The vertical component of the velocity is given by š¢įµ§ = š¢ sin š where š is the angle of projection. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.comRead more
The vertical component of the velocity is given by š¢įµ§ = š¢ sin š where š is the angle of projection. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding.
Displacement is the straight-line distance between the initial and final positions and is always less than or equal to the actual distance traveled. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding. For moreRead more
Displacement is the straight-line distance between the initial and final positions and is always less than or equal to the actual distance traveled. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding.
In non-uniform circular motion, there is tangential acceleration due to the change in speed and centripetal acceleration due to the change in direction. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding. ForRead more
In non-uniform circular motion, there is tangential acceleration due to the change in speed and centripetal acceleration due to the change in direction. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding.
Using s = 1/2gtĀ² , where s= 80m and g = 10 m/s Ā²: 80 = 1/2(10)tĀ² 80 = 5tĀ² tĀ² = 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 = 1/2gtĀ² , where s= 80m and g = 10 m/s Ā²:
80 = 1/2(10)tĀ²
80 = 5tĀ²
tĀ² = 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.
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.
Displacement is given by s = vt, where v = 10m/s and t = 5s: s= 10 Ć 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 Ć 5 = 50m.
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.
Using s = ut +1/2atĀ², where u = 0, a = 4m/sĀ² and t = 5s: s = 0 +1/2(4)(5Ā²) = 1/2(4)(25) = 50 Ć 2 = 100m. 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.tiwaRead more
Using s = ut +1/2atĀ², where u = 0, a = 4m/sĀ² and t = 5s:
s = 0 +1/2(4)(5Ā²) = 1/2(4)(25) = 50 Ć 2 = 100m.
This question related to Chapter 2 physics Class 11th NCERT. From the Chapter 2. Motion in Straight Line. Give answer according to your understanding.
The second equation of motion is s=ut+ 1/2atĀ², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2atĀ². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chaptRead more
The second equation of motion is s=ut+ 1/2atĀ², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2atĀ².
The second equation of motion is s = ut +1/2atĀ². Substituting u = 0, s = 20m and t = 2s: 20 = 0 +1/2a(2Ā²) 20 = 2a a = 10 m/sĀ². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
The second equation of motion is s = ut +1/2atĀ². Substituting u = 0, s = 20m and t = 2s:
20 = 0 +1/2a(2Ā²)
20 = 2a
a = 10 m/sĀ².
If the horizontal component of a projectiles velocity is š¢ cos ā”š what is the vertical component?
The vertical component of the velocity is given by š¢įµ§ = š¢ sin š where š is the angle of projection. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding. For more please visit here: https://www.tiwariacademy.comRead more
The vertical component of the velocity is given by š¢įµ§ = š¢ sin š where š is the angle of projection. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-3/
The magnitude of displacement in projectile motion is always:
Displacement is the straight-line distance between the initial and final positions and is always less than or equal to the actual distance traveled. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding. For moreRead more
Displacement is the straight-line distance between the initial and final positions and is always less than or equal to the actual distance traveled. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-3/
A particle moves along a circular path with increasing speed. The net acceleration of the particle is:
In non-uniform circular motion, there is tangential acceleration due to the change in speed and centripetal acceleration due to the change in direction. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding. ForRead more
In non-uniform circular motion, there is tangential acceleration due to the change in speed and centripetal acceleration due to the change in direction. This question related to Chapter 3 physics Class 11th NCERT. From the Chapter 3. Motion in Plane. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-3/
An object is dropped from a height of 80 m. Assuming no air resistance, how long will it take to reach the ground? ( š = 10 ām/sĀ² )
Using s = 1/2gtĀ² , where s= 80m and g = 10 m/s Ā²: 80 = 1/2(10)tĀ² 80 = 5tĀ² tĀ² = 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 = 1/2gtĀ² , where s= 80m and g = 10 m/s Ā²:
80 = 1/2(10)tĀ²
80 = 5tĀ²
tĀ² = 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:
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/
A 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 Ć 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 Ć 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/
An object moves with uniform acceleration of 4 m/sĀ². If it starts from rest, what will be its displacement after 5 seconds?
Using s = ut +1/2atĀ², where u = 0, a = 4m/sĀ² and t = 5s: s = 0 +1/2(4)(5Ā²) = 1/2(4)(25) = 50 Ć 2 = 100m. 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.tiwaRead more
Using s = ut +1/2atĀ², where u = 0, a = 4m/sĀ² and t = 5s:
s = 0 +1/2(4)(5Ā²) = 1/2(4)(25) = 50 Ć 2 = 100m.
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/
If an object starts from rest and moves with uniform acceleration, what will be its displacement after š” seconds?
The second equation of motion is s=ut+ 1/2atĀ², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2atĀ². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chaptRead more
The second equation of motion is s=ut+ 1/2atĀ², where u is the initial velocity, a is the acceleration, and t is the time. For objects starting from rest (u =0), the displacement simplifies to s = 1/2atĀ².
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
A particle covers 20m in the first 2 seconds of its motion from rest under uniform acceleration. What is the magnitude of its acceleration?
The second equation of motion is s = ut +1/2atĀ². Substituting u = 0, s = 20m and t = 2s: 20 = 0 +1/2a(2Ā²) 20 = 2a a = 10 m/sĀ². For more please visit here: https://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/
The second equation of motion is s = ut +1/2atĀ². Substituting u = 0, s = 20m and t = 2s:
20 = 0 +1/2a(2Ā²)
20 = 2a
a = 10 m/sĀ².
For more please visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-2/