Here we have perfectly inelastic collision. Perfectly inelastic collision is type of collision during which two objects collide, stay connected and momentum is conserved. Formula used for conservation of momentum is:
In case of perfectly inelastic collision v'1 and v'2 are same.
We are given information:
m₁=0.5kg
m₂=0.8kg
v₁=3m/s
v₂=2m/s
v'₁=v'₂=x
0.5*3 + 0.8*2 = 0.5*x + 0.8*x
1.5 + 1.6 = 1.3x
3.1 = 1.3x
x = 2.4 m/s
Answer: a. 198.6J b. - 198.6J
Explanation: Parameters given:
m = 15kg
g = 9.8m/s²
∅ = 12°
a. Work done by the force Fp on the cart if the ramp is 6.5m long.
Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N
Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J
b. The work done by the force mg on the cart.
Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance
= 15kg x -9.8m/s² x Sin12° x 6.5m
= - 198.6J
Answer:The charge passing through the circuit always passes through an appliance (which acts as a resistor) or through another resistor, which limits the amount of current that can flow through a circuit. Appliances are designed to keep current at a relatively low level for safety purposes.
Explanation:
Answer:
Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s
Explanation:
given information:
moment inertia of arm and leg when in, I₁ = 0.9 kgm²
moment inertia of arm and leg when extended, I₂ = 2.9 kgm²
angular speed when in, ω₁ = 4.5 rev/s
so, her angular speed (in rev/s) when her arms and one leg open outward is
L₁ = L₂
I₁ω₁ = I₂ω₂
ω₂ = I₁ω₁/I₂
= 0.9 x 4.5/2,9
= 1.4 rev/s
Here’s 3 examples:
MCB is more sensitive to current than fuse.
In case of MCB, the faulty zone of electrical circuit can be easily identified.
With MCB it is very simple to resume to the supply.
~Emmy
