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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a) since force = mass * acceleration
f= 900 * 0 (because constant speed = 0 acceleration)
similarly b) f = 0
Answer: 996m/s
Explanation:
Formula for calculating velocity of wave in a stretched string is
V = √T/M where;
V is the velocity of wave
T is tension
M is the mass per unit length of the wire(m/L)
Since the second wire is twice as far apart as the first, it will be L2 = 2L1
Let V1 and V2 be the speed of the shorter and longer wire respectively
V1 = √T/M1... 1
V2 = √T/M2... 2
Since V1 = 249m/s, M1 = m/L1 M2 = m/L2 = m/2L1
The equations will now become
249 = √T/(m/L1) ... 3
V2 = √T/(m/2L1)... 4
From 3,
249² = TL1/m...5
From 4,
V2²= 2TL1/m... 6
Dividing equation 5 by 6 we have;
249²/V2² = TL1/m×m/2TL1
{249/V2}² = 1/2
249/V2 = (1/2)²
249/V2 = 1/4
V2 = 249×4
V2 = 996m/s
Therefore the speed of the wave on the longer wire is 996m/s
Answer:
Lithium has the lowest electronegativity.
Explanation:
Electronegativity measures the tendency of an atom to attract the bonding pair of electrons. As we move left to right in a period, the number of shell remains same but the number of electron increases(negative charge increases. Hence, electronegativity also increases.
As Lithium is the left most element in this period. It has the lowest electronegativity value which is equal to 0.98.
Answer:
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