Answer:
= 26.94 m/s
Explanation:
given,
angle of inclination = 30°
mass of the sleigh = 200 kg
coefficient of kinetic friction = 0.2
height of inclination = 10 m
pull force be = 5000 N
now,.
T - f_s - mg sin \theta = m aT−f
s
−mgsinθ=ma
T - \mu N - mg sin \theta = m aT−μN−mgsinθ=ma
T - \mu mg - mg sin \theta = m aT−μmg−mgsinθ=ma
a = \dfrac{T}{m} - \mu g - g sin \thetaa=
m
T
−μg−gsinθ
a = \dfrac{5000}{200} - \0.2\times 9.8 - 9.8 \times sin 30^0a=
200
5000
−\0.2×9.8−9.8×sin30
0
a = 18.14\ m/s^2a=18.14 m/s
2
L = \dfrac{10}{sin 30}L=
sin30
10
L = 20 m
v² = u² + 2 as
v² = 0 + 2 x 18.14 x 20
v = 26.94 m/s
Answer:
Explanation:
From the data it appears that A is the middle point between two charges.
First of all we shall calculate the field at point A .
Field due to charge -Q ( 6e⁻ ) at A
= 9 x 10⁹ x 6 x 1.6 x 10⁻¹⁹ / (2.5)² x 10⁻⁴
= 13.82 x 10⁻⁶ N/C
Its direction will be towards Q⁻
Same field will be produced by Q⁺ charge . The direction will be away
from Q⁺ towards Q⁻ .
We shall add the field to get the resultant field .
= 2 x 13.82 x 10⁻⁶
= 27.64 x 10⁻⁶ N/C
Force on electron put at A
= charge x field
= 1.6 x 10⁻¹⁹ x 27.64 x 10⁻⁶
= 44.22 x 10⁻²⁵ N
Answer:
4.4 rad/s
Explanation:
When the 2 skaters is spinning with a distance of 1.5m, their rotation radius is half of that distance, which is 1.5/2 = 0.75m.
Then their moments of inertia, given that their mass being 51 kg, is (treating them as point mass particle):

When they change from linear motion to rotational motion, their energy must be conserved:





Answer:
b) se duplica
Explanation:
The disk is moving with constant angular velocity, let's call it
.
The linear velocity of a point on the disk is given by

where r is the distance of the point from the axis of rotation.
In this problem, the object is moved at a distance twice as far as the initial point, so

Therefore, the new linear velocity is

So, the velocity has doubled, and the correct answer is
b) se duplica
Answer:
4400 N
Explanation: (c)
P= 
F= PA
P= pressure
A= area
P=800x
and A= 5.5 x 

Force F= 800x
x0.0055= 4400 N
force exerted by the gas on the piston is 4400 N