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Solnce55 [7]
3 years ago
10

A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3300 N and t

he mass of the truck is 1600 kg, what's the radius of the curve?
Physics
2 answers:
kozerog [31]3 years ago
6 0
Well, first of all, the truck's velocity is constantly changing, not 'uniform'.
Velocity consists of speed and direction.  So, even if the truck's speed is
constant, its direction keeps changing as long as it's on a circular curve,
so its velocity is constantly changing.

The force needed to keep a mass moving in a circle is

                                 F = (mass) x (speed)² / (radius)
                      
                             3300 N  =  (1600 kg) (13 m/s)² / R

                           3300 kg-m/s²  =  (1600 kg) (169 m²/s²) / R

                           R  =  (1600 kg) · (169 m²/s²) / (3300 kg·m/s²)

                               =  (1600 · 169 / 3300)  meters

                               =        81.9  meters     
Lelu [443]3 years ago
5 0

The Correct answer to this question for Penn Foster Students is: 81.94m

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Explanation:

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If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?.
Oksanka [162]

The fraction of the tensile strength which is the stress on the femur is 1.4%.

<h3>What is Tensile strength?</h3>

This is defined as the amount of load or stress that a material can handle before it stretches and breaks.

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5 0
2 years ago
A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor
Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm  

Hence, 1.16 cm were cut from the end of the second pipe

6 0
3 years ago
What is the kinetic engery of a baseball five feet above the ground when it has already fallen 32 feet
mylen [45]
The answer is 25 ...
6 0
4 years ago
The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t
maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s

Work done is equal to the change in kinetic energy. It can be given as follows :

W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J

So, the required work done is 32.99 J.

3 0
3 years ago
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