You may have a cold if you do not feel well, depends on the symptoms
The image will form in the vicinity of F. Its nature will be small and inverted
Ignoring the air resistance it will take about 3 seconds for the object to reach the ground.We know that the acceleration due to gravity is 10m/s2.
We also know that the final velocity is 30 m/s while the initial velocity is 0 m/s
we can use the formulae for acceleration to calculate the time taken/
(final - initial velocity)/timetaken=10
(30-0)/timetaken=10
timetaken =30/10=3 seconds
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
Answer:
45000 K .
Explanation:
Given :
A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure
We need to find the temperature in which 1 litre of the same gas weigh 1 gram
in pressure 75 atm.
We know, by ideal gas equation :

Here , n is no of moles , 
Putting initial and final values and dividing them :


Hence , this is the required solution.