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2 years ago

7

: A small block with mass 0.130 kg is attached to a string passing through a hole in a frictionless, horizontal surface. The blo

ck is originally revolving in a circle with a radius of 0.800 m about the hole with a tangential speed of 4.00 m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 30.0 N. What is the radius of the circle when the string breaks

1 answer:

LenKa [72]2 years ago

6 0

Answer:

r = 0.0173 m = 1.73 cm

Explanation:

Here, the centripetal force of the block will be providing the required breaking tension in the string:

$Tension = Centripetal Force\\T = F_c\\\\T = \frac{mv^2}{r} \\\\r = \frac{mv^2}{T}\\$

where,

r = radius = ?

m = mass of block = 0.13 kg

v = tangential spee of block = 4 m/s

T = Breaking Strength = 30 N

Therefore,

$r = \frac{(0.13\ kg)(4\ m/s)^2}{30\ N}$

<u>r = 0.0173 m = 1.73 cm</u>

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A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

2 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

2 years ago

PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer this correctly!!!! (60pts)

mr Goodwill [35]

The density of an object determines whether it will float or sink in another substance. An object will float if it is less dense than the liquid it is placed in. An object will sink if it is more dense than the liquid it is placed in.
So since the boat has a lower density than the water, it will float.
So the answer is choice B

3 0

2 years ago

You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this state

posledela

Answer:

a) y₂ = 49.1 m , t = 1.02 s , b) y = 49.1 m , t= 1.02 s

Explanation:

a) We will solve this problem with the missile launch kinematic equations, to find the maximum height, at this point the vertical speed is zero

$v_{y}$ ² = $v_{oy}$ ² - 2 g (y –yo)

The origin of the coordinate system is on the floor and the ball is thrown from a height

y-yo = $v_{oy}² /2 g
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 y - 0 = 5.10 m
 
The height from the ground is the height that rises from the reference system plus the depth of the ground from the reference system
 y₂ = 5.1 + 44
 y₂ = 49.1 m
Let's use the other equation to find the time
 [tex]v_{y}$ = $v_{oy}$ - g t

t = $v_{oy}$ / g

t = 10 / 9.8

t = 1.02 s

b) the maximum height

y- 44.0 = $v_{y}$ ² / 2 g

y - 44.0 = 5.1

y = 5.1 +44.0

y = 49.1 m

The time is the same because it does not depend on the initial height

t = 1.02 s

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3 years ago

Use the data provided to calculate the gravitational potential energy of each cylinder mass. Round your answers to the nearest t

Goryan [66]

Answer: 14.7kJ, 29.4kJ, 44.1kJ

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

<em />

In the case of the Earth, in which the gravitational field is considered constant, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g=9.8m/s^{2}$ the acceleration due gravity and $h=500m$ the height of the object.

Knowing this, let's begin with the calculaations:

For m=3kg

$U_{p}=(3kg)(9.8m/s^{2})(500m)$

$U_{p}=14700J=14.7kJ$

For m=6kg

$U_{p}=(6kg)(9.8m/s^{2})(500m)$

$U_{p}=29400J=29.4kJ$

For m=9kg

$U_{p}=(9kg)(9.8m/s^{2})(500m)$

$U_{p}=44100J=44.1kJ$

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3 years ago