Answer:
Explanation:
We shall first calculate the velocity at height h = 575 m .
acceleration a = 2.2 m /s²
v² = u² + 2 a s
u is initial velocity , v is final velocity , s is height achieved
v² = 0 + 2 x 2.2 x 575
v = 50.3 m /s
After 575 m , rocket moves under free fall so g will act on it downwards
If it travels further by height H
from the relation
v² = u² - 2 g H
v = 0 , u = 50.3 m /s
H = ?
0 = 50.3² - 2 x 9.8 H
H = 129.08 m
Total height attained by rocket
= 575 + 129.08
= 704.08 m .
Answer:
Angle with the +x axis is θ = 79.599degree
Then the velocity of owner = 1.235m/s
Explanation:
Given that the mass of dog is m1 =26.2 kg
velocity of dog is u1 = 3.02 m/s (north)
mass of cat is m2 = 5.3 kg
velocity is u2 = 2.74 m/s (east )
Mass of owner is M = 65.1 kg
Consider the east direction along +x axis andnorth along +y
momentum of dog is Py = m1 x u1
= 79.124 kg.m/s (j)
momentum of cat is Px = m2 x u2
= 14.522 kg.m/s (i)
Then the net magnitude of momentum is P = (Px2 + Py2)1/2
= 80.445
Angle with the +x axis is θ =tan-1(Py / Px ) = 79.599 degree
Then the velocity of owner is v = P / M = 1.235 m/s
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
One of the main uses of this device is to test a beam of charged particles
What is cathode ray tube?
The cathode ray tube determines the charge flowing in a gas. When an electric field is set up using metal electrodes, the cathode ray tends to bend towards the positive electrode.
This concludes that a charge and the electrodes present helps determine the charge of the beam of charged particle.
Thus, one of the main uses of this device is to test a beam of charged particles.
Learn more about cathode ray tube.
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uhhhh idk cheif...
thats a big oof right there.
ged is always an option