Answer:
40mph
Explanation:
20miles .5 miles away if you go 20 mph you get there in a hour soo if you double to 40 mpg you get there in 0.5
Answer:
(a) 110 rev/ min
(b) 5/6
Explanation:
As per the conservation of linear momentum,
L ( initial ) = L ( final )
I' ω' = ( I' + I'' ) ωf
I' is the rotational inertia of first wheel and I'' is the rotational inertia of second wheel which is at rest.
(a)
So, ωf = I' ω' / ( I' + I'' )
As I'' = 5I'
ωf = I' ω' / ( I' + 5I' )
ωf = ω'/ 6
now we know ω' = 660 rev / min
therefore ωf = 660/6
= 110 rev/ min
(b)
Initial kinetic energy will be K'
K' = I'ω'² / 2
and final K.E. will be K'' = ( I' + I'' )ωf² / 2
K'' = ( I' + 5I' ) (ω'/ 6)²/ 2
K'' = 6I' ω'²/72
K'' = I' ω'²/ 12
therefore the fraction lost is
ΔK/K' = ( K' - K'' ) / K'
= {( I'ω'² / 2) - (I' ω'²/ 12)} / ( I'ω'² / 2)
= 5/6
Answer:
V₂ = -22 V
Explanation:
Electric potential and field are related
ΔV = - E d
where ΔV is the potential difference between the plates, E the electric field and d the separation between the plates
In this exercise we are given the parcionero d = 4 mm = 0.004 m, the potential of one of the plates V1 = -6V and the value of the electric field E = 4000 V / m
V₂- V₁ = - E d
V₂ = - Ed + V₁
V₂ = - 4000 0.004 + (- 6)
V₂ = -16 - 6
V₂ = -22 V
According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:

where F is the magnitude of the force, m is the mass of the object and a its acceleration.
In this problem, the object is the insect, with mass

. The acceleration of the insect is

, therefore we can calculate the force exerted by the car on the insect:

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.
Answer:

Explanation:
First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.
Since
and
, we can rewrite the first equation as:

Now, we solve for
and calculate it:

This means that the crate's coefficient of kinetic friction on the floor is 0.18.