Usually, tall trees collapse in stroms?<br>

A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia

Sloan [31]

Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

6 0

3 years ago

You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

lina2011 [118]

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

$L = \frac{\mu N^2 A}{l}$

Here,

$\mu$ = Permeability at free space

N = Number of loops

A = Cross-sectional Area

l = Length

Replacing with our values we have,

$L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}$

$L = 0.00002442H$

$L = 24.42\mu H$

Therefore the Inductance is $24.42\mu H$

3 0

3 years ago

How do physicists distinguish one type of electromagnetic wave from another?

GenaCL600 [577]

A wave is characterized by the cyclic occurrences of crests and troughs. Wavelengthis defined as the distance between two consecutive troughs or two crests and the Frequency is defined as the number of cycles that pass through a point per second

7 0

3 years ago

A chunk of paraffin (wax) has a mass of 50.4 grams and a volume of 57.9 cm3. What is the density of

ExtremeBDS [4]

$Formula\ for\ density:\\\\
p=\frac{m}{V}\\p-density,\\m-mass,\\V-volume\\\\
Data:\\
m=50,4grams\\
V=57,9cm^3\\\\
p=\frac{50,4g}{57,9cm^3}=0,87\frac{g}{cm^3}\\\\Density\ of\ paraffin\ is\ equal\ to\ 0,87\frac{g}{cm^3}.
$

6 0

4 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$