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Andrej [43]
3 years ago
6

Statement A: Area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Physics
1 answer:
pochemuha3 years ago
7 0

Explanation:

Formula for calculating the area of a  rectangle A = Length *width

For statement A;

Given area of a rectangle with measured length = 2.536 mm and width = 1.4 mm.

Area of the rectangle = 2.536mm * 1.4mm

Area of the rectangle = 3.5504mm²

The rule of significant figures states that we should always convert the answer to the least number of significant figure amount the given value in question. Since 1.4mm has 2 significant figure, hence we will convert our answer to 2 significant figure.

Area of the rectangle = 3.6mm² (to 2sf)

For statement B;

Given area of a rectangle with measured length = 2.536 mm and width = 1.41 mm.

Area of the rectangle = 2.536mm * 1.41mm

Area of the rectangle = 3.57576mm²

Similarly, Since 1.41mm has 3 significant figure compare to 2.536 that has 4sf, hence we will convert our answer to 3 significant figure.

Area of the rectangle = 3.58mm² (to 3sf)

Based on the conversion, it can be seen that 3.6mm²  is greater than 3.58mm², hence the area of rectangle in statement A is greater than the area of the rectangle in statement B.

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A skier of mass 103 kg comes down a slopeof constant angle 32◦with the horizontal.What is the force on the skier parallel tothe
LuckyWell [14K]

Answer:

534.9 N

Explanation:

The skier weight is his mass times gravitational acceleration g

W = mg = 103 * 9.8 = 1009.4 N

This weight can be divided into 2 components, one perpendicular and the other parallel to the 32-degree slope. The parallel component would equal to

Wsin(32^0) = 1009.4sin(32^0) = 534.9 N

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3. The velocity of sound is 332 m/s. Answer the following questions:
Angelina_Jolie [31]

Answer:

20 Hz, 20000 Hz

0.0166 m, 16.6 m

Explanation:

The minimum frequency that a human ear can hear is 20 Hz

The maximum frequency that a human ear can hear is 20000 Hz.

v = Velocity of sound = 332 m/s

Wavelength is given by

\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{332}{20}\\\Rightarrow \lambda=16.6\ \text{m}

The longest wavelength that can be heard by the human ear is 16.6 m

\lambda=\dfrac{332}{20000}\\\Rightarrow \lambda=0.0166\ \text{m}

The shortest wavelength that can be heard by the human ear is 0.0166 m.

3 0
2 years ago
A spring scale hung from the ceiling stretches by 6.4 cm when a 2.0 kg mass is hung from it. The 2.0 kg mass is removed and repl
dolphi86 [110]
In the first case, the force acting on the spring is the weight of the mass:
F=mg=(2.0 kg)(9.81 m/s^2)=19.6N
This force causes a stretching of x=6.4 cm=0.064 m on the spring, so we can use these data to find the spring constant:
k= \frac{F}{x}= \frac{19.6 N}{0.064 m}=306.3 N/m

In the second case, the first mass is replaced with a second mass, whose weight is
F=mg=(2.5 kg)(9.81 m/s^2)=24.5 N
And since we know the spring constant, we can calculate the new elongation of the spring:
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5 0
3 years ago
A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)
wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

Plug in the given values for each point and solve.

(a)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C

(d)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C

7 0
3 years ago
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