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Nadya [2.5K]
3 years ago
7

A ball of mass 0.35 kg hangs straight down on a string of length 13 cm. It is then swung upward, keeping the string taut, until

the string makes an angle of 70° with respect to the vertical. Find the change in the gravitational potential energy of the Earth-ball system.

Physics
1 answer:
Anni [7]3 years ago
6 0

Answer:

PE=0.29J

Explanation:

According to the description, there is a angle and in point swung upward of 70°

So,

Y=13*10^{-2}*cos(70) \\Y=0.0444m

Appling the equation of Potential Energy we have,

PE=mgh\\PE=(0.35)(9.8)(0.13-0.0444)\\PE=0.29J

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3 years ago
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When a 360 nF air capacitor is connected to a power supply, the energy stored in the capacitor is 1.85 x 10-5 J. While the capac
GaryK [48]

Answer:

(a) Approximately 10.1\; {\rm V}.

Explanation:

Let C denote the capacitance of a capacitor. Let V be the potential difference (voltage) between the two plates of this capacitor. The energy E stored in this capacitor would be:

\displaystyle E = \frac{1}{2}\, C\, (V^{2}).

Rearrange this equation to find an expression for the potential difference V in terms of capacitance C and energy E:

\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}.

\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}

The capacitance C of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}.

Given that the energy stored in this capacitor is E = 1.85 \times 10^{-5}\; {\rm J}, the potential difference across the capacitor plates would be:

\begin{aligned}V &= \sqrt{\frac{2 \times 1.85 \times 10^{-5}\; {\rm J}}{3.60 \times 10^{-7}\; {\rm F}}} \\ &\approx 10.1\; {\rm V}\end{aligned}.

7 0
3 years ago
Match these items.
aleksandrvk [35]

Answer:

Here's your answer :

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hope it helps!

3 0
3 years ago
A car turns a certain curve of radius 24.98 m with constant linear speed of
Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

 = 3525.19 kg

7 0
3 years ago
"Giant Swing", the seat is connected to two cables as shown in the figure (Figure 1) , one of which is horizontal. The seat swin
Bingel [31]
The horizontal force is m*v²/Lh, where m is the total mass. The vertical force is the total weight (233 + 840)N. 

<span>Fx = [(233 + 840)/g]*v²/7.5 </span>

<span>v = 32.3*2*π*7.5/60 m/s = 25.37 m/s </span>

<span>The horizontal component of force from the cables is Th + Ti*sin40º and the vertical component of force from the cable is Ta*cos40º </span>

<span>Thh horizontal and vertical forces must balance each other. First the vertical components: </span>

<span>233 + 840 = Ti*cos40º </span>

<span>solve for Ti. (This is the answer to the part b) </span>

<span>Horizontally </span>

<span>[(233 + 840)/g]*v²/7.5 = Th + Ti*sin40º </span>

<span>Solve for Th </span>

<span>Th = [(233 + 840)/g]*v²/7.5 - Ti*sin40º </span>

<span>using v and Ti computed above.</span>
3 0
3 years ago
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