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umka2103 [35]
2 years ago
15

A cart was pulled for a distance of 1 kilometer and the amount of work accomplished equaled 40,000 joules. With what force was t

he work accomplished?

Physics
1 answer:
goblinko [34]2 years ago
3 0

Answer:

40 N

Explanation:

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A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N
crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

Unbalanced Force = Horizontal Component Applied Force - Frictional Force

Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

f = frictional force  = ?

F = Applied force  = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Therefore,

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

<u>f = 485.62 N</u>

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3 years ago
Which of these BEST describes the term ecliptic? A. the circle where "Earth meets the sky" B. the change in azimuth and altitude
jeyben [28]
I'll say B ....hope that helps...
4 0
3 years ago
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You push on a box with a force of 300 N directly north. Another person pushes the box with a
tangare [24]

Answer:

resultant \\  \: F =  \sqrt{ {300}^{2} +  {600}^{2}  }  \\  =  \sqrt{450000}  \\  = 670.82 \: newtons

7 0
3 years ago
A bridge to be fabricated of steel girders is designed to be 500 m long and 12 m wide at ambient temperature (assumed 20°C). Exp
andrew-mc [135]

Answer:

23

21.7391304348 m

Explanation:

L = Initial length = 500 m

\Delta T = Change in temperature = 40-(-35)

\alpha = Coefficient of thermal expansion = 12\times 10^{-6}\ /^{\circ}C

Change in length is given by

\Delta L=L\alpha \Delta T\\\Rightarrow \Delta L=500\times 12\times 10^{-6}\times (40-(-35))\\\Rightarrow \Delta L=0.45\ m

The change in length is 0.45 m

The number of joints would be

n=\dfrac{0.45}{0.02}\\\Rightarrow n=22.5\\\Rightarrow n\approx 23

The number of joints is 23

Each bridge section length would be

\dfrac{L}{n}=\dfrac{500}{23}=21.7391304348\ m

The length of each bridge section would be 21.7391304348 m

4 0
3 years ago
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
3 years ago
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