Answer:78 N
Explanation:Impulse=Force*time
=39*2 =78 Ns
Answer:
1⁺ ion
Explanation:
Metals in the first group on the periodic table will prefer to form 1⁺ ion. This is because the 1 valence electron in their orbital.
Most metals are electropositive and would prefer to lose electrons than to gain it.
Like all metals, the group 1 elements called the alkali metals would prefer to lose and electron.
On losing an electron the number of protons is then greater than the number of electrons. This leaves a net positive charge.
Answer:
82.7 m
Explanation:
u= 22m/s
a= 2.4 m/s^2.
t= 3.2 secs
Therefore the distance travelled can be calculated as follows
S= ut + 1/2at^2
= 22 × 3.2 + 1/2 × 2.4 × 3.2^2
= 70.4 + 1/2×24.58
= 70.4 + 12.29
= 82.7 m
Hence the distance travelled by the truck is 82.7 m
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Height of baby carriage from ground = 21m
Mass of carriage with baby = 1.5 kg
The carriage has potential energy by virtue of its height.
Potential energy = mgh = 1.5×10×21 = 315 J
Hence, potential energy of the carriage is 315 Joule.
