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yawa3891 [41]
2 years ago
11

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

roll toward each other and collide. The velocity is measured before and after each collision. The collected data is shown below. A 5 column table with 3 rows. The first column is unlabeled with entries Trial 1, Trial 2, Trial 3, Trial 4. The second column is labeled Initial Velocity Ball A (meters per second) with entries positive 1, positive 0.5, positive 2, positive 0.5. The third column is labeled Initial Velocity Ball B (meters per second) with entries negative 2, negative 1.5, positive 1, negative 1. The third column is labeled Final Velocity Ball A (meters per second) with entries negative 2, negative 0.5, positive 1, positive 1.5. The fourth column is labeled Final Velocity Ball B (meters per second) with entries negative 1, negative 0.5, negative 2, negative 1.5. Which trial shows the conservation of momentum in a closed system? Trial 1 Trial 2 Trial 3 Trial 4
Physics
2 answers:
Studentka2010 [4]2 years ago
4 0

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

kvasek [131]2 years ago
4 0

Answer: Trial 2

Explanation:

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.. A 15.0-kg fish swimming at 1.10 m>s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effec
stira [4]

Answer:

(a) 0.846 m/s

(b) 2.097J

Explanation:

Parameters given:

Mass of big fish, M = 15 kg

Mass of small fish, m = 4.5 kg

Initial speed of big fish, U = 1.1 m/s

Initial speed of small fish, u = 0 m/s (it is stationary)

(a) We apply the principle of conservation of momentum:

Total initial momentum = Total final momentum

Since both fish have the same final speed, V, (the small fish is in the mouth of the big fish), we have:

MU + mu = (M + m)*V

(15 * 1.1) + (4.5 * 0) = ( 15 + 4.5) * V

16.5 = 19.5V

=> V = 16.5/19.5

V = 0.846 m/s

The speed of the large fish after the meal is 0.846 m/s.

(b) We need to find the change in Kinetic energy of the entire system to find the total mechanical energy dissipated.

Initial Kinetic energy:

KEini = (½ * M * U²) + (½ * m * u²)

KEini = (½ * 15 * 1.1²) + (½ * 4.5 * 0²)

KEini = 9.075 J

Final Kinetic Energy:

KEfin = (½ * M * V²) + (½ * m * V²)

KEfin = (½ * 15 * 0.846²) + (½ * 4.5 * 0.846²)

KEfin = 5.368 + 1.610 = 6.978 J

Change in kinetic energy will be:

KEfin - KEini = 9.075 - 6.978

ΔKE = 2.097 J

The energy dissipated in eating the meal is 2.097 J

5 0
2 years ago
What type of equilibrium is guaranteed by each condition of equilibrium
Ostrovityanka [42]

Answer: Conditions for equilibrium require that the sum of all external forces acting on the body is zero (first condition of equilibrium), and the sum of all external torques from external forces is zero (second condition of equilibrium). These two conditions must be simultaneously satisfied in equilibrium

Explanation: Hope this helped

5 0
1 year ago
A 4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft>s. if the attached cord is pulled
Leya [2.2K]
Position #1:
radius, r₁ = 3 ft
Tangential speed, v₁ = 6 ft/s

By definition, the angular speed is
ω₁ = v₁/r₁ = (3 ft/s) / (3 ft) = 1 rad/s

Position #2:
Radius, r₂ = 2 ft

By definition, the moment of inertia in positions 1 and 2 are respectively
I₁ = (4 lb)*(3 ft)² = 36 lb-ft²
I₂ = (4 lb)*(2 ft)² = 16 lb-ft²

Because momentum is conserved,
I₁ω₁ = I₂ω₂
Therefore the angular velocity in position 2 is
ω₂ = (I₁/I₂)ω₁
      = (36/16)*1 = 2.25 rad/s
The tangential velocity in position 2 is
v₂ = r₂ω₂ = (2 ft)*(225 rad/s) = 4.5 ft/s

At each position, there is an outward centripetal force.
In position 1, the centripetal force is
F₁ = m*(v²/r₂) = (4)*(6²/3) = 48 lbf
In position 2, the centripetal force is
F₂ = (4)*(4.5²/2) = 40.5 lbf

The radius diminishes at a rate of 2 ft/s.
Therefore the force versus distance curve is as shown below.

The work done is the area under the curve, and it is
W = (1/2)*(48.0+40.5 ft)*(3-2 ft) = 44.25 ft-lb

Answer:  44.25 ft-lb


6 0
2 years ago
A glass beaker of unknown mass contains of water. The system absorbs of heat and the temperature rises as a result. What is the
kakasveta [241]

From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

volume of water = 74 mL

Mass of water = 74 g

specific heat of glass = 0.18 cal/g ∙ °C

specific heat of water = 1.0 cal/g ∙ C°

Mass of glass =  x g

Total heat gained by the system = 2000.0cal

Temperature rise = 20.0°C

Heat gained by system = Heat gained  by glass + Heat gained by water

Heat gained by glass = x ×  0.18 × 20

Heat gained by water = 74  ×  1.0 × 20

Hence;

2000 =  (x ×  0.18 × 20) + ( 74  ×  1.0 × 20)

2000 - 1480 =  (x ×  0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.​

Learn more: brainly.com/question/1446583

3 0
1 year ago
A vehicle has an initial velocity of v0 when a tree falls on the roadway a distance xf in front of the vehicle. The driver has a
Korvikt [17]

Answer:

v^2=v_o^2-2\times a\times (v_o.t)

Explanation:

Given:

Initial velocity of the vehicle, v_o

distance between the car and the tree, x_f

time taken to respond to the situation, t

acceleration of the car after braking, a

Using equation of motion:

v^2=u^2+2a.s ..............(1)

where:

v= final velocity of the car when it hits the tree

u= initial velocity of the  car when the tree falls

a= acceleration after the brakes are applied

s= distance between the tree and the car after the brakes are applied.

s=v_o\times t

Now for this situation the eq. (1) becomes:

v^2=v_o^2-2\times a\times (v_o.t) (negative sign is for the deceleration after the brake is applied to the car.)

5 0
1 year ago
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