PHYSICS COLLEGE

A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its kinetic energy at circled A? 1.2635 Correct: Your answer is correct. J (b) What is its speed at circled B? 4.54 Correct: Your answer is correct. m/s (c) What is the net work done on the particle by external forces as it moves from circled A to circled B?

Answers

Answer 1

Answer:

Answer:

a). $E_(kA)=1.2635 J$

b). $V_(B)=4.535(m)/(s)$

c). Δ $E_(t)=8.4635 J$

Explanation:

ΔE=kinetic energy

a).

$E_(kA)=(1)/(2)*m*v_(A) ^(2) \n v_(A)=1.9 (m)/(s)\n m=0.70kg\nE_(kA)=(1)/(2)*0.70kg*(1.9 (m)/(s))^(2) \nE_(kA)=1.2635 J$

b).

$E_(kB)=(1)/(2)*m*v_(B) ^(2)$

$V_(B)^(2)=(E_(kB)*2)/(m) \nV_(B)=\sqrt{(E_(kB)*2)/(m)} \nV_(B)=\sqrt{(7.2J*2)/(0.70kg)} \nV_(B)=4.53 (m)/(s)$

c).

net work= EkA+EkB

$E_(t)=E_(kA)+ E_(kB)\nE_(t)=1.2635J+7.2J\nE_(t)=8.4635J$

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A physics student standing on the edge of a cliff throws a stone vertically downward with an initial speed of 10.0 m/s. The instant before the stone hits the ground below, it is traveling at a speed of 30.0 m/s. If the physics student were to throw the rock horizontally outward from the cliff instead, with the same initial speed of 10.0 m/s, what is the magnitude of the velocity of the stone just before it hits the ground? Ignore any effects of air resistance.

Answers

Answer:

vf = 30 m/s : (the magnitude of the velocity of the stone just before it hits the ground)

Explanation:

Because the stone moves with uniformly accelerated movement we apply the following formulas:

vf²=v₀²+2*g*h Formula (1)

Where:

h: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

g: acceleration due to gravity in m/s²

Free fall of the stone

Data

v₀ = 10 m/s

vf = 30.0 m/s

g = 9,8 m/s²

We replace data in the formula (1) to calculate h:

vf²=v₀²+2*g*h

(30)² = (10)² + (2)(9.8)*h

(30)²- (10)²= (2)(9.8)*h

h =( (30)²- (10)²) /( 2)(9.8)

h = 40.816 m

Semiparabolic movement of the stone

Data

v₀x = 10 m/s

v₀y = 0 m/s

g = 9.8 m/s²

h= 40.816 m

We replace data in the formula (1) to calculate vfy :

vfy² = v₀y² + 2*g*h

vfy² = 0 + (2)(9.8)( 40.816)

$v_(fy)=√(2*9.8*40.816) = 28.284 (m)/(s)$

$v_(f)=\sqrt{v_(ox)^2+v_(fy)^2}=√((10)^2+(28.284)^2) = 30(m)/(s)$

The magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

The given parameters;

initial vertical velocity of the stone, $v_y_0$ = 10 m/s

final vertical velocity of the stone, $v_y_f$ = 30 m/s

The height traveled by the stone before it hits the ground is calculated as;

$v_y_f^2 = v_y_0^2 + 2gh\n\nh = (v_y_f^2- v_y_0^2)/(2g) \n\nh = ((30)^2 - (10)^2)/(2* 9.8) \n\nh = 40.82 \ m$

If the the stone is projected horizontally with initial velocity of 10 m/s;

the initial vertical velocity = 0

Final vertical velocity of the stone is calculated as follow;

$v_y_f^2 = v_y_0^2 + 2gh\n\nv_y_f^2 = 0 + 2* 9.8* 40.82\n\nv_y_f^2 = 800.07\n\nv_y_f = √(800.07) \n\nv_y_f = 28.28 \ m/s$

The horizontal velocity doesn't change.

the final horizontal velocity, $v_x_f$ = initial horizontal velocity = 10 m/s

The resultant of the final velocity of the stone before it hits the ground;

$v _f= √(v_x_f^2 + v_y_f^2) \n\nv_f = √(10^2 + 28.28^2) \n\nv_f= 29.99 \ m/s \approx 30 \ m/s$

Thus, the magnitude of the velocity of the stone just before it hits the ground is 30 m/s.

Learn more here:brainly.com/question/13533552

Could you please solve it with shiwing the full work1-

A high powered projectile is fired horizontally from the top of a cliff at a speed of 638.6 m/s. Determine the magnitude of the velocity (in m/s) after 5 seconds.

Take gravitational acceleration to be 9.81 m/s2.

2-

A man throws a ball with a velocity of 20.9 m/s upwards at 33.2° to the horizontal. At what vertical distance above the release height (in metres) will the ball strike a wall 13.0 m away ?

Take gravitational acceleration to be 9.81 m/s2.

3-

A particle is moving along a straight path and its position is defined by the equation s = (1t3 + -5t2 + 3) m, where t is measured in seconds. Determine the average velocity (in m/s) of the particle when t = 5 seconds.

4-

A particle has an initial speed of 26 m/s. The particle undergoes a deceleration of a = (-9t) m/s2, where t is measured in seconds. Determine the distance (in metres) the particle travels before it stops. When t = 0, s = 0.

Answers

Answer:

1.V= 640.48 m/s :total velocity in t= 5s

2. Y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. v =25m/s

4. s= (-1.5t³+26t ) m

Explanation:

1. Parabolic movement in the x-y plane , t=5s

V₀=638.6 m/s=Vx :Constant velocity in x

Vy=V₀y +gt= 0+9.8*5 = 49 m/s : variable velocity in y

$v=\sqrt{v_(x) ^(2) +v_(y) ^(2) }$

$v=\sqrt{ 638.6^(2) +49 ^(2) }$

V= 640.48 m/s : total velocity in t= 5s

2. $v_(ox) =v_(o) cos33.2=20.9*cos33,2= 17.49 m/s$

$v_(oy)=v_(o)*sin33,2 =20.9*sin33,2=11.44 m/s$

x=v₀x*t

13=v₀x*t

13=17.49*t

t=13/17.49=0.743s : time for 13.0 m away

th=v₀y/g=11.44/9.8= 1,17s :time for maximum height

at t=0.743 sthe ball is going up ,then g is negative

y=v₀y*t - 1/2 *g¨*t²

y=11.44*0.743 -1/2*9.8*0.743²

y= 5.79m : vertical distance above the height of release (in meters) where the ball will hit a wall 13.0 m away

3. s = (1t3 + -5t2 + 3) m

v=3t²-10t=3*25-50=75-50=25m/s

at t=0, s=3 m

at t=5s s=5³-5*5²+3

4. a = (-9t) m/s2

a=dv/dt=-9t

dv=-9tdt

v=∫ -9tdt

v=-9t²/2 + C1 equation (1)

in t=0 , v₀=26m/s ,in the equation (1) C1= 26

v=-9t²/2 + 26=ds/dt

ds=( -9t²/2 + 26)dt

s= ∫( -9t²/2 + 26)dt

s= -9t³/6+26t+C2 Equation 2

t = 0, s = 0 , C2=0

s= (-9t³/6+26t ) m

s= (-1.5t³+26t ) m

Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?

Answers

Answer:

$v=3.564\ m.s^(-1)$

$\Delta v =2.16\ m.s^(-1)$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

Using the equation of motion:

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2* (9.8* (1.8)/(3))* 3$

$v_J=5.94\ m.s^(-1)$

Now using the law of conservation of momentum at the bottom of the slide:

Sum of initial momentum of kids before & after collision must be equal.

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30* 5.94+0=(30+20)v$

$v=3.564\ m.s^(-1)$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

Now we find the force along the slide due to the body weight:

$F=m_J.g.sin\theta$

$F=30* 9.8* (1.8)/(3)$

$F=176.4\ N$

Hence the net force along the slide:

$F_R=71.4\ N$

Now the acceleration of John:

$a_j=(F_R)/(m_J)$

$a_j=(71.4)/(30)$

$a_j=2.38\ m.s^(-2)$

Now the new velocity:

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2* 2.38* 3$

$v_J_n=3.78\ m.s^(-1)$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^(-1)$

Find T1, the magnitude of the force of teTo practice Problem-Solving Strategy 11.1 Equilibrium of a Rigid Body. A horizontal uniform bar of mass 2.6 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to the end of the bar, and string 2 is attached a distance 0.7 m from the other end. A monkey of mass 1.3 kg walks from one end of the bar to the other. Find the tension T1 in string 1 at the moment that the monkey is halfway between the ends of the bar.nsion in string 1, at the moment that the monkey is halfway between the ends of the bar. Express your answer in newtons using three significant figures. View Available Hint(s)

Answers

Answer:

$T_2 = 24.95 N$

$T_1 = 13.3 N$

Explanation:

As we know that total torque on the rod must be zero when monkey is at mid point of the rod

So we have

torque due to Tension at other end = torque due to weight of monkey + rod

so we will have

$T_2 (3 - 0.7) = (Mg + mg)(1.5)$

here we know that

M = 2.6 kg

m = 1.3 kg

$T_2(2.3) = (2.6 + 1.3)(9.81)(1.5)$

$T_2 = 24.95 N$

Now similarly we can say that

$T_1 + T_2 = (m + M)g$

$T_1 + 24.95 = (2.6 + 1.3)(9.81)$

$T_1 = 13.3 N$

A 2.4 kg toy oscillates on a spring completes a cycle every 0.56 s. What is the frequency of this oscillation?

Answers

Answer:

$f=1.79Hz$

Explanation:

The period is defined as the time taken by an object to complete a cycle in a simple harmonic motion. As the frequency of the motion increases, the period decreases. Therefore, they are inversely proportional. The frequency does not depend on the mass of the object.

$f=(1)/(T)\nf=(1)/(0.56 s)\nf=1.79Hz$

Guitar string has an overall length of 1.22 m and a total mass of 3.5 g before being strung on a guitar. Once it is used on the guitar, there is a distance of 70 cm between fixed end points. The guitar string is tightened to a tension of 255 N.What is the frequency of the fundamental wave on the guitar string?