Answer:
2.4 m/s". 1
Explanation:
A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1
Answer:
A) v = 40 m / s, B) v_average = 20 m / s
Explanation:
For this exercise we will use the kinematics relations
A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero
v = vo + a t
v = 0 + 8 5
v = 40 m / s
B) the average velocity can be found with the relation
v_average = vf + vo / 2
v-average = 0+ 40/2
v_average = 20 m / s
Answer:
a) F = 1.26 10⁵ N, b) F = 2.255 10³ N, c) F_ {soil} = 3078 N
Explanation:
For this exercise we will use the relationship between momentum and moment
I = Δp
F t = p_f -p₀
a) with stiff legs, final speed is zero, initial velocity is down
Ft = 0-p₀
F = m v / t
let's calculate
F = 84.0 6.82 / 4.56 10⁻³
F = 1.26 10⁵ N
b) bending the legs
let's calculate
F = 84.0 6.82 / 0.254
F = 2.255 10³ N
c) It is requested to calculate the force of the ground on the man
∑ F = F_soil -W
F_soil = F + W
F_ {soil} = 2.255 103 + 84 9.8
F_ {soil} = 3078 N
Answer:
3.5434 eV
Explanation:
For a particle with kinetic energy E and mass m , the wavelength associated is given by the following relation,
E = 
Putting the values we get
E = 
=1.063 x 10⁻²¹ J
= .0066 eV.
Energy of¹light in terms of eV
= 1244 / 350 = 3.55 eV.
Work function = 3.55 - 0.0066 = 3.5434 eV.
Answer:
h = 18.41 m
Explanation:
Given that,
Mass of a test rocket, m = 11 kg
Its fuel gives it a kinetic energy of 1985 J by the time the rocket engine burns all of the fuel.
According to the law of conservation of energy,
PE = KE = mgh
h is height will the rocket rise
So, the rocket will rise to a height of 18.41 m.
