Tag: weightlessness

Quizzes (0) Questions (54)

Questions Related to weightlessness

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A 50.0 kg boy is sitting on an amusement park ride where he accelerates straight upward from rest to a speed 30.0 m/s in 3.0 s. What is his mass as he accelerates upward?

  1. 990.0 kg

  2. 100.0 kg

  3. 50.0 kg

  4. 5.00 kg

  5. 0 kg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass of an object always remains constant whether the object is accelerating or not. Apparent weight of the object changes due to acceleration.

Hence the mass of the boy is $50$ kg even he accelerates upward.
Thus option C is correct.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A plumb bob is hung from the ceiling of a train compartment. the train moves on an inclined track of inclination $30^\circ $ with horizntal. The acceleration of train up the plane is $a=\,g/2$. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is-

  1. $30^\circ $
  2. ${\tan ^{ - 1}}\left( {2/\sqrt 3 } \right)$
  3. ${\tan ^{ - 1}}\left( {\sqrt 3/2 } \right)$
  4. ${\tan ^{ - 1}}\left( 2 \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A man of mass 'm' stands on a weighing machine in a lift

List - I  List-II
(a) Lift moves up with uniform acceleration a  (d) mg
(b) Lift moves down with uniform acceleration a (e) m(g$+$a)
(c) Lift moves down with uniform velocity (f) m(g-a)
  1. $a\rightarrow e,b\rightarrow f,c\rightarrow d,$
  2. $a\rightarrow d,b\rightarrow f,c\rightarrow e,$
  3. $a\rightarrow d,b\rightarrow e,c\rightarrow f,$
  4. $a\rightarrow f,b\rightarrow d,c\rightarrow e,$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the lift moves $UP$ with uniform acceleration $a$ $Pseudo$ force on the man is $ma$ downward.

Net downward force is $mg+ma$. Hence, $e$

When the lift moves $down$ with uniform acceleration $a$ $Pseudo$ force on the man is $ma$ upward.
Net downward force is $mg+ma$. Hence, $f$


When the lift moves down with $uniform \ velocity$, only force acting is gravity.

Hence net force on man is $mg$. Hence, $d$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A passenger in a train hangs a stone from a string and holds it vertically. It is observed that the string is vertical as seen from the man.
Which of the following could be true about the motion of the train?
I. The train is at rest.
II. The train is moving with a constant velocity.
III. The train is increasing its speed.
IV. The train is decreasing its speed.

  1. I only

  2. III only

  3. I or II, but not III or IV

  4. III or IV, but not I or II

  5. IV only
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the train is either at rest or moving with constant velocity, then the acceleration of the train is zero and hence there is no pseudo force acting on the stone in the horizontal direction which makes the string tied to the stone to stay in vertical direction.

If the train either increases or decreases its speed, then the stone experiences a pseudo force in the horizontal direction due the acceleration of train and thus the string gets tilted by some angle with the vertical direction. 
Thus option C is correct.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

When a car accelerates rapidly forward, the heads of people in the car seem to jerk backward.
This is best explained by which of the following?

  1. The action is the car accelerating forward; the reaction is the heads accelerating backward

  2. Large forward velocities cause large backward accelerations

  3. The inertia of people's heads is not as big as the inertia of the car

  4. The people's heads initially have no force to accelerate them forward. The heads momentarily remain at rest until the necks and /or head rests apply forward forces to the heads to accelerated then forward

  5. The forward momentum of the car results in an equal backward momentum of the people's head to conserve momentum
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

This happens due to inertia of rest . initially both car and people are in rest , when car accelerates rapidly , feet of the people also accelerates with car because they are in contact with the floor of car , but heads of the people do not accelerates quickly because they want to be in rest due inertia of rest therefore people feel a jerk .

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A bird weight 2 kg and is inside a cage of 1kg. If it starts flying then the weight of the bird and cage assembly is

  1. 1 kg

  2. 2 kg

  3. 3 kg

  4. 4 kg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since cage is at rest no pseudo force will be acting on bird hence weight of bird will be $2kg$ in air, therefore weight of bird and cage assembly will be  equal to 2kg and 1kg i.e equal to $3kg$.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A bird is sitting in a wire cage hanging from the spring balance. Let the reading of the spring balance be $a$. If the bird flies about inside the cage, the reading of the spring balance is $b$. Which of the following is true?

  1. $a=b$
  2. $a>b$
  3. $a$

  4. Cannot be obtained
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Reading shown on the spring balance is weight of bird + weight of cage. When bird starts flying normal force exerted by bird on the case become zero. When bird flies his weight is carried by air but case in made of wire so no additional force of weight of bird act on the cage so reading on the cage is less when bird is flying.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

If the position of two like parallel forces shifted by one-fourth of the distance between the forces when the two forces are interchanged. The ratio of the two forces is:

  1. $1:2$
  2. $2:3$
  3. $3:4$
  4. $3:5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let forces be F1 and F2 at distance d. The resultant acts at x = F2*d / (F1+F2). If forces are swapped, new position x' = F1*d / (F1+F2). The shift is |x - x'| = d * |F1-F2| / (F1+F2) = d/4. Thus, 4|F1-F2| = F1+F2. If F1 > F2, 4F1 - 4F2 = F1 + F2 => 3F1 = 5F2 => F1/F2 = 5/3. If F2 > F1, 4F2 - 4F1 = F1 + F2 => 3F2 = 5F1 => F1/F2 = 3/5. Option D matches 3:5.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A particle is observed from two frames $S _{1}$ and $S _{2}$. The frame $S _{2}$ moves with respect to $S _{1}$ with an acceleration $a$. Let $F _{1}$ and $F _{2}$ be the pseudo forces on the particle when seen from $S _{1}$ and $S _{2}$ respectively. Which of the following are not possible?

  1. $F _{1} = 0, F _{2}\neq 0$
  2. $F _{1} \neq 0, F _{2} = 0$
  3. $F _{1} \neq 0, F _{2}\neq 0$
  4. $F _{1} = 0, F _{2} = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pseudo force is defined as F_p = -m * a_frame. If a_frame = 0 (inertial frame), then F_p = 0. If the frame is non-inertial (a != 0), the pseudo force is non-zero. It is impossible for a frame to be both inertial and non-inertial simultaneously, so F1=0 and F2=0 is impossible if S2 is accelerating relative to S1.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A cyclist riding with a speed of $27kmph$. As he approaches a circular turn on the road of radius $80m$, he applies brakes and reduces his speed at the constant rate of $0.50m/s$ every second. The net acceleration of the cyclist on the circular turn is 

  1. $0.5 \quad m/s^2$
  2. $0.86\quad m/s^2$
  3. $0.56 \quad m/s^2$
  4. $1 \quad m/s^2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ v = 27 Km/hr$ 

$  = \cfrac{27 \times 1000}{3600} = 7.5 m/sec$
Centripetal acceleration = $ \cfrac{(7.5)^2}{80} = 0.7 m/s^2$
Tangential acceleration = $ -0.5m/s^2$ 
$ \therefore a _{net} = \sqrt{(0.7)^2 + (0.5)^2} = 0.86m/s^2$