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Amiraneli [1.4K]
3 years ago
14

A string under a tension of 50.4 N is used to whirl a rock in a horizontal circle of radius 2.51 m at a speed of 21.1 m/s. The s

tring is pulled in and the speed of the rock increases. When the string is 1.22 m long and the speed of the rock is 51.6 m/s, the string breaks. What is the breaking strength (in newtons) of the string?
Physics
1 answer:
Leokris [45]3 years ago
5 0

Answer:

619.8 N

Explanation:

The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:

T=m\frac{v^2}{r}

where

T is the tension

m is the mass of the rock

v is the speed

r is the radius of the circular path

At the beginning,

T = 50.4 N

v = 21.1 m/s

r = 2.51 m

So we can use the equation to find the mass of the rock:

m=\frac{Tr}{v^2}=\frac{(50.4)(2.51)}{21.1^2}=0.284 kg

Later, the radius of the string is decreased to

r' = 1.22 m

While the speed is increased to

v' = 51.6 m/s

Substituting these new data into the equation, we find the tension at which the string breaks:

T'=m\frac{v'^2}{r'}=(0.284)\frac{(51.6)^2}{1.22}=619.8 N

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<span>C. Mao Zedong
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3 years ago
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which of the following statements best explains how consumers determine growth in technological areas
Mrac [35]
If these were the missing choices:

a)      Consumers fill out questionnaires concerning their need for new products.

b)      Consumers vote for politicians who decide which kind of research to support

c)       Consumers decide what to buy and what not to buy

d)      Consumers influence the decisions of private foundations by deciding where to donate money.


My answer would be: c)       <span>Consumers decide what to buy and what not to buy</span>

Every growth is based on the demand of the people. If a good or service is needed then its demand will increase. If a good or service is not needed then its demand will decrease until such time that said good or service will be eliminated. 

6 0
3 years ago
Convert <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%280.779mg%29%28min%29%7D%7BL%7D" id="TexFormula1" title="\frac{(0.779mg)(mi
Orlov [11]

The number converted is 0.0467 \frac{(kg)(s)}{m^3}

Explanation:

In order to convert from the original units to the final units, we have to keep in mind the following conversion factors:

1 kg = 1000 g = 10^6 mg

1 min = 60 s

1 m^3 = 1000 L

The original unit that we have is

\frac{mg\cdot min}{L}

Therefore, it can be rewritten as:

=\frac{mg \frac{1}{10^6 mg/kg}\cdot min\cdot  60 s/min}{L\frac{1}{1000L/m^3}}=0.06 \frac{(kg)(s)}{m^3}

Therefore, since the initial number was 0.779, the final value is

0.779\cdot 0.06 \frac{(kg)(s)}{m^3}=0.0467 \frac{(kg)(s)}{m^3}

#LearnwithBrainly

5 0
3 years ago
A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

      a₁/a₂ t² = t² -2 4 t + 16

      t² (1- 2.0 / 4.0) - 8 t +16

      t² 0.5 - 8 t +16 = 0

      t² -16 t + 32 = 0

Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

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nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

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v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

x

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x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

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3 years ago
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