V ( initial ) = 20 m/s
h = 2.30 m
h = v y * t + g t ² / 2
d = v x * t
1 ) At α = 18°:
v y = 20 * sin 18° = 6.18 m/s
v x = 20 * cos 18° = 19.02 m/ s
2.30 = 6.18 t + 4.9 t²
4.9 t² + 6.18 t - 2.30 = 0
After solving the quadratic equation ( a = 4.9, b = 6.18, c = - 2.3 ):
t 1/2 = (- 6.18 +/- √( 6.18² - 4 * 4.9 * (-2.3)) ) / ( 2 * 4.9 )
t = 0.3 s
d 1 = 19.02 m/s * 0.3 s = 5.706 m
2 ) At α = 8°:
v y = 20* sin 8° = 2.78 m/s
v x = 20* cos 8° = 19.81 m/s
2.3 = 2.78 t + 4.9 t²
4.9 t² + 2.78 t - 2.3 = 0
t = 0.46 s
d 2 = 19.81 * 0.46 = 9.113 m
The distance is:
d 2 - d 1 = 9.113 m - 5.706 m = 3.407 m
GOOD LUCK AND HOPE IT HELPS U
Answer:
The equilibrium temperature is
21.97°c
Explanation:
This problem bothers on the heat capacity of materials
Given data
specific heat capacities
copper is Cc =390 J/kg⋅C∘,
aluminun Ca = 900 J/kg⋅C∘,
water Cw = 4186 J/kg⋅C∘.
Mass of substances
Copper Mc = 235g
Aluminum Ma = 135g
Water Mw = 825g
Temperatures
Copper θc = 255°c
Water and aluminum calorimeter θ1= 16°c
Equilibrium temperature θf =?
Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water
McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)
Substituting our data into the expression we have
235*390(255-θf)=
(135*900+825*4186)(θf-16)
91650(255-θf)=(3574950)(θf-16)
23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6
Collecting like terms and rearranging
23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf
8.2*10^6=3.66*10^6θf
θf=80.5*10^6/3.6*10^6
θf =21.97°c
Answer: They are put in front for defense so so they can block the opponents from getting the ball
Explanation:
Answer:
35 288 mile/sec
Explanation:
This is a problem of special relativity. The clocks start when the spaceship passes Earth with a velocity v, relative to the earth. So, out and back from the earth it will take:
If we use the Lorentz factor, then, as observed by the crew of the ship, the arrival time will be:
Then the amount of time wil expressed as a reciprocal of the Lorentz factor. Thus:
solving for v, gives = 35 288 miles/s
