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3 years ago

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A 730 N student stands in the middle of a frozen pond having a radius of 5.8 m. He is unable to get to the other side because of

a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 2.6 kg physics textbook horizontally toward the north shore at a speed of 6.1 m/s. The acceleration of gravity is 9.81 m/s². How long does it take him to reach the south shore? Answer in units.

1 answer:

dedylja [7]3 years ago

7 0

Answer:27.35 s

Explanation:

Given

weight of man =730 N

mass of man $m_1=\frac{730}{9.8}=74.48 kg$

radius of pound r=5.8 m

mass of book $m_2=2.6 kg$

velocity of book $v_2=6.1 m/s$

let $v_1$ be the velocity of man

conserving momentum

$0=m_1v_1+m_2v_2$

$0=74.48\times v_1+2.6\times 6.1$

$v_1=\frac{15.86}{74.48}=0.212 m/s$

time taken by man to reach south end

$t=\frac{5.8}{0.212}=27.35 s$

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A 95kg fullback (football player for those not into sports) moving south with a speed of 5.0 m/s has a perfectly inelastic colli

Lunna [17]

Answer:

a. $v=3.11mls$ , $29.4^{0}$

b. $K.E =-697.8J$

Explanation:

To calculate the values in the question, a deep understanding of perfect inelastic collision is important.

When two bodies undergo inelastic collision, two important parameters must be well understood i.e

Momentum: the momentum is always conserved in perfectly inelastic collision. i.e the total momentum after collision is the sum of the individual momentum before collision

Kinetic energy: Kinetic energy is not conserved due to dissipative force.

a.To calculate the velocity, we first find the total momentum before collision

Momentum of player 1 $p_{1} =mv=95kg*5m/s\\p_{1} =475kgm/s\\$

Momentum of player 2 $p_{2} =mv=90kg*3m/s\\p_{1} =270kgm/s\\$

Hence the total momentum $p_{12}=p_{1}+p_{2}\\$

Note, since the direction of movement before collision is due south and due north respectively we have to represent the velocity using the rectangular coordinate

Hence $p_{12}=(m_{1}+m_{2})v=p_{1}i+p_{2}j\\$

$(95+90)v=475i+270j\\$

$v=2.57i+1.45j\\$

solving for the resultant velocity, we have

$v=\sqrt{2.75^{2} +1.45^{2}}\\ v=3.11mls$

To calculate the direction of movement, we have

$\alpha =tan^{-1}=\frac{v_{j} }{v_{i}}\\ \alpha =tan^{-1}=\frac{1.45}{2.57}\\\alpha =29.4^{0}$

b. to calculate the decrease in total kinetic energy, before collision, the total kinetic was

$K.E_{initial} =\frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}.\\K.E_{initial} =((1/2)*95*5^{2})+((1/2)*90*3^{2})\\K.E_{initial} =1187.5+405\\K.E_{initial} =1592.5J\\$

And the final kinetic energy after collision is

$K.E_{final} =\frac{1}{2}(m_{1}+m_{2} )v^{2}\\ K.E_{final} =\frac{1}{2}(95+90)* 3.11^{2}\\ K.E_{final} =894.7J$

The decrease in Kinetic energy is

$K.E =K.E_{final}- K.E_{initial}=894.7-1592.5$

$K.E =-697.8J$

The negative sign indicate a decrease in Kinetic energy

4 0

3 years ago

A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/

german

Answer:

<h2>128.61 Watts</h2>

Explanation:

Average power done by the torque is expressed as the ratio of the workdone by the toque to time.

Power = Workdone by torque/time

Workdone by the torque = $\tau \theta$ = $I\alpha * \theta$

I is the rotational inertia = 16kgm²

$\theta = angular\ displacement$

$\theta = 2 rev = 12.56 rad$

$\alpha \ is \ the\ angular\ acceleration$

To get the angular acceleration, we will use the formula;

$\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}$

$\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}$

Workdone by the torque = 16 * 1.28 * 12.56

Workdone by the torque = 257.23 Joules

Average power done by the torque = Workdone by torque/time

= 257.23/2.0

= 128.61 Watts

8 0

3 years ago

1. Two forces act on a box as follows: F1 = 100 N at 01 = 170° and F2 = 75 N

lilavasa [31]

Answer:

a) F = 64.30 N, b) θ = 121.4º

Explanation:

Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components

let's use trigonometry

Force F1

sin 170 = F_{1y} / F₁

cos 170 = F₁ₓ / F₁

F_{1y} = F₁ sin 170

F₁ₓ = F₁ cos 170

F_{1y} = 100 sin 170 = 17.36 N

F₁ₓ = 100 cos 170 = -98.48 N

Force F2

sin 30 = F_{2y} / F₂

cos 30 = F₂ₓ / F₂

F_{2y} = F₂ sin 30

F₂ₓ = F₂ cos 30

F_{2y} = 75 sin 30 = 37.5 N

F₂ₓ = 75 cos 30 = 64.95 N

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = -98.48 +64.95

Fₓ = -33.53 N

Y axis

F_y = F_{1y} + F_{2y}

F_y = 17.36 + 37.5

F_y = 54.86 N

a) the magnitude of the resultant vector

let's use Pythagoras' theorem

F = Ra Fx ^ 2 + Fy²

F = Ra 33.53² + 54.86²

F = 64.30 N

b) the direction of the resultant

let's use trigonometry

tan θ’= F_y / Fₓ

θ'= $tan^{-1} \frac{F_y}{F_x}$

θ'= tan⁻¹ (54.86 / (33.53)

θ’= 58.6º

this angle is in the second quadrant

The angle measured from the positive side of the x-axis is

θ = 180 -θ'

θ = 180- 58.6

θ = 121.4º

5 0

2 years ago

What is the most potentially destructive way to deal with stress while at college? a. texting friends back home b. watching TV c

diamong [38]

Answer:

Option (C)

Explanation:

Recreational drugs are usually referred to those drugs or medicines that are legally or illegally used without any prescription given by the doctors. These drugs include marijuana, cocaine, alcohol, nicotine, heroin, and many others. These drugs provide relaxation for a shorter period of time, but using of these substances, it makes an individual addicted to drugs.

So, recreational drugs are one of the most potentially destructive ways to handle stress when a student is in college.

Thus, the correct answer is option (C).

7 0

3 years ago

In Young’s two slit experiment, the first dark fringe above the central bright fringe occurs at an angle of 0.44˚. What is the r

melomori [17]

Answer:

d / λ = 26.7

Explanation:

In Young's double slit experiment, constructive interference is described by the expression

d sin θ = m λ

In the case of destructive interference we must add half wavelength (λ/2)

d siyn θ = (m + ½) λ

Let's clear

d / λ = (m + ½) / sin θ

Let's calculate

d / λ = (2+ ½) / sin 5.4

d / λ = 5 / (2 sin 5.4)

d / λ = 26.7

8 0

2 years ago