हिंदी

A Force Acts for 10 S on a Stationary Body of Mass 100 Kg, After Which the Force Ceases to Act. the Body Moves Through a Distance of 100 M in the Next 5 S. Calculate : the Magnitude of the Force

Advertisements
Advertisements

प्रश्न

A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate : The magnitude of the force

योग
Advertisements

उत्तर

Mass, m = 100 kg

Distance moved, s = 100 m

Initial velocity, u = 0

Force, F = ma

Or, F = (100) (2) N.

Or, F = 200 N.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Laws of Motion - Exercise 3 (C) [पृष्ठ ७०]

APPEARS IN

सेलिना Concise Physics [English] Class 9 ICSE
अध्याय 3 Laws of Motion
Exercise 3 (C) | Q 7.3 | पृष्ठ ७०

संबंधित प्रश्न

A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)


Suppose you are running fast in a field and suddenly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration?


If the tension in the cable supporting an elevator is equal to the weight of the elevator, the elevator may be
(a) going up with increasing speed
(b) going down with increasing speed
(c) going up with uniform speed
(d) going down with uniform speed


A block of mass 0.2 kg is suspended from the ceiling by a light string. A second block of mass 0.3 kg is suspended from the first block by another string. Find the tensions in the two strings. Take g = 10 m/s2.


A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s2.


How can Newton's first law of motion be obtained from the second law of motion?


A pebble is thrown vertically upwards with a speed of 20 m s-1. How high will it be after 2 s? (Take g = 10 m s-2)


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 depth of water surface


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


The motion of a particle of mass m is given by x = 0 for t < 0 s, x(t) = A sin 4 pt for 0 < t < (1/4) s (A > o), and x = 0 for t > (1/4) s. Which of the following statements is true?

  1. The force at t = (1/8) s on the particle is – 16π2 Am.
  2. The particle is acted upon by on impulse of magnitude 4π2 A m at t = 0 s and t = (1/4) s.
  3. The particle is not acted upon by any force.
  4. The particle is not acted upon by a constant force.
  5. There is no impulse acting on the particle.

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×