Answer:
Explanation:
Given:
Initial velocity of the vehicle, 
distance between the car and the tree, 
time taken to respond to the situation, 
acceleration of the car after braking, 
Using equation of motion:
..............(1)
where:
final velocity of the car when it hits the tree
initial velocity of the car when the tree falls
acceleration after the brakes are applied
distance between the tree and the car after the brakes are applied.
Now for this situation the eq. (1) becomes:
(negative sign is for the deceleration after the brake is applied to the car.)
Q = mCΔT, where Q = Amount of energy required, m = mass of the blcok, C = specific heat, ΔT = change in temperature.
Using the given values;
Q = 12*913*(118-20) = 1073688 J = 1073.688 kJ.
The correct answer in B.
Suppose the spring begins in a compressed state, so that the block speeds up from rest to 2.6 m/s as it passes through the equilibrium point, and so that when it first comes to a stop, the spring is stretched 0.20 m.
There are two forces performing work on the block: the restoring force of the spring and kinetic friction.
By the work-energy theorem, the total work done on the block between the equilbrium point and the 0.20 m mark is equal to the block's change in kinetic energy:
or
where <em>K</em> is the block's kinetic energy at the equilibrium point,
Both the work done by the spring and by friction are negative because these forces point in the direction opposite the block's displacement. The work done by the spring on the block as it reaches the 0.20 m mark is
Compute the work performed by friction:
By Newton's second law, the net vertical force on the block is
∑ <em>F</em> = <em>n</em> - <em>mg</em> = 0 ==> <em>n</em> = <em>mg</em>
where <em>n</em> is the magnitude of the normal force from the surface pushing up on the block. Then if <em>f</em> is the magnitude of kinetic friction, we have <em>f</em> = <em>µmg</em>, where <em>µ</em> is the coefficient of kinetic friction.
So we have
For this task we need to use Kepler's third law:
T = 2*pi*
where T is orbital period in seconds, r is Venus's semi-major axis, G is gravitational constant and M is mass of the sun.
Distance from earth to sun is 1AU so if we know earths period and distance from the sun we can calculate Venus period.
Te-earths period
Tv-venus period
From the text we know that Re/Rv = 1/0.723
Which means that:
Tv = 0.614 Te
We can answer this problem using Ampere’s Law:
<span>Bh = μoNI </span>
Where:
B = Magnetic Field
h = coil length
<span>μo = permeability =4π*10^-7 T·m/A </span>
N = number of turns
I = current
It is given that B=0.0015T, I=1.0A, h=10 cm =
0.1m<span>
Use Ampere's law to find # turns:
Which can be rewritten as:
<span>N = Bh/μoI </span>
N = (0.0015)(0.1)/(4π*10^-7)(1.0)
N = 119.4
</span>
<span>Answer:
119.4 turns</span>
