Answer:
Option c is correct
Explanation:
There are two types of collisions-elastic collision and inelastic collision.
In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.
Total energy is always conserved in both types. Thus, option a is incorrect.
In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.
Answer:
The last option, 20 N and 2.04 kg
Explanation:
work = (force)(distance)
work = 120 joules
distance: 6 m
rearrange to find force:
120=(6)F
F= 120/6 = 20 Newtons.
Assuming its lifted from Earth's surface, the force of gravity will be 9.81 m/s^2. Let's find mass:
F=mg
m=F/g
m=(20)/(9.81)= 2.038 kg
Answer:
Explanation:
Take sum of torques at the point the step touches the wheel, that eliminates two torques
Σ
Since we are looking for when the wheel just starts to rise up N-> 0 so no torque due to normal force
The perpendicular lever arm for the F force is R-h
And the T of gravity according to the image
Σ
Answer:
Explanation:
A ) angular velocity ω = 2π / T
= 2 x 3.14 / 60
= .10467 rad / s
linear velocity v = ω R
= .10467 x 50
= 5.23 m / s
centripetal force = m v² / R
= mg v² / gR
= 834 x 5.23² / 9.8 x 50
= 46.55 N
B )
apparent weight
= mg - centripetal force
= 834 - 46.55
= 787.45 N
C ) apparent weight
= mg + centripetal force
= 834 + 46.55
= 880.55 N.
D )
For apparent weight to be zero
centripetal force = mg
mg = mv² / R
v² = gR
= 9.8 x 50
= 490
v = 22.13 m /s
time period of revolution
= 2π R /v
2 x 3.14 x 50 / 22.13
= 14.19 s
Answer:
Lone pairs cause bond angles to deviate away from the ideal bond angles
Explanation:
Bonded electrons are stabilized and clustered between the bonding electrons meaning they are much closer together. Non-bonding electrons however are not being shared between any atoms which allows them to roam a little further spreading the charge density over a larger space and therefore interfering with what would be an expected bond angle
