हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

A Spy Jumps from an Airplane with His Parachute. the Spy Accelerates Downward for Some Time When the Parachute Opens. the Acceleration is Suddenly Checked and the Spy Slowly Falls to the

Advertisements
Advertisements

प्रश्न

A spy jumps from an airplane with his parachute. The spy accelerates downward for some time when the parachute opens. The acceleration is suddenly checked and the spy slowly falls to the ground. Explain the action of the parachute in checking the acceleration.

टिप्पणी लिखिए
Advertisements

उत्तर

Air applies a velocity-dependent force on the parachute in upward direction when the parachute opens. This force opposes the gravitational force acting on the spy. Hence, the net force in the downward direction decreases and the spy decelerates.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Newton's Laws of Motion - short answers [पृष्ठ ७७]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 5 Newton's Laws of Motion
short answers | Q 13 | पृष्ठ ७७

संबंधित प्रश्न

The below figure shows the position-time graph of a particle of mass 4 kg.

  1. What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
  2. What is the impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)


Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to

  1. A,
  2. B along the direction of string. What is the tension in the string in each case?

A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 ms–2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.


A person drops a coin. Describe the path of the coin as seen by the person if he is in

  1. a car moving at constant velocity and
  2. in a free falling elevator.

A car accelerates on a horizontal road due to the force exerted by.


A particle of mass 50 g moves in a straight line. The variation of speed with time is shown in the following figure. Find the force acting on the particle at t = 2, 4 and 6 seconds.


The force of buoyancy exerted by the atmosphere on a balloon is B in the upward direction and remains constant. The force of air resistance on the balloon acts opposite the direction of velocity and is proportional to it. The balloon carries a mass M and is found to fall to the earth's surface with a constant velocity v. How much mass should be removed from the balloon so that it may rise with a constant velocity v?


In the following figure shows a uniform rod of length 30 cm and mass 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light.


Write the mathematical form of Newton's second law of motion. State the conditions if any.


The linear momentum of a body of mass m moving with velocity v is  : 


A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. Calculate : The velocity acquired by the body


A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The time when echo is heard after the pebble is dropped.


Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.


An electron of mass 9 × 10−31 kg is moving with a linear velocity of 6 × 107 ms−1. Calculate the linear momentum of electron.


Prove mathematically F = ma


Name the physical entity used for quantifying the motion of a body.


Use Newton's second law to explain the following:
While catching a fast moving ball, we always pull our hands backwards.


A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 ms−1, q = 4 ms−2 and r = 5 ms−3. The force acting on the body at t = 2 seconds is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×