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2 years ago

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How should the student use the data collected and the known quantities from the experiment to determine the initial total mechan

ical energy of the block-ramp-Earth system for all trials in the experiment

1 answer:

fiasKO [112]2 years ago

7 0

It should be noted that the student should use K = 1/2mv² with the initial speed of the block for one trial.

<h3>Method of using the data collected.</h3>

From the complete information, the student wants to use the data collected and the known quantities from the experiment to determine the initial total mechanical energy of the block-ramp-Earth system for all trials in the experiment.

In this case, it's important to use K = 1/2mv² with the block's initial speed for one trial due to the fact that the initial speed is the same in all the trials.

Learn more about experiments on:

brainly.com/question/17274244

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Jupiter's moon Io has active volcanoes (in fact, it is the most volcanically active body in the solar system) that eject materia

Anton [14]

Answer:

91.64 km

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

Explanation:

According to Newton Law of gravitation:

$g=\frac{Gm}{r^2}$

Where:

G is gravitational constant= $6.67*10^{-11} m^3/kg.s^2$

For Moon lo g is:

$g_M=\frac{6.67*10^{-11}*8.93*10^{22}}{(1821*10^3)^2m^2} \\g_M=1.7962 m/s^2$

According to law of conservation of energy

Initial Energy=Final Energy

$K.E_i+mgh_i=K.E_f+mgh_f$

$\frac{1}{2}m(v_0)^2+mgh_o= \frac{1}{2}m(v_f)^2+mgh_f\\At\ maximum\ height\ v_f=0\\\frac{1}{2}m(v_0)^2+0=mgh_f\\v_0=\sqrt{2gh_f}$

For Jupiter's moon Io:

Velocity is given by:

$v_0_M=\sqrt{2g_Mh_f_M}$

For Earth Velocity is given by:

$v_0_E=\sqrt{2g_Eh_f_E}$

Now:

$v_o_M=v_o_E$

$\sqrt{2g_Mh_f_M}=\sqrt{2g_Eh_f_E}\\h_f_E=\frac{g_Mh_f_M}{g_E}$

$g_E=9.8 m/s^2$

$g_m=1.7962 m/s^2, As\ Calculated\ above$

$h_f_E=\frac{1.7962*500*10^3m}{9.8} \\h_f_E=91642.85 m\\h_f_E=91.64Km$

91.64 km high material would go on earth if it were ejected with the same speed as on Io.

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3 years ago

A bicycle travels 141 m along a circular track of radius 30 m. What is the angular displacement in radians of the bicycle from i

Veseljchak [2.6K]

Answer: 4.7rad

Explanation:

Angular displacement =s/r

Where s=distance traveled

r=radius

Angular displacement =141m/30m

Angular displacement =4.7rad.

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3 years ago

How is fitness walking beneficial?

jonny [76]

Answer:

It can relieve stress and improve mood.

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3 years ago

Read 2 more answers

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of t

muminat

<h2><em><u>⇒</u></em>Answer:</h2>

In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

Step-by-Step Solution:

Solution 35PE

This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.

Step 1 of 3<

/p>

The legs have an extension of 0.600 m in the crouch position.

So, m

The person is at rest initially, so the initial velocity will be zero.

The acceleration is m/s2

Acceleration m/s2

Let the final velocity be .

Step 2 of 3<

/p>

Substitute the above given values in the kinematic equation ,

m/s

Therefore, the final velocity or jumping speed is m/s

Explanation:

3 0

3 years ago

Read 2 more answers

A uniform rod is hung at one end and is partially submerged in water. If the density of the rod is 5/9 that of water, find the f

VashaNatasha [74]

Answer:

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

Explanation:

Data given:

Density of the rod = 5/9 of the density of the water.

Let's denote density of Water with w

And density of rod with r

So,

r = 5/9 x w

Required:

Fraction of the length of the rod above water.

Let's denote total length of the rod with L

and length of the rod above with = y

Let's denote the density of rod = r

And density of water = w

So, the required is:

Fraction of the length of the rod above water = y/L

y/L = ?

In order to find this, we first need to find out the all type of forces acting upon the rod.

We know that, a body will come to equilibrium if the net torque acting upon a body is zero.

As, we know

F = ma

Density = m/v

m = Density x volume

Volume = Area x length = X ( L-y)

So, let's say X is the area of the cross section of the rod, so the forces acting upon it are:

F = mg

F = (Density x volume) x g

g = gravitational acceleration

F1 = X(L-y) x w x g (Force on the length of the rod submerged in water)

where,

X (L-y) = volume

w = density of water.

Another force acting upon it is:

F = mg

F2 = X x L x r x g

Now, the torques acting upon the body:

T1 + T2 = 0

F1 ( y + $(\frac{L-y}{2})$ ) g sinФ - F2 x ( $\frac{L}{2}$ ) x gsinФ = 0

plug in the equations of F1 and F2 into the above equation and after simplification, we get:

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . r

where, w is the density of water and r is the density of rod.

As we know that,

r = 5/9 x w

So,

$(L^{2} - y^{2} ) . w$ = $L^{2}$ . 5/9 x w

Hence,

$(L^{2} - y^{2} )$ = $\frac{5L^{2} }{9}$

$\frac{L^{2} - y^{2} }{L^{2} }$ = $\frac{5}{9}$

Taking $L^{2}$ common and solving for $\frac{y}{L}$ , we will get

$\frac{y}{L}$ = 0.66

Hence, the fraction of the length of the rod above water = $\frac{y}{L}$ = 0.66

and fraction of the length of the rod submerged in water = 1 - $\frac{y}{L}$ = 1 - 0.66 = 0.34

8 0

3 years ago