:
We aim to find the length of the driveway, thus utilizing the following equations
W=ΔK.E where W represents work and ΔK.E indicates the change in kinetic energy
Moreover,
also
W = F.d where F is the force and d denotes distance
Given that
= 4000 N indicating this frictional force
m = 2100 Kg
θ= 20.0°
V=3.8 m/s representing the car's speed at the bottom of the driveway
W=Δ K.E
= 15162 J
As the x component of gravity is
= mg sinФ
thus
= (2100)(9.8)sin(20.0°) results in
= 7038.77 N
And the Net force is
=
-
= 7038.77 - 4000 = 3038.77 N
So, the length of the driveway equals W / (
) = 15162/3038.77 = 4.98 m
Thus, this is the length of the driveway.
Incomplete query. The complete inquiry is as follows
Calculate the torque exerted on the shaft of a vehicle transmitting 225 hp at a rotation speed of 3000 rpm.
Response:
Torque=0.51 Btu
Analysis:
Given Information
Power=225 hp
Revolutions =3000 rpm
To determine
T( torque )=?
Process
As an object is moved by force over a distance, work is performed on that object. Similarly, when torque rotates an object through an angle, work is also accomplished.
1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t):
Next, we will compute the speed at t=2 s and t=4 s:
The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending.
2) The maximum height of the projectile occurs when its speed equals zero:
Thus, we have
And solving yields
3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s:
4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to
This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is
, which indicates the landing time of the projectile.
5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:
, carrying a negative sign to denote a downward direction.
