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2 years ago

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Uzupełnij zdania właściwymi sformułowaniami. Wyobraź sobie, że między linę a siodełko karuzeli łańcuchowej wmontowany jest siłom

ierz. Jeśli na postoju nie dotykasz nogami do ziemi, to siłomierz wskazuje A / B. Kiedy karuzela się kręci, na siłomierzu odczytasz C / D. A. Twój ciężar wraz z siodełkiem C. Wartość wytrzymałości liny B. Twój ciężar D. Wartość siły dośrodkowej

1 answer:

iragen [17]2 years ago

3 0

Explanation:

Here's a clearer rendering of the question requirements;

Complete the sentences with the correct wording. Imagine that a force gauge is mounted between the rope and the chain carousel saddle. If you do not touch your feet to the ground when the vehicle is stationary, the dynamometer indicates A / B. When the carousel turns, you will read C / D on the dynamometer.

A. Your weight with the saddle

C. Rope strength value

B. Your weight

D. Centripetal force value

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Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim

AnnZ [28]

Answer:

$\boxed{v=\frac {V}{\sqrt {2}}}$

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that $f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}$

Now, the speed of the father will be $V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})$ while for the child the speed will be $v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})$

The ratio of the father’s speed to the child’s speed will be

$\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}$

8 0

2 years ago

A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im

geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

Em₀ = Em_f

½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

v_f ² = v₀² + 2g (y₁ -y₂)

we calculate

v_f ² = 20² + 2 9.8 (10 -15)

v_f = √302

v_f = 17.4 m / s

8 0

2 years ago

An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

kolbaska11 [484]

Answer:

$mg=200.4 N$ .

Explanation:

This problem can be solved using Newton's law of universal gravitation: $F=G\frac{m_{1}m_{2}}{r^{2}}$ ,

where F is the gravitational force between two masses $m_{1}$ and $m_{2}$ , $r$ is the distance between the masses (their center of mass), and $G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2})$ is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

$mg=8*10^{2}$ ,

$m=(8*10^{2})/g$ ,

since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

Now we can use Newton's law of universal gravitation

$F=G\frac{Mm}{r^{2}}$ ,

where $m$ is the mass of the astronaut and $M$ is the mass of the earth. From Newton's second law we know that

$F=ma$ ,

in this case the acceleration is the gravity so

$F=mg$ , (<u>becarefull, gravity at this point is no longer</u> $9.8m/s^{2}$ <u>because we are not in the surface anymore</u>)

and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

We need to remember that the mass of the earth is $M=5.972*10^{24}kg$ and its radius is $6.37*10^{6}m$ .

The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

$mg=200.4 N$ .

5 0

2 years ago

A gas station owner suspects that he is being overcharged for gasoline deliveries by a gasoline supplier. The overcharge seems p

Tems11 [23]

Answer:

Explanation:

delta V = v * alpha * delta T

= V * 0.00053 * (92.2 - 55.0)

= 0.019716 V

percentage that the owner

= [delta V / V] * 100

= [0.019716 V / V] * 100

= 1.9716 %

4 0

2 years ago

Suppose you want to make a scale model of a hydrogen atom. You choose, for the nucleus, a small ball bearing with a radius of 1.

RoseWind [281]

Answer:

A) x _electron = 0.66 10² m , B) x _Eart = 1.13 10² m , C) d_sphere = 1.37 10⁻² mm

Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

$R_{E}$ = 6.37 10⁶ m

tex]R_{Sum}[/tex] = 6.96 10⁸ m

Distance = 1.5 10¹¹ m

x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5

x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶

d_sphere = 1.37 10⁻² mm

5 0

2 years ago