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Elena L [17]
3 years ago
11

A bag weighing 20 N is pushed horizontally a distance of 35 m across a

Physics
1 answer:
topjm [15]3 years ago
6 0

Answer:

350J

Explanation:

Given parameters:

Weight of bag  = 20N

Distance moved horizontally  = 35m

Force applied  = 10N

Unknown:

Work done on the bag  = ?

Solution:

Work done is the force applied to move a body through given distance.

  Work done  = Force applied x distance

So;

 Work done  = 10 x 35  = 350J

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The weight (W) of an object is the product of its mass (m) in kilograms and g, the acceleration due to gravity in meters/second2
Nookie1986 [14]
The weights in newtowns for the given masses are

<span> masses        22.1,                 33.5,             41.3,           59.2,            78
 weights        216.58N           328.3N       404.74N      580.16N     764.4N

e.g,   for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>
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3 years ago
What best describes the motion of the medium particles in a longitudinal wave? The medium does not move. The medium moves in all
Citrus2011 [14]

In the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

<h3>What is a longitudinal wave?</h3>

A longitudinal wave is a wave that is transversing along the length. When the displacement of medium and travel of wave is the same in that condition wave is known as the longitudinal wave.

It requires some medium to travel. A mechanical and sound wave is an example of a longitudinal wave.

Hence in the motion of the medium particles in a longitudinal wave, the medium vibrates parallel to the direction of the wave.

To learn more about the longitudinal wave refer to the link;

brainly.com/question/8497711

3 0
2 years ago
The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and
Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

\eta = 7.07 mole

Explanation:

Part 1

below equation is used to determine the type Gas by determining \gamma value

\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}

\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)

substituing value to get final temperature

T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}

T_f = 330.0 K

Part 3

determine number of moles by using following formula

\eta =\frac{PV}{RT}

\eta =\frac{1.013*10^{5}*0.151}{8.314*260}

\eta = 7.07 mole

4 0
3 years ago
Lourdes mixes several ingredients in a bowl, creating a cake batter. She holds the bowl up and turns it upside down, causing the
sasho [114]

friction and gravity ....................

4 0
3 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
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