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nikitadnepr [17]
3 years ago
9

A 55 meter rope is lying on the floor and has has a weight of 4040 N in total. How much work is required to lift up one end of t

he rope to a height of 33 meters
Physics
1 answer:
Kazeer [188]3 years ago
6 0

Answer:

Explanation:

Given

Length of rope L=55\ m

Weight of rope W=4040\ N

weight density \lambda =\frac{4040}{55}=73.45\ N/m

Work done to lift rope 33 m

W=\int_{0}^{33}\lambda hdh

W=\int_{0}^{33}73.45hdh

W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0

W=39.993\ kJ      

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the gas in a balloon has P=100000 pa and v=0.0279 m^3. if the pressure increases to 120000 pa at constant temperature, what is t
mash [69]

Answer:

New volume of the baloon is 0.02325m^3

Explanation:

To answer this question we need to know the ideal gas law, which says:

p•V = n•R•T

p is pressure, V is volume, n is amount of substance (in moles), R is constant value and T is temperature.

Since it's stated that n and T are constant, and we know that R is a constant too, that means that p•V = constant value. Basically, that means that p1•V1 (pressure and volume before the pressure increase) equals to p2•V2 (pressure and volume after the pressure increase).

That means that:

100000 Pa • 0.0279 m^3 = 120000 Pa • V2. Next, V2= 100000•0.0279/120000. So, V2=0.02325m^3.

6 0
3 years ago
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Some waves can only travel through matter. What is this matter called?
ANTONII [103]
A water wave is an example of a mechanical wave. A wave that can travel only through matter is called a mechanical wave.
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3 years ago
Please help me asdrtyuio
marta [7]

Answer:

9. The Sun's Gravity

10. The core is the densest layer

4 0
3 years ago
Light of wavelength 441.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
pshichka [43]

Answer: R = 0.131 m

Explanation: The formulae for the distance between a fringe and the first (center) is given by

y = R×mλ/d

Where y = distance between first and nth fringe = 4mm = 4×10^-3m

λ = wavelength of light = 441.1nm = 441.1×10^-9m

R = distance between slits and screen =?

d = distance between slits = 0.29mm = 2.9×10^-4m

4×10^-3 = R ×2 ×441.4×10^-9/ 2.9×10^-5

Hence R = (4×10^-3) ×(2.9×10^-5)/2×441.4×10^-9

R = 1.16 × 10^-7/8.828×10^-7

R = 1.16/8.828

R = 0.131 m

8 0
3 years ago
Find the magnitude of the sum
shusha [124]

Answer:

\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m

Explanation:

<u>Sum of Vectors in the Plane</u>

Given two vectors

\displaystyle \vec{v_1}\ ,\ \vec{v_2}

They can be expressed in their rectangular components as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

The sum of both vectors can be done by adding individually its components

\displaystyle \vec{v_1}+\vec{v_2}=

If the vectors are given as a magnitude and an angle (M\ ,\ \theta ), each component can be found as

\displaystyle \vec{v_1}=

\displaystyle \vec{v_2}=

The first vector has a magnitude of 3.14 m and an angle of 30°, so

\displaystyle \vec{v_1}=

\displaystyle \vec{v_1}=

The second vector has a magnitude of 2.71 m and an angle of -60°, so

\displaystyle \vec{v_2}=

\displaystyle \vec{v_2}=

The sum of the vectors is

\displaystyle \vec{v_1}+\vec{v_2}=

\displaystyle \vec{v_1}-\vec{v_2}=

Finally, we compute the magnitude of the sum

\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{(4.08)^2+(-0.78)^2}

\displaystyle |\vec{v_1}+\vec{v_2}|=\sqrt{17.25}

\displaystyle |\vec{v_1}+\vec{v_2}|=4.15m

3 0
3 years ago
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