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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?

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>

6 0

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

Read 2 more answers

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