All categories

2 years ago

Please help 50 points

Physics

1 answer:

PilotLPTM [1.2K]2 years ago

7 0

Answer:

3a 0.7m/s

3b partially inelastic

4a 8.33 m/s

4b completely inelastic

5. puck: 10.59 m/s octopus: 10.59m/s

6. car: 117.44 m/s truck: 17.44m/s

7a - 1 m/s , the red cart travels to the left

7b) elastic

Explanation:

For all these questions, you have find momentum (P=mv) always remember initial P is always equal to final P

Initial P:

mass of first ball x velocity of first ball + mass of second ball x velocity of second ball

0.5*3.5 + 0.5*0 = 1.75 kg.m/s

final P: also 1.75Kg.m/s

let x be the velocity for first ball

0.5*2.8+0.5*x=1.75

0.5x=1.75-1.4

x=0.35/0.5

x=0.7m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is partially inelastic, partially because the ball did not stick together

Initial P:

2575*11 + 825*0 = 28,325 kg.m/s

final P: also 28,325 Kg.m/s

let x be the velocity for both vehicles

(2575+825)x=28325

x=28325/3400

x=8.33 m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi does not equal to KEf

so the collision is completely inelastic, completely because the vechicles slides off together

Initial P:

0.115*35 + 0.265*0 = 4.025 kg.m/s

final P: also 4.025Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the puck and octopus,

(0.115+0.265)x=4.025

x=4.025/0.38

x=10.59

6. Initial P:

565*25+785*12 = 23,545 kg.m/s

final P: also 23,545 Kg.m/s

since they slides off together, the velocity will be the same, so let x be the velocity for the car and truck,

(565+785)x=23545

x=23545/1350

x=17.44

Initial P:

0.25*2 + 0.75*0 = 0.5 kg.m/s

final P: also 0.5 Kg.m/s

let x be the velocity for red cart

0.75*1+0.25*x=0.5

0.25x=0.5-0.75

x=-0.25/0.25

x=-1m/s

b) collision is elastic when KEi = KEf, so calculate and see if that's true

using 1/2mv² we know KEi equals to KEf

so the collision is elastic

You might be interested in

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.58 m. A child appl

lozanna [386]

Answer:

The value is $KE = 259.6 \ J$

Explanation:

From the question we are told that

The weight of the horizontal solid disk is $W = 805 \ N$

The radius of the horizontal solid disk is $r = 1.58 \ m$

The force applied by the child is $F = 49.5 \ N$

The time considered is $t = 2.95 \ s$

Generally the mass of the horizontal solid disk is mathematically represented as

$m_h = \frac{W}{ g}$

=> $m_h = \frac{805}{ 9.8 }$

=> $m_h = 82.14 \ N$

Generally the moment of inertia of the horizontal solid disk is mathematically represented as

$I = \frac{1}{2} * m * r^ 2$

=> $I = \frac{1}{2} * 82.14 * 1.58^ 2$

=> $I = 102.5 \ kg \cdot m^2$

Generally the net torque experienced by the horizontal solid disk is mathematically represented as

$T = I * \alpha = F * r$

=> $\alpha = \frac{ F * r }{ I }$

=> $\alpha = \frac{ 49.5 * 1.58 }{ 102.53 }$

=> $\alpha = 0.7628$

Gnerally from kinematic equation we have that

$w = w_o + \alpha t$

Here $w_o$ is the initial angular velocity velocity of the horizontal solid disk which is $w_o = 0\ rad/s$

$w = 0 + 0.7628 * 2.95$

=> $w = 2.2503 \ rad/s$

Generally the kinetic energy is mathematically represented as

$KE = \frac{1}{2} * I * w^2$

=> $KE = \frac{1}{2} * 102.53 * 2.2503 ^2$

=> $KE = 259.6 \ J$

8 0

2 years ago

Prank text my sister, I wanna see her reaction. ‪(346) 298-3870‬

dlinn [17]

Answer:

she is going to be mad dude

4 0

2 years ago

In the picture above, what is the period of the pendulum? (Remember to include a unit)

-BARSIC- [3]

69 i agree with her hope this helps

6 0

2 years ago

In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra

shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

0.4726•I + Mf = 68.5832 equation 1

Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

8 0

3 years ago

1-Calculate Req

ArbitrLikvidat [17]

Answer:

1. 21.66 Ohms

2. 3.38 A

3. 6.7 V

Explanation:

1. Req = 6+2 = 8 Ohms (2 and 6 are in a series circuit)

Req = 1/8 +1/4 = 3/8 = 8/3 = 2.66 Ohms (8 and 4 are parallel, so we will add them using this equation)

Req = 2.66 + 1 + 9 + 3 + 6 = 21.66 Ohms

2. I = V/R = 9/2.66 = 3.38 A (In a series circuit, the current is the same across the resistors, so we will add them and divided them by 9 volts)

3. V = IR = 3.38 x 2 = 6.7 V (In a series circuit, the voltage is different, so each resistor will have a different voltage.)

I hope this helps. I am not an expert in physics but its ok :)

Note: If the answer benefited u, mark me as the brainliest answer if u can, thx.

3 0

2 years ago