Answer:
Part a)

Part b)

Part C)

Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Explanation:
Part a)
As we know that car A moves by distance 6.1 m after collision under the frictional force
so the deceleration due to friction is given as



now we will have




Part b)
Similarly for car B the distance of stop is given as 4.4 m
so we will have


Part C)
By momentum conservation we will have



Part d)
Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.
Answer: clean drinking water I THINK
Explanation:
Answer:
here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
Explanation:
As we know that gravitational field is defined as the force experienced by the satellite per unit of mass
so we will have

now in order to find the acceleration of the satellite we know by Newton's II law

so we will have

so here we can say that acceleration of the satellite is same as the gravitational field due to Earth at that location
Answer:
2.86×10⁻¹⁸ seconds
Explanation:
Applying,
P = VI................ Equation 1
Where P = Power, V = Voltage, I = Current.
make I the subject of the equation
I = P/V................ Equation 2
From the question,
Given: P = 0.414 W, V = 1.50 V
Substitute into equation 2
I = 0.414/1.50
I = 0.276 A
Also,
Q = It............... Equation 3
Where Q = amount of charge, t = time
make t the subject of the equation
t = Q/I.................. Equation 4
From the question,
4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs
Q = 7.899×10⁻¹⁹ C
Substitute these value into equation 4
t = 7.899×10⁻¹⁹/0.276
t = 2.86×10⁻¹⁸ seconds