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2 years ago

Suppose a woman raises a 65 N object in 2m in 4 seconds.

Physics

1 answer:

Novosadov [1.4K]2 years ago

8 0

Answer:

$\huge\boxed{\sf P.E = 130\ Joules}$

$\huge\boxed{\sf P = 32.5\ Watts}$

Explanation:

Given Data:

Weight = W = 65 N

Height = h = 2 m

Time = t = 4 secs

Required:

Power = P = ?

Work Done in the form of Potential Energy = P.E. = ?

Formula:

P.E. = Wh

P = P.E. / t

Solution:

P.E. = (65)(2)

P.E = 130 Joules

P = P.E. / t

P = 130 / 4

P = 32.5 Watts

$\rule[225]{225}{2}$

Hope this helped!

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2 years ago

Anyone know how to do this?

Gala2k [10]

Answer:

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Explanation:

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2 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the minimum value of x= 0.17 m.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$

$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

Hence, the maximum distance is 5.002 m

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

$F= 6.38$ × $10^{-26} N$

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

F= 7.37× $10^{-29$ N

4 0

2 years ago

Determine the magnitude and direction of the resultant force of the following free body diagram.

Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force ( $\vec F$ ), in newtons, by vectorial sum:

$\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}$ (1)