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zlopas [31]
3 years ago
15

‏Calculate the kinetic energies of the following objects moving at the given speeds : ( a ) a 91 kg football linebacker running

at 8.9 m / s ; ( b ) a 4,9 g bullet at 807 m / s ; ( C ) Carrier Moritz , 91,400 metric tons at 23 knots . ( 1 metric ton - 1000 kg , 1 knot - 0.514 m / s )
Physics
1 answer:
damaskus [11]3 years ago
5 0

Answer:

Explanation:

KE = ½mv²

a) KE = ½(91)(8.9²) = 3,604.055 = 3.6 kJ

b) KE = ½(0.0049)(807²) = 1,595.56005 = 1.6 kJ

c) KE = ½(91 400 000)(23(0.514))² = 6,387,017,558.8 = 6.4 GJ

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A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result
Virty [35]

Hello!

A 10-kg cart moving at 5 m/s collides with a 5-kg cart at rest and causes it to move 10 m/s. Which principle explains the result? A) law of differential mass B) law of conservation of momentum C) law of unequal forces D) law of accelerated collision

We have the following data¹:

ΔP (momentum before impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 5 m/s

mB (mass) = 5 kg

vB (velocity) = 0 m/s

Solving:

ΔP = mA*vA + mB*vB

ΔP = 10 kg*5 m/s + 5 kg*0 m/s

ΔP = 50 kg*m/s + 0 kg*m/s

Δp = 50 kg*m/s ← (momentum before impact)

We have the following data²:

ΔP (momentum after impact) = ?  

mA (mass) = 10 kg

vA (velocity) = 0 m/s

mB (mass) = 5 kg

vB (velocity) = 10 m/s

Solving:

Δp = mA*vA + mB*vB

Δp = 10 kg*0 m/s + 5 kg*10 m/s

Δp = 0 kg*m/s + 50 kg*m/s

Δp = 50 kg*m/s ← (momentum after impact)

*** Then, which principle explains the result ?

Law of conservation of momentum, <u>since the total momentum of body A and B before impact is equal to the total momentum of body A and B after impact.</u>

Note:  Bodies of different masses and velocities may have the same kinetic energy, if proportionality between the units is maintained it can occur that they have the same kinetic energy.

Answer:

B) law of conservation of momentum

_______________________

\bf\red{I\:Hope\:this\:helps,\:greetings ...\:Dexteright02!}\:\:\ddot{\smile}

5 0
3 years ago
A train is accelerating at a rate of 2 km/hr/s.  If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?
Mademuasel [1]
Here it is. *WARNING* VERY LONG ANSWER

________________________________________... 
<span>11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) </span>

<span>The change in PE =mgh=5*9.8*12=588 J </span>
<span>______________________________________... </span>
<span>12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. </span>

<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>

<span>Their KE as they crossed the line=(1/2)Mv^2 </span>

<span>Their KE as they crossed the line=0.5*550*(15.98)^2 </span>

<span>Their KE as they crossed the line is 70224.11 J </span>

<span>______________________________________... </span>
<span>13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m </span>

<span>She trips and drops the spare tire of mass = m = 10.0 kg, </span>

<span>The tire rolls down the hill with an intial speed = u = 2.00 m/s. </span>

<span>The height of top of the next hill = h = 5.00 m </span>

<span>Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 </span>

<span>Initial total mechanical energy =mgH+(1/2)mu^2 </span>

<span>Suppose the final speed at the top of second hill is v </span>

<span>Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 </span>

<span>As mechanical energy is conserved, </span>

<span>Final total mechanical energy =Initial total mechanical energy </span>

<span>mgh+(1/2)mv^2=mgH+(1/2)mu^2 </span>

<span>v = sq rt [u^2+2g(H-h)] </span>

<span>v = sq rt [4+2*9.8(20-5)] </span>

<span>v = sq rt 298 </span>

<span>v =17.2627 m/s </span>

<span>The speed of the tire at the top of the next hill is 17.2627 m/s </span>
<span>______________________________________... </span>
<span>14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. </span>

<span>a.)The mass of bean = m = 2.0 g </span>

<span>Height up to which the been jumps = h = 1.0 cm from hand </span>

<span>Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg </span>

<span>b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s </span>
<span>_____________________________ </span>
<span>15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. </span>

<span>The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s </span>
<span>__________________________________ </span>
<span>16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, </span>

<span>The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 </span>
<span>______________________________________... </span>

<span>EDIT </span>
<span>1.) A train is accelerating at a rate = a = 2.0 km/hr/s. </span>

<span>Acceleration </span>

<span>Initial velocity = u = 20 km/hr, </span>

<span>Velocity after 30 seconds = v = u + at </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr </span>

<span>Velocity after 30 seconds = v = 80 km/hr=22.22 m/s </span>
<span>_______________________________- </span>
<span>2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. </span>

<span>His acceleration = a =11.1/9=1.233 m/s^2 </span>

<span>Distance he covered = s = (1/2)at^2=49.95 m</span>
7 0
3 years ago
Plane-polarized light is incident on a single polarizing disk, withthe direction of E0 parallel to thedirection of the transmiss
Alenkinab [10]

Answer;

Cos²စ= I/Io

So

A. Cos²စ = 1/2.2

Cosစ= √1/2.2

စ = cos^-1 0.68

= 47.2°

B.

Cosစ = √1/5.2

စ = ,cos^-1 0.4385

= 64°

C.

Cosစ = √1/12

စ = cos^-1 0.2886

= 73.2°

6 0
3 years ago
A skydiver jumps out of a plane and immediately begins falling toward the Earth.
worty [1.4K]

Answer:

5.2m/s^2

Explanation:

A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.

When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.

Vf=Vo+a.t (1)\\{Vf^{2}-Vo^2}/{2.a} =X(2)\\X= VoT+0.5at^{2} (3)\\X=(Vf+Vo)T/2 (4)

Where

Vf = final speed

Vo = Initial speed

T = time

A = acceleration

X = displacement

In conclusion to solve any problem related to a body that moves with constant acceleration we use the 4 above equations and use algebra to solve

for this case we can use the ecuation number 3

x=100m

t=6.2s

Vo=0m/s

X= VoT+0.5at^{2}\\X= 0.5at^{2}\\a=\frac{X}{0.5t^2} \\a=\frac{100}{0.5(6.2)^2}=5.2m/s^2

7 0
3 years ago
A 900 kg vehicle moves around a curve with an incline of 20\circ∘ at a speed of 12.5 m/s. If the curve has a radius of 50 meters
valentina_108 [34]

Answer:

The normal force experienced by the car is approximately 8223.2 N

Explanation:

The question relates to banking of road where the centripetal force for the circular motion of the vehicle is provided by the horizontal component of the normal reaction

The mass of the vehicle that moves around the curve, m = 900 kg

The incline of the curve, θ = 20°

The speed with which the vehicle moves around the curve, v = 12.5 m/s

The radius of the curve, R = 50 meters

We have;

N \cdot sin(\theta) = \dfrac{m \cdot v^2}{R}

Where;

θ = The angle of inclination of the road = 20°

N = The normal force experienced by the car

m = The mass of the car = 900 kg

v = The velocity with which the car is moving = 12.5 m/s

R = The radius of the curve around which the vehicle moves = 50 m

\therefore N = \dfrac{m \cdot v^2}{R \cdot sin(\theta)} = \dfrac{900 \times (12.5)^2}{50 \times sin(20^{\circ})}  = 8223.1998754586828969046217875927

The normal force experienced by the car = N ≈ 8223.2 N.

6 0
3 years ago
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