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Softa [21]
3 years ago
6

A ball of silly putty hits andsticks to a bowling ball that was initially at rest. After thecollision, the total kinetic energy

of the bowling ball and sillyputty is
1. the same as the kinetic energy of thesilly putty before the collision
2. more than thekinetic energy of the silly putty before thecollision

3. less than the kinetic energy of thesilly putty before the collision
well since the momentum of the putty isgreat enough to move the bowling ball then wouldnt the totalkinetic energy of the bowling ball and puuty be more than thekinetic energy of the putty before the collision
Physics
1 answer:
Jobisdone [24]3 years ago
7 0

Answer:

3. less than the kinetic energy of thesilly putty before the collision.

Explanation:

This is because kinetic energy is dependent on the mass and velocity of an object. Mathematically, it is given as:

K. E. = ½*m*v²

Where m = mass

v = velocity

In the case of the silly putty, we know that the masses of the ball of silly putty and the bowling ball are conserved, hence, the kinetic energy depends solely on the velocity at which the object moves.

After the collision with the bowling ball, because of how heavy a bowling ball is, the speed of the silly putty and bowling ball will definitely be less than the speed of the silly putty before collision, i. e. u > v.

Hence, the kinetic energy after collision will be less than the kinetic energy before collision.

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Karen runs sets in basketball practice. She starts from a line runs 2.0 m, returns to the line, runs 4.0 m, to the line, runs 6.
Aleonysh [2.5K]

D. distance = 23 m, displacement = + 1 m

Explanation:

Let's remind the difference between distance and displacement:

- distance is a scalar, and is the total length covered by an object, counting all the movements in any direction

- displacement is a vector connecting the starting point and the final point of a motion, so its magnitude is given by the length of this vector, and its direction is given by the direction of this vector.

In this case, the distance covered by Karen is given by the sum of all its movements:

distance = 2.0 m + 2.0 m+4.0 m+4.0 m +6.0 + 5.0 m=23.0 m

The displacement instead is given by the difference between the final point (1.0 m in front of the starting line) and the starting point (the starting line, 0 m):

displacement = +1.0 m-0 m=+1 m

8 0
3 years ago
in which type of wave are vibrations at right angles to the direction in which the wave is travelling​
Pani-rosa [81]

Answer:

longitudinal waves have those properties

7 0
3 years ago
A 15 kg kangaroo jumps with an upward acceleration of 3 m/s2 by pushing hard off the ground.
patriot [66]

Explanation:

The net force would be upwards since the kangaroo would have to overcome gravity to jump

3 0
3 years ago
Read 2 more answers
A slender rod is 90.0 cm long and has mass 0.120 kg. A small 0.0200 kg sphere is welded to one end of the rod, and a small 0.070
likoan [24]

Given Information:

length of slender rod = L = 90 cm = 0.90 m

mass of slender rod = m = 0.120 kg

mass of sphere welded to one end = m₁ = 0.0200 kg

mass of sphere welded to another end = m₂ = 0.0700 kg (typing error in the question it must be 0.0500 kg as given at the end of the question)

Required Information:

Linear speed of the 0.0500 kg sphere = v = ?

Answer:

Linear speed of the 0.0500 kg sphere = 1.55 m/s

Explanation:

The velocity of the sphere can by calculated using

ΔKE = ½Iω²

Where I is the moment of inertia of the whole setup ω is the speed and ΔKE is the change in kinetic energy

The moment of inertia of a rigid rod about center is given by

I = (1/12)mL²

The moment of inertia due to m₁ and m₂ is

I = (m₁+m₂)(L/2)²

L/2 means that the spheres are welded at both ends of slender rod whose length is L.

The overall moment of inertia becomes

I = (1/12)mL² + (m₁+m₂)(L/2)²

I = (1/12)0.120*(0.90)² + (0.0200+0.0500)(0.90/2)²

I = 0.0081 + 0.01417

I = 0.02227 kg.m²

The change in the potential energy is given by

ΔPE = m₁gh₁ + m₂gh₂

Where h₁ and h₂ are half of the length of slender rod

L/2 = 0.90/2 = 0.45 m

ΔPE = 0.0200*9.8*0.45 + 0.0500*9.8*-0.45

The negative sign is due to the fact that that m₂ is heavy and it would fall and the other sphere m₁ is lighter and it would will rise.

ΔPE = -0.1323 J

This potential energy is then converted into kinetic energy therefore,

ΔKE = ½Iω²

0.1323 = ½(0.02227)ω²

ω² = (2*0.1323)/0.02227

ω = √(2*0.1323)/0.02227

ω = 3.45 rad/s

The linear speed is

v = (L/2)ω

v = (0.90/2)*3.45

v = 1.55 m/s

Therefore, the linear speed of the 0.0500 kg sphere as its passes through its lowest point is 1.55 m/s.

8 0
3 years ago
A uniform, 4.5 kg, square, solid wooden gate 2.0 mm on each side hangs vertically from a frictionless pivot at the center of its
True [87]

Answer:

The angular velocity is  w = 1.43\  rad/sec

Explanation:

From the question we are told that

   The  mass of wooden gate  is m_g = 4.5 kg

    The  length of side is  L = 2 m

    The mass of the raven is  m_r = 1.2 kg

     The initial speed of the raven is u_r = 5.0m/s

     The final speed of the raven is   v_r = 1.5 m/s

From the law of  conservation of angular momentum we express this question mathematically as

       Total initial angular momentum  of both the Raven and  the Gate =  The Final angular momentum of both the Raven and the Gate  

The initial angular momentum of the Raven is m_r * u_r * \frac{L}{2}

Note: the length is half because the Raven hit the gate at the mid point

The initial angular momentum of the Gate is  zero

Note: This above is the generally formula for angular momentum of  square objects

  The final angular velocity  of the Raven is  m_r * v_r * \frac{L}{2}

   The  final angular velocity of the Gate  is   \frac{1}{3} m_g L^2 w

Substituting this formula

  m_r * u_r * \frac{L}{2}  =   \frac{1}{3} m_g L^2 w + m_r * v_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r * v_r * \frac{L}{2} -   m_r * u_r * \frac{L}{2}

  \frac{1}{3} m_g L^2 w   =    m_r *  \frac{L}{2} * [u_r - v_r]

Where w is the angular velocity

     Substituting value  

   \frac{1}{3} (4.5)(2)^2  w   =    1.2 *  \frac{2}{2} * [5 - 1.5]

     6w = 4.2

       w = \frac{6}{4.2}

            w = 1.43\  rad/sec

5 0
3 years ago
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