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Zielflug [23.3K]
3 years ago
14

Work done by a gravitational force by lowering the bucket into the well is

Physics
1 answer:
katovenus [111]3 years ago
3 0

Answer:

when we lower a bucket into a well to fetch water, the work done by gravity is positive since force and displacement are in the same direction.

Explanation:

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What does 3F means?<br><br>Please help me with this.....​
Debora [2.8K]

Answer: Fitness, Fun & Friendship

Explanation:

5 0
3 years ago
La tensión en newtons necesaria para que una onda transversal cuya longitud de onda es 3.33 cm vibre a razón de 625 ciclos por s
NemiM [27]

Answer:

9.34 N

Explanation:

First of all, we can calculate the speed of the wave in the string. This is given by the wave equation:

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

For the waves in this string we have:

f=625 Hz, since it completes 625 cycles per second

\lambda=3.33 cm = 0.033 m is the wavelength

So the speed of the wave is

v=(625)(0.0333)=20.6 m/s

The speed of the waves in a string is related to the tension in the string by

v=\sqrt{\frac{T}{\mu}} (1)

where

T is the tension in the string

\mu=\frac{m}{L} is the linear density

In this problem:

m=16.5 g = 16.5\cdot 10^{-3} kg is the mass of the string

L = 0.75 m is the its length

Solving the equation (1) for T, we find the tension:

T=\mu v^2 = \frac{m}{L} v^2 = \frac{16.5\cdot 10^{-3}}{0.75}(20.6)^2=9.34 N

8 0
3 years ago
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
kifflom [539]

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

     

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = \sqrt[4]{P/Ae}

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

4 0
3 years ago
What is the KE if a 10kg mass traveling at 5 m/s
11111nata11111 [884]

Answer: KE = 25 J

Explanation: You must use the formula

KE = 1/2 m v²

to solve this problem.

KE = 1/2 (10 Kg) (5 m/s)

KE = 1/2 (50 kgm/s)

KE = 25 J

7 0
2 years ago
Before starting this problem, review Conceptual Example 3 in your text. Suppose that the hail described there comes straight dow
bulgar [2K]

Answer:

0.9 N

Explanation:

The force exerted on an object is related to its change in momentum by:

F=\frac{\Delta p}{\Delta t}

where

F is the force exerted

\Delta p is the change in momentum

\Delta t is the time interval

The change in momentum can be rewritten as

\Delta p = m(v-u)

where

m is the mass

u is the initial velocity

v is the final velocity

So the formula can be rewritten as

F=\frac{m(v-u)}{\Delta t}

In this problem we have:

\frac{m}{\Delta t}=0.030 kg/s is the mass rate

u=-15 m/s is the initial velocity

v=+15 m/s is the final velocity

Therefore, the force exerted by the hail on the roof is:

F=(0.030)(+15-(-15))=0.9 N

6 0
3 years ago
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