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3 years ago

Number 3,4, 6, and 8

Physics

1 answer:

siniylev [52]3 years ago

3 0

1. When the object is waiting to be released, it is storing a lot of potential energy. When it is released, the potential energy that was once stored is converted into kinetic energy.

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Power can be defined as

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3 years ago

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A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the

Ann [662]

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s

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3 years ago

Which of the following is not an example of bias

neonofarm [45]

Answer:

Explanation:

The writers are from a trusted source that is considered impartial. The message is backed up by scientific investigation, and the writers have no monetary gain from their message.

5 0

3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

A 10-g bullet moving horizontally with a speed of 1.8 km/s strikes and passes through a 5.0-kg block initially at rest on a hori

UNO [17]

Answer:

6.4 J

Explanation:

m = mass of the bullet = 10 g = 0.010 kg

v = initial velocity of bullet before collision = 1.8 km/s = 1800 m/s

v' = final velocity of the bullet after collision = 1 km/s = 1000 m/s

M = mass of the block = 5 kg

V = initial velocity of block before collision = 0 m/s

V' = final velocity of the block after collision = ?

Using conservation of momentum

mv + MV = mv' + MV'

(0.010) (1800) + (5) (0) = (0.010) (1000) + (5) V'

V' = 1.6 m/s

Kinetic energy of the block after the collision is given as

KE = (0.5) M V'²

KE = (0.5) (5) (1.6)²

KE = 6.4 J

4 0

4 years ago