A joule is used to measure a. power. b. energy. c. torque. d. pressure.

"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its

prisoha [69]

Answer:

1.3m/s

Explanation:

Data given,

Mass,m=1.0kg,

Amplitude,A=0.10m,

Frequency,f=2.0Hz.

From the equation of a simple harmonic motion, the displacement of the object at a given time is define as

$x=Acos\alpha \\$

we can express the velocity by the derivative of the displacement,

Hence

$V=-Awsin\alpha \\$

at equilibrium, the velocity becomes

$V=wA\\w=2\pi f$

Hence if we substitute values we arrive at

$V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s$

4 0

3 years ago

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Four seconds after being launched, what is the height of a ball that starts from a height of 12 m with an initial upward velocit

MrMuchimi

Answer:

15.24 m/s in the downward direction

Explanation:

Given that the initial upward velocity of the ball is 24 m/s.

Assuming that the upward direction is positive.

As gravitational force acts in the downward direction and the direction of acceleration is the same as the direction of force, so the acceleration due to gravity will be negative.

Now, from the equation of motion, when an object is launched with initial velocity u, the final velocity, v, of an object after time t is v=u+at.

Given that u=24 m/s, t=4 seconds, $g=-9.81 m/s^2$ .

So, the final velocity is

$v= 24 + (-9.81)\times 4 \\\\\Rightarrow v= 24-9.81\times 4$

$\Rightarrow v=-15.24$ m/s

Here, the negative sign means the final velocity is in the downward direction.

Hence, the velocity after 4 seconds is 15.24 m/s in the downward direction.

8 0

3 years ago

In a hydroelectric power plant, 65 m3 /s of water flows from an elevation of 90-m to a turbine-generator, where electricity is g

Mariulka [41]

The electric output of the plant is 48.19 MW

First we need to calculate the water power, it is given by the formula

WP=ρQgh

Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head

Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW

Now the overall efficiency of the hydroelectric power plant is given as

η= $\frac{electric power}{water power}$

Plugging the values in the above equation

0.84=EP/57.38

EP=48.19 MW

Therefore, the electric output of the plant is 48.19 MW.

3 0

2 years ago

If you remove large amounts of heat from a liquid, what could happen

snow_lady [41]

A liquid becomes a solid when energy is removed. The energy content decreases, and the speed of the particles decrease.

3 0

3 years ago

An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming

Vlad [161]

Answer:

Explanation:

Given

mass of archer $m=0.3\ kg$

Average force $F_{avg}=201\ N$

extension in arrow $x=1.3\ m$

Work done to stretch the bow with arrow

$W=F\cdot x$

$W=201\times 1.3=261.3\ m$

This work done is converted into kinetic Energy of arrow

$W=\frac{1}{2}mv^2$

where v= velocity of arrow

$261.3=\frac{1}{2}\times 0.3\times v^2$

$v=\sqrt{1742}$

$v=41.73\ m/s$

(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

$W=mgh$

$261.3=0.3\times 9.8\times h$

$h=\frac{261.3}{0.3\times 9.8}$

$h=88.87\ m$

4 0

3 years ago