The net torque exerted by the children on the branch of the tree is 1382 N-m.
The torque exert by the kids is calculated as
T= 45.6*9.8*1.28*cos27.5°+36*9.8*(2.25-1.28)cos27.5°
T=1382 N-m
Hence, The net torque exerted by the children on the branch of the tree is 1382 N-m.
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
Answer:
<em>480m/s²</em>
Explanation:
Acceleration is the change in velocity of a body with respect to time;
Acceleration = change in velocity/change in time
change in velocity = 29,500 - 24,000
change in velocity= 5,500
Change in time = 55 - 5
change in time = 50secs
Substitute into the formula;
spaceships acceleration = 24000/50
spaceships acceleration = 480 m/s²
<em>Hence the spaceships acceleration is 480m/s²</em>
Answer:B
Explanation:
Given
Wavelength of light 
Screen distance 
First fringe is at a distance 
No of lines per mm is given by N
where d=slit width
From N-slits Experiment
Position of bright fringe is given by
Put the value of 
in eq. 1
Therefore 
for 
Answer:
The correct question is:
"Find the energy each gains"
The energy gained by a charged particle accelerated through a potential difference is given by
where
q is the charge of the particle
is the potential difference
For a proton,
And since 
The energy gained by the proton is
For an alpha particle,
Therefore, the energy gained is
Finally, for a singly ionized helium nucleus (a helium nucleus that has lost one electron)
So the energy gained is the same as the proton:
