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An object is dropped from a height of 80 m. Assuming no air resistance, how long will it take to reach the ground? ( 𝑔 = 10 m/s² )
Using s = 1/2gt² , where s= 80m and g = 10 m/s ²: 80 = 1/2(10)t² 80 = 5t² t² = 16 t = 4s This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.com/ncRead more
Using s = 1/2gt² , where s= 80m and g = 10 m/s ²:
80 = 1/2(10)t²
80 = 5t²
t² = 16
t = 4s
This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
Which of the following is a correct statement about acceleration?
Acceleration is defined as the rate of change of velocity with respect to time. It can be positive, negative, or zero depending on whether the velocity is increasing, decreasing, or constant. This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. GiRead more
Acceleration is defined as the rate of change of velocity with respect to time. It can be positive, negative, or zero depending on whether the velocity is increasing, decreasing, or constant.
This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of 14 m/s to the heavier block in the direction of the lighter block. The velocity of the centre of mass is
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocityRead more
The velocity of a center of mass for a pair of blocks, given their respective masses and the velocities, will be determined for this problem as well. We therefore have two different blocks: heavier with a mass of 10 kg and light with a mass of 4 kg. The bigger block is thus given a certain velocity of 14 m/s towards the minor block, given that the little block is assumed to be moving on a horizontal frictionless ground.
It is the average position of the entire mass in the system weighted by their respective velocities. The velocity of the center of mass for this system would depend on the contributions from each block’s mass and its respective velocity. In this case, because the light block has no movement at all it contributes nothing to the velocity of the center of mass. Its movement is dominated by that of the heavier block since the former has more mass and higher speed.
By adding both blocks’ masses and velocities together, the center of mass’s velocity is found to be 10 m/s. This represents the motion of the system in a frictionless surface. The center of mass’s velocity is only changed when acted on by an outside force according to the principles of physics.
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See lessA body moves in a straight line with a uniform velocity of 10m/s for 5 seconds. What is the total displacement?
Displacement is given by s = vt, where v = 10m/s and t = 5s: s= 10 × 5 = 50m. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
Displacement is given by s = vt, where v = 10m/s and t = 5s:
s= 10 × 5 = 50m.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
What does the area under the velocity-time graph of an object represent?
The area under the velocity-time graph represents the displacement of an object. It is calculated by integrating the velocity with respect to time or finding the area geometrically in cases of simple shapes. For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/cRead more
The area under the velocity-time graph represents the displacement of an object. It is calculated by integrating the velocity with respect to time or finding the area geometrically in cases of simple shapes.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/