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1 year ago

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A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. T

he stone is whirled in a horizontal circle on a frictionless tabletop; the other end of the string remains fixed. (a) Draw a free-body diagram of the stone. (b) Find the maximum speed the stone can attain without the string breaking.

1 answer:

AleksandrR [38]1 year ago

5 0

Answer:

$v=8.2158m/s$

Explanation:

(a) Free-body diagram attached.

(b) The stone attached with the string experiences both centripetal (towards the center) and centrifugal (away from the center) forces. The tension of the string counters the centrifugal force until it breaks.

We know that,

Centrifugal force = $\frac{mv^2}{r}$

where,

$m$ = mass of the stone

$v$ = velocity of the stone

$r$ = length of the string

To find the maximum speed attained by the stone without the string breaking, we must equate:

$\frac{mv^2}{r} =60$

or, $v=\sqrt \frac{{r\times 60}}{m} } =\sqrt{\frac{0.90\times60}{0.80} } =8.2158m/s$

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Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, $m=40\ kg$

Radius of the barrel is, $R=2.90\ m$

Number of revolutions are, $n =10$

Time taken for 10 revolutions is, $t=29.3\ s$

Therefore, the time period of the child is given as:

$T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s$

Now, angular velocity is related to time period as:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s$

Now, centripetal force acting on the child is given as:

$F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N$

Therefore, the centripetal force acting on the child is 39400.56 N.

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1 year ago

I take 1.0 kg of ice and dump it into 1.0 kg of water and, when equilibrium is reached, I have 2.0 kg of ice at 0°C. The water w

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Answer:

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Explanation:

In the whole process one kg of water at 0°C loses heat to form one kg of ice and heat lost by them is taken up by ice at −160°C . Now see whether heat lost is equal to heat gained or not.

heat lost by 1 kg of water at 0°C

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= 80000 cals

heat gained by ice at −160°C to form ice at 0°C

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= 1 x .5 x 1000 x 160

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A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'

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Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3

= 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

$Y =\frac{mgL}{A\Delta L}$

$\Delta L =\frac{mgL}{A\times Y}$

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

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1 year ago

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el

Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is $Q =2.094 C$

Explanation:

From the question we are told that

The diameter of the wire is $d = 0.205cm = 0.00205 \ m$

The radius of the wire is $r = \frac{0.00205}{2} = 0.001025 \ m$

The resistivity of aluminum is $2.75*10^{-8} \ ohm-meters.$

The electric field change is mathematically defied as

$E (t) = 0.0004t^2 - 0.0001 +0.0004$

Generally the charge is mathematically represented as

$Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt$

Where A is the area which is mathematically represented as

$A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2$

So

$\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega$

Therefore

$Q = 120 \int\limits^{t}_{0} { E(t) } \, dt$

substituting values

$Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt$

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.$

From the question we are told that t = 5 sec

$Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.$

$Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }$

$Q =2.094 C$

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2 years ago

A helicopter flies 250 km on a straight path in a direction 60° south of east. The east component of the helicopter’s displaceme

GaryK [48]

Given that,

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since,

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East component = 125 km

8 0

2 years ago

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