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3 years ago

Using kinetic energy to determine height

1 answer:

7 0

For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters. Notice that gravitational potential energy has the same units as kinetic energy, kg m2 / s2.

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Metallic bonds are important to how paperclips are both made and used. Which statement best describes how metallic bond properti

Vikentia [17]

The statement that best describes how metallic bond properties are important for making and using paperclips is Ductility is important for making paperclips, and malleability is important for using them.

Ductility is defined as the ability of the solid material to stretch under tensil stress. The metal of the paper clip is ductile when paperclips are made because the metal is stretch until it forms a wire.

Malleability is defined as the ability of the solid material to deform under pressure. Paperclips are very malleable. It can be manipulated into forming different shapes with the used of our hand strength.

3 0

4 years ago

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A satellite is in elliptical orbit with a period of 7.20 104 s about a planet of mass 7.00 1024 kg. At aphelion, at radius 5.1 1

soldi70 [24.7K]

Answer:

1.64x $10^{-4}$ rad/s

Explanation:

Given:

satellite is in elliptical orbit with a period T= 7.20 x $10^{4}$ s

Mass of planet m= 7.00x $10^{24}$ kg

Satellite's angular speed ωa- = 4.987 10-5 rad/s

At aphelion, at radius Ra= 5.1 x 10^7 m

Assuming torque is zero for this system, therefore, the conservation of angular momentum can be defines as

Lp = La

Ip ωp = Ia ωa--->eq(1)

Where,

'a' represents aphelion and 'p' represents perihelion

ω represents angular velocity

and I represents the rotational inertia

Since, I= 1/2 mR²

Here R is the radius at aphelion / perihelion

m = mass of planet

eq(1)=> ωp= Ia ωa/ Ip

ωp= (1/2 m Ra² ωa) / 1/2 m Rp²

ωp= (Ra/Rp)² ωa --->eq(2)

In order to find Rp, we use : 2a= Rp + Ra

where a represents semimajor axis

with the help of Kepler's third law for elliptical orbits

a= ∛(GmT²/4π²)

a= ∛ 6.67x $10^{-11}$ x 7.00x $10^{24}$ x (7.20 x $10^{4}$ )² / 4π²

a= 3.97 x $10^{7}$ m

2a= Rp + Ra

Rp= 2a-Ra

Rp= 2 x 3.97 x $10^{7}$ - 5.1x $10^{7}$

Rp= 2.84 x $10^{7}$ m

Substituting all the required values in eq 2, we have

ωp= (Ra/Rp)² ωa

ωp= (5.1x $10^{7}$ /2.84 x $10^{7}$ )² x 4.987 x $10^{-5$

ωp= 3.224 x 4.987 x $10^{-5$

ωp= 1.64x $10^{-4}$ rad/s

Therefore, angular speed at perihelion is 1.64x $10^{-4}$ rad/s

3 0

3 years ago

A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.

mash [69]

Answer:

9.6 m

Explanation:

This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .

a = 3.6 t + 5.6

d²x / dt² = 3.6 t + 5.6

Integrating on both sides

dx /dt = 3.6 t² / 2 + 5.6 t + c

where c is a constant.

dx /dt = 1.8 t² + 5.6 t + c

when t = 0 , velocity dx /dt is zero

Putting these values in the equation above

0 = 0 +0 + c

c = 0

dx /dt = 1.8 t² + 5.6 t

Again integrating on both sides

x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁

x = 0.6 t³ + 2.8 t² + c₁

when t =0, x = 0

c₁ = 0

x = 0.6 t³ + 2.8 t²

when t = 1.6

x = .6 x 1.6³ + 2.8 x 1.6²

= 2.4576 + 7.168

= 9.6256

9.6 m

5 0

3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

a)Kinematics equation for the first ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

What's the meaning of quantum entanglement?

spayn [35]

Answer:

the phenomenon whereby a pair of particles are generated in such a way that the individual quantum states of each are indefinite until measured, and the act of measuring one determines the result of measuring the other, even when at a distance from each other.

3 0

2 years ago

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