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SOLUTION of the Above Given Question..🙌✨
Join➵PW Arjuna JEE - Class 11
SOLUTION of the Above Given Question..🙌✨
Join➵PW Arjuna JEE - Class 11
[ JEE PYQ ]
A particle moving in a straight line covers half the distance with speed 6 m/s. The other half is covered in two equal time intervals with speeds 9 m/s and 15 m/s respectively. The average speed of the particle during the motion is
A particle moving in a straight line covers half the distance with speed 6 m/s. The other half is covered in two equal time intervals with speeds 9 m/s and 15 m/s respectively. The average speed of the particle during the motion is
Anonymous Quiz
28%
9.2 m/s
34%
8.8 m/s
25%
10 m/s
13%
8 m/s
[ THREE DIMENSIONAL GEOMETRY ]
What is the distance between the points (2, –1, 3) and (–2, 1, 3)?
What is the distance between the points (2, –1, 3) and (–2, 1, 3)?
Anonymous Quiz
66%
2√5 units
16%
25 units
14%
4√5 units
4%
√5 units
Answer: A. 2√5 units
Explanation: Let the points be P (2, – 1, 3) and Q (– 2, 1, 3)
By using the distance formula,
PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 1, z1 = 3
x2 = – 2, y2 = 1, z2 = 3
PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
Therefore, the required distance is 2√5 units.
Explanation: Let the points be P (2, – 1, 3) and Q (– 2, 1, 3)
By using the distance formula,
PQ = √[(x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2]
So here,
x1 = 2, y1 = – 1, z1 = 3
x2 = – 2, y2 = 1, z2 = 3
PQ = √[(-2 – 2)2 + (1 – (-1))2 + (3 – 3)2]
= √[(-4)2 + (2)2 + (0)2]
= √[16 + 4 + 0]
= √20
= 2√5
Therefore, the required distance is 2√5 units.
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Let's Understand a Very Basic Topic :- EXPONENTS ( POWER )
Many Students Have Doubts in this Topic.. So, Sir is Here To Explain to All Arjunas 👊
Ye Videos me Detailed Explanation Hai so, watch it carefully fir koi bhi doubt nhi hoga🤩
BY PAWAN SIR ♥️
Many Students Have Doubts in this Topic.. So, Sir is Here To Explain to All Arjunas 👊
Ye Videos me Detailed Explanation Hai so, watch it carefully fir koi bhi doubt nhi hoga🤩
BY PAWAN SIR ♥️
Graham's law deals with the relation between:
Anonymous Quiz
30%
Pressure and volume
36%
Density and rate of diffusion
25%
Rate of diffusion and volume
8%
Rate of diffusion and viscosity
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Ye Dimag me hona chahiye question paper attempt krte time ⏰