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1 year ago

15

URGENT!!! sorry this is my first time using brainly! but here is my question: Two ropes are attached to a wagon, one horizontal

to the west with a tension force of 30 N, and the other east and at an angle of 30° northward and a tension force of 40 N. Find the components of the net force on the cart. Show all your work. Thank you in advance!

1 answer:

Elan Coil [88]1 year ago

3 0

The components of the net force on the cart is determined as 67.66 N.

<h3>Component of net force on the cart</h3>

The component of net force on the cart is determined by resolving the forces into x and y -components.

T1 = 30 N

T2 = 40 N

T1x = -30cos(0) = 30 N

T1y = 30sin(0) = 0

T2x = 40 x cos(30) = 34.64 N

T2y = 40 x sin(3) = 20 N

∑X = 30 N + 34.64 N = 64.64 N

∑Y = 0 + 20 N = 20 N

<h3>Resultant force</h3>

R = √(64.64² + 20²)

R = 67.66 N

Learn more about net force here: brainly.com/question/25239010

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F = 50 N

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3 years ago

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The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass $1.67 \times 10^(-27)$ kg, charge +e = $+1.60 \times 10^(-19)$ C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed $2.50 \times 10^6$ m/s. The proton comes momentarily to rest at a distance $5.31 \times 10^(-13)$ m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are $5.31 \times 10^{-13}$ m apart?

Explanation:

The given data is as follows.

Mass of proton = $1.67 \times 10^{-27}$ kg

Charge of proton = $1.6 \times 10^{-19} C$

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Distance traveled = $5.31 \times 10^{-13} m$

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$(K.E + P.E)_{initial}$ = $(K.E + P.E)_{final}$

$(\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)$

where, $\frac{kq_{p}q_{t}}{r} = U = Electric potential energy$

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

Putting the given values into the above formula as follows.

U = $(\frac{1}{2}m_{p}v^{2}_{p})$

= $(\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})$

= $5.218 \times 10^{-15} J$

Therefore, we can conclude that the electric potential energy of the proton and nucleus is $5.218 \times 10^{-15} J$ .

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