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Nana76 [90]
3 years ago
8

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

nless tabletop. If the maximum tension that the string can withstand without breaking is 350 N. What is the maximum speed the mass can have without breaking the string?
Physics
1 answer:
xenn [34]3 years ago
6 0

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string  

For without breaking the string tension must be equal to the centripetal force

So T=\frac{mv^2}{r}

So 350=\frac{0.5\times v^2}{1}

v^2=700

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

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A boy whirls a stone in a horizontal circle of radius 1.8 m and at height 1.8 m above level ground. The string breaks, and the s
umka2103 [35]

Answer:

Acceleration is 148.33\ m/s^{2}

Solution:

As per the question:

Radius of the circle, R = 1.8 m

Height above the ground, h = 1.8 m

Horizontal distance, x = 9.9 m

Now,

The magnitude of the centripetal acceleration can be calculated as:

a_{c} = \frac{v^{2}}{R}        

where

v = velocity

R = radius

a_{c} = centripetal acceleration

Now, if we consider the vertical component of motion only, then considering the initial velocity, 'u' = 0, from kinematic eqn:

h = ut + \frac{1}{2}gt^{2}                

h = 0.t + \frac{1}{2}gt^{2}                

t = \sqrt{\frac{2h}{g}}                

t = \sqrt{\frac{2\times 1.8}{9.8}}

t = 0.606 s

Now, for the horizontal component of velocity:

x = vt

v =\frac{x}{t}

v =\frac{9.9}{0.606} = 16.34\ s

Now, we know that the centripetal acceleration is given by:

a_{c} = \frac{16.34^{2}}{1.8} = 148.33\ m/s^{2}        

                                                                                                                                                                                                                                           

6 0
3 years ago
Forecasting the time of, location, and magnitude of a seismic event does not prevent
Drupady [299]
<span>Forecasting the time of, location, and magnitude of a seismic event does not prevent
the event from happening, but it can help us reduce the destruction caused by
A) earthquakes.</span>
4 0
3 years ago
E
IgorC [24]

Answer:

Speed = 300 m/s

Explanation:

Given the following data;

Frequency = 30 Hz

Wavelength = 10 m

To find the speed of the wave;

Mathematically, the speed of a wave is given by the formula:

Speed = wavelength * frequency

Substituting into the formula, we have;

Speed = 10 * 30

Speed = 300 m/s

6 0
3 years ago
50 Points | Earth &amp; Space Science
babymother [125]
When a product fails to perform as warranted, this is called a) contractual liability. O b) product malfunction. c) malicious manufacture. d) breach of warranty
4 0
3 years ago
Three small objects are arranged along a uniform rod of mass m and length L. one of mass m at the left end, one of mass m at the
yKpoI14uk [10]

Answer:

See answer below

Explanation:

Hi there,

To get started, recall the Center of Mass formula for two masses:

x_c_m = \frac{m_1x_1+m_2x_2}{m_1+m_2}  where m is mass and x is displacement <em>from the center of the shape.</em>

Since masses at the center of a geometric shape have a displacement (x) value of 0, as the mass is already of the center, and does not affect Xcm. So, we can disregard the central mass, hence we use the above formula for two masses.

We can arbitrarily define left to be a negative (-) displacement, and vice versa for right direction. We proceed with the formula:x_c_m=\frac{(-L/2)m+(L/2)2m}{m+2m} =\frac{(L/2)(-m+2m)}{3m} \\ x_c_m=\frac{L(m)}{6m} =\frac{L}{6}

Since we defined left (-) and right (+), we notice the center of mass is (+) value. This makes sense, as there is slightly more mass on the right side. Hence, you should place a support 1/6 of the rod's length away from the rod's center.

Study well and persevere.

thanks,

4 0
4 years ago
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