Answer:
103.57 Km/h
Explanation:
From the question given above, the following data were obtained:
Distance = 725 Km
Time = 7 hours
Speed =?
Speed can be defined as the distance travelled per unit time. Mathematically, it is expressed as:
Speed = Distance /time
With the above formula, we can calculate how fast he will drive (i.e the speed) in order to get there on time. This is illustrated below:
Distance = 725 Km
Time = 7 hours
Speed =?
Speed = Distance /time
Speed = 725 / 7
Speed = 103.57 Km/h
Thus, to get there on time, he will drive with a speed of 103.57 Km/h
Answer:
The value is 
Explanation:
From the question we are told that
The magnitude of the horizontal force is 
The mass of the crate is 
The acceleration of the crate is 
Generally the net force acting on the crate is mathematically represented as
Here 
is force of kinetic friction (in N) acting on the crate
So
=>
Here's a fun and useful factoid:
The ratio of the voltages on a transformer is the same
as the ratio of the number of turns in each winding.
So the ratio of (345 to the secondary turns) is (115V to 24V).
That's a proportion.
(115/24) = (345/x)
I'll bet you can take it and solve it from here.
Just cross-multiply in the proportion and etc. etc.
Answer is: <span>1/4 its old kinetic energy .
</span>V₁ = 10 m/s.
V₂ = 5 m/s.
m₁ = m₂ = m.
E₁ = 1/2 · m₁ · V₁², E₁ = 1/2 · m · (10 m/s)² = 50 · m.
E₂ = 1/2 · m₂ · V₂², E₂ = 1/2 · m · (5 m/s)² = 12,5 · m.
E₂/E₁ = 12,5m / 50m = 0,25.
V - speed of semi-truck.
m - mass of semi-truck.
E - kinetic energy of semi-truck.
Answer:
Option C is correct.
The magnitude of the field is reduced to half at twice the distance,
Explanation:
The magnetic field produced in a long, straight conductor carrying a current I at distance r is given by
B = μ₀I/2πr
Where μ₀ is the constant permeability of free space.
If we increase the distance by twice then
B = μ₀I/2π(2r)
B = μ₀I/2πr(2)
B = B/2
Therefore, the magnitude of magnetic field is reduced by B/2 at twice the distance.
