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Naddika [18.5K]

2 years ago

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A boy does 465 J of work pulling an empty wagon along level ground with a force of 111 N [31o below the horizontal]. A frictiona

l force of 155 N opposes the motion and is actually slowing the wagon down from an initial high velocity. The distance the wagon travels is

1 answer:

vovikov84 [41]2 years ago

6 0

Can you list the answer's

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A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

2 years ago

A 5.09 × 1014-hertz electromagnetic wave is traveling through a transparent medium. The main factor that determines the speed of

sergiy2304 [10]

We are given an electromagnetic wave with a frequency of 5.09 x 10^14 Hz and travelling through a transparent medium. If the medium was vacuum, the speed of the wave would be equal to the speed of light. Otherwise, the main factor that would determine the speed of the wave is its wavelength.

6 0

2 years ago

A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 38

oksano4ka [1.4K]

Answer:

the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

Explanation:

Given

$Initial Temperature T_1=387 KFinal Temperature T_2=774 K$

The RMS velocity of molecules in a gas is given by

$V_{rms}=\sqrt{\dfrac{3k_bT}{m}}$

where T=temperature

$k_b=constant$

For T = 387K

$V_1=\sqrt{\frac{3k_b\cdot 387}{m}}----1$

For T = 774

$V_2=\sqrt{\frac{3k_b\cdot 774}{m}}----(2)$

dividing eqn 1 and eqn 2

$\frac{V_2}{V_1}=\sqrt{\frac{774}{387}}$

$\frac{V_2}{V_1}=\sqrt{2}$

Thus,the ratio is $\frac{V_2}{V_1}=\sqrt{2}$

5 0

2 years ago

Read 2 more answers

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

Vedmedyk [2.9K]

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

5 0

2 years ago

Two window washers, Bob and Joe, are on a 3.00 m long, 395 N scaffold supported by two cables attached to its ends. Bob weighs 8

WINSTONCH [101]

Answer:

- the forces on the left hand side is 1.038 kN

- the forces on the right hand side is 1.483 kN

Explanation:

Given the data in the question, as illustrated in the image below;

Length of the scaffold = 3 m

weight of the scaffold = 395 N

Weight of Bob = 805 N and stands 1 m from the left end

weight of washing equipment = 500N and on sits 2 m from the left end

Weight of Joe = 820 N and stand 0.500 m from the right end

so the force on the left cable will be;

$T_{left$ = $\frac{1}{3m}$ [ (805 N)( (3-1) m) + ( 395 N )( $\frac{3}{2} m$ ) + ( 500 N )(1m ) + ( 820 N)( 0.500m ) ]

$T_{left$ = $\frac{1}{3m}$ [ 1610 + 592.5 + 500 + 410 ]

$T_{left$ = $\frac{1}{3m}$ [ 3112.5 ]

$T_{left$ = 1037.5 N

$T_{left$ = 1.038 kN

Therefore, the forces on the left hand side is 1.038 kN

On the right hand side;

$T_{Right$ = ( 805 N + 395 N + 500 N + 820 N ) - 1037.5 N

$T_{Right$ = 2520 N - 1037.5 N

$T_{Right$ = 1482.5 N

$T_{Right$ = 1.483 kN

Therefore, the forces on the right hand side is 1.483 kN

5 0

1 year ago