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3 years ago

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A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. If the block is released from rest

at the top of the incline, what is its speed at the bottom?

1 answer:

Ludmilka [50]3 years ago

6 0

(if it's frictionless the length doesn't even matter :) )
It would have the same kinetic energy down as the potential energy up. That is, $mgh=\frac{mv^2}{2}$ or $2gh=v^2$ (the mass doesn't even matter). The result is $\sqrt{2gh}$ , so only the height matters really. It is almost 9 (it is $\sqrt{80}=4\sqrt{5}$ ).

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1. You serve a volleyball with a mass of 2.1kg. The ball leaves your

Darina [25.2K]

The kinetic energy is 945 joules.

Kinetic energy is the energy that an object has as a result of motion. It is defined as the effort required to accelerate a mass-determined body from rest to the indicated velocity.

The speed of an object or particle, which is a scalar quantity, is the size of the change in its location over time or the size of the change in its position per unit of time.

The mass of the volleyball is 2.1 kg.

The speed of the ball when the ball leaves the hand is 30 m/s.

m = 2.1 kg

v = 30 m/s

The kinetic energy of an object is given as:

KE = (1/2 ) × m × v²

KE = (1 / 2) × 2.1 kg × ( 30 m/s)²

KE = (1 / 2) × 2.1 kg × 30 m/s × 30 m/s

KE = 2.1 kg × 15 m/s × 30 m/s

KE = 945 J

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1 year ago

And object has traveled 200 m in 20 seconds what is its average speed

BabaBlast [244]

It is going 10m speed

8 0

3 years ago

A particular roller coaster has a mass of 3500 kg, a height of 4.0 m, and a velocity of 12 m/s. What is the kinetic energy? If n

Alenkinab [10]

Explanation:

given

m=3500kg,h=4.0m,v=12m/s

ke=½mv²

½×3500×12²=3500×144÷2

=252,000j or 252j

8 0

2 years ago

What is 1113.28 in a sig fig?

Afina-wow [57]

Answer:

6 significant figure

Explanation:

The digits 111328 all are 6 figures with no figure being zero, neither zero after the other digits. In this case, all the numbers are significant and since they are only six numbers, then this is a six significant figure. In case we add another zero after digit 8, the zero is not significant but if added either infront of 8 or 2, the zero becomes significant.

4 0

3 years ago

I need answers and solvings to these questions

den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) $15.7 rad/s^2$

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f$ is the final angular velocity

$\omega_i$ is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

$\omega_i = 0$ (since it starts from rest)

$\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s$ is the final angular velocity

t = 2 s

Substituting, we find

$\alpha = \frac{31.4-0}{2}=15.7 rad/s^2$

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

$a=-\omega^2 x$

where

a is the acceleration

x is the displacement

$\omega$ is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

$a=-100 x$

which implies that

$\omega^2 = 100$

And so

$\omega = \sqrt{100}=10 rad/s$

Also, the angular frequency is related to the frequency $f$ by

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

Therefore, the frequency of this simple harmonic oscillator is

$f=\frac{10}{2\pi}=1.6 Hz$

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

$mg=kx$

where

m is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

$\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5$

The angular frequency of the spring is given by

$\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz$

And therefore, its period is

$T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s$

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the hose, with $r_1 = 5 mm$ being the radius of the hose

$v_1 = 4 m/s$ is the speed of the petrol in the hose

$A_2 = \pi r_2^2$ is the cross-sectional area of the nozzle, with $r_2$ being the radius of the nozzle

$v_2 = 16 m/s$ is the speed in the nozzle

Solving for $r_2$ , we find the radius of the nozzle:

$\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm$

So, the diameter of the nozzle will be

$d_2 = 2r_2 = 2(2.5)=5.0 mm$

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

where

$F_1$ is the force that must be applied to the small piston

$A_1 = \pi r_1^2$ is the area of the first piston, with $r_1= 2 cm$ being its radius

$F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N$ is the force applied on the bigger piston (the weight of the car)

$A_2 = \pi r_2^2$ is the area of the bigger piston, with $r_2= 15 cm$ being its radius

Solving for $F_1$ , we find

$F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N$

So, the closest answer is B) 266.7 N.

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5 0

3 years ago