Solution :-
Given :
Distance 1 = 30 km
Distance 2 = 70 km
We know that speed = distance/time
and, Average speed = total distance/total time taken
When the train acquired a speed of 30 km/hr, the time taken = 30/30 = 1 hour
Average speed = 9distance 1 + distance 2)/(time 1 + time 2)
AS time 2 or t2 is time taken for the second part of the journey of 70 km
⇒ 40 = 100/(1 + t2)
⇒ 40 + 40t2 = 100
⇒ 40t2 = 100 - 40
⇒ 40t2 = 60
⇒ t2 = 60/40
⇒ t2 = 1.5
So, t2 or time taken to travel the second part of the journey is 1.5 hours.
Speed of the second part of the journey = distance 2/time 2
⇒ 70/1.5
⇒ 46.666 km/hr or 46.7 km/hr.
Hence the answer is = 46.666 km/hr or 46.7 km/hr.
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Convection is the movement caused within a fluid by the tendency of hotter and therefore less dense material to rise, and colder, denser material to sink under the influence of gravity, which consequently results in transfer of heat
Answer: 10 m/s^2
Explanation:
1) The second law of Newton gives the definition and formula to calculate the net force:
Net force acting on an object = mass * acceleration.
2) From that, when you know the net force acting of the object and its mass, you can solve for the acceleration:
acceleration = Net force / mass
acceleration = 50 N / 5 kg = 10 m/s^2, which is the answer.
Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.
f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
Answer:
1.925 μC
Explanation:
Charge: This can be defined as the product of the capacitance of a capacitor and the voltage. The S.I unit of charge is Coulombs (C)
The formula for the charge stored in a capacitor is given as,
Q = CV ................... Equation 1
Where Q = charge, C = Capacitor, V = Voltage.
Note: 1 μF = 10⁻⁶ F
Given: C = 0.55 μF = 0.55×10⁻⁶ F, V = 3.5 V.
Substitute into equation 1
Q = 0.55×10⁻⁶×3.5
Q = 1.925×10⁻⁶ C.
Q = 1.925 μC
Hence the charge on the plate = 1.925 μC
