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2 years ago

8

A force f = bx 3 acts in the x direction, where the value of b is 3.7 n/m3. how much work is done by this force in moving an obj

ect from x = 0.00 m to x = 2.7 m?

1 answer:

Elodia [21]2 years ago

8 0

The answer would be 21.6 but rounded up it would be 22J.

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A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0

2 years ago

An ambulance driving 35.0 m/s emits a sound wave with a wavelength of 80.0 centimeters. As it drives away from a hospital, which

katen-ka-za [31]

Apparent frequency heard by the staff: 389 Hz

Explanation:

The phenomenon described in this situation is called Doppler effect.

Doppler effect occurs when there is a source emitting a wave in relative motion with respect an observer. In such situation, the frequency of the wave as perceived by the observer ("apparent frequency") is shifted from the real frequency of the sound ("proper frequency"). In particular:

- The observer perceives a higher frequency if the source is moving towards them

- The observer perceives a lower frequency if the source is moving away from them

The formula to calculate the apparent frequency in the Doppler effect is

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the proper frequency

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if they are moving towards the source, negative if moving away)

$v_s$ is the velocity of the source (positive if it is moving away, negative if moving towards the observer)

First of all, in this problem we have to calculate the proper frequency of the sound wave emitted from the ambulance; we have:

v = 343 m/s (speed of sound wave)

$\lambda=80 cm = 0.80 m$ (wavelength)

So the proper frequency is

$f=\frac{v}{\lambda}=\frac{343}{0.80}=429 Hz$

Now we can calculate the apparent frequency heard by the staff at the hospital when the ambulance moves away; we have:

$v_s = +35.0 m/s$ (velocity of the ambulance)

$v_o = 0$ (velocity of the staff)

Substituting,

$f'=\frac{343+0}{343+35}(429)=389 Hz$

Learn more about frequency and wavelength:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 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

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

Naily [24]

Answer:

700J

Explanation:

8 0

2 years ago

An electric power plant uses energy from burning coal to generate steam at 450∘C. The plant is cooled by 20∘C water from a nearb

Goryan [66]

Answer:

40 MJ (D)

Explanation:

Quantity of heat (Qh) = 100 MJ

temperature of steam (Th) = 450°c = 450 + 273 = 723 K

emperature of water (TI) = 20 °c = 20 + 273 = 293 k

efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th

$\frac{100x10x^{6}-Qi }{100x10^{6}} = \frac{723-293}{723}$

$100x10^{6}$ - Qi= 0.5947 x $100 x 10 ^{6}$

$100x10^{6}$ - (0.5947 x $100x10^{6}$ ) = Qi

Qi = 40.5 MJ equivalent to 40 MJ (D)

6 0

2 years ago