Forces at a Distance

A solid cylinder with a mass of 2.72 kg and a radius of 0.083 m starts from rest at a height of 4.20 m and rolls down a 88.7 ◦ s

Bingel [31]

Explanation:

According to the law of conservation of energy ,

Potential energy = kinetic energy

$mgh = \frac{1}{2} \times mv^{2} + \frac{1}{2} \times I \times \omega^{2}$

I = $\frac{mr^{2}}{2}$

$\omega = \frac{v}{r}$

$mgh = [\frac{1}{2} \times mv^{2}] + [\frac{1}{2} \times (\frac{mr^{2}}{2}) \frac{v^{2}}{r^{2}}]$

mgh = $[\frac{1}{2} \times mv^{2}] + [\frac{1}{4} \times mv^{2}]$

$g \times h = \frac{3}{4} \times v^{2}$

$9.8 \times 4.2 = \frac{3}{4} \times v^{2}$

v = 7.4 m/s

thus, we can conclude that the translational speed of the cylinder when it leaves the incline is 7.4 m/s.

5 0

2 years ago

So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.

bekas [8.4K]

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at² Subtract ViT from both sides

D - ViT = 0.5at² Divide both sides by 0.5t²

$\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }$ I wrote this step out a little more to show how your fraction will cancel

$\frac{D - ViT}{0.5t^{2} }$ = a I like to flip these around so the single variable is on the right

a = $\frac{D - ViT}{0.5t^{2} }$

7 0

2 years ago

In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1

valentina_108 [34]

Answer:

The velocity of the photo electron is $6.11\times 10^5\ m/s$ .

Explanation:

Given that,

Supplied energy, $E_s=5.3\times 10^{-19}\ J$

Minimum energy of the electron to escape from the metal, $E_e=3.6\times 10^{-19}\ J$

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

$5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J$

The formula of kinetic energy is given by :

$K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s$

So, the velocity of the photo electron is $6.11\times 10^5\ m/s$ .

4 0

2 years ago

A rock is thrown off a 50.0 m high cliff. How fast must the rock leave the cliff top to land on level ground below, 90 m from th

blagie [28]

Answer:

The rock must leave the cliff at a velocity of 28.2 m/s

Explanation:

The position vector of the rock at a time t can be calculated using the following equation:

r = (x0 + v0x · t, y0 + 1/2 · g · t²)

Where:

r = position vector at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).

Please, see the attached figure for a graphical description of the problem. Notice that the origin of the frame of reference is located at the edge of the cliff so that x0 and y0 = 0.

When the rock reaches the ground, the position vector will be (see r1 in the figure):

r1 = (90 m, -50 m)

Then, using the equation of the vector position written above:

90 m = x0 + v0x · t

-50 m = y0 + 1/2 · g · t²

Since x0 and y0 = 0:

90 m = v0x · t

-50 m = 1/2 · g · t²

Let´s use the equation of the y-component of the vector r1 to find the time it takes the rock to reach the ground and with that time we can calculate v0x:

-50 m = 1/2 · g · t²

-50 m = -1/2 · 9.81 m/s² · t²

-50 m / -1/2 · 9.81 m/s² = t²

t = 3.19 s

Now, using the equation of the x-component of r1:

90 m = v0x · t

90 m = v0x · 3.19 s

v0x = 90 m / 3.19 s

v0x = 28.2 m/s

8 0

2 years ago

Please help me .....explain an experiment of phenomenon of rainfall

barxatty [35]

Answer:

Hindi ko alam pasensya ka ha godbless

8 0

2 years ago