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Troyanec [42]
3 years ago
12

How much potential energy does a 40n block medicine ball gain when it is lifted 5.m

Physics
2 answers:
MAVERICK [17]3 years ago
5 0
By definition, the gain in PE (potential energy) is
ΔPE = m*g*h

Given:
mg = 40 N  (Note that m*g = weight)
h = 5 m

ΔPE = (40 N)*(5 m) =200 J

Answer: 200 J

Ksivusya [100]3 years ago
5 0
The answer is 200J I hope it helps ^_^
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Cheryl is riding on the edge of a merry-go-round, 2m from the center, which is rotating with an increasing angular speed. Cheryl
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In circular motion we know that there is two type of acceleration that Cheryl experience

1. Tangential acceleration

2. Centripetal acceleration

here given that

a_t = 3 m/s^2

for centripetal acceleration we know that

a_c = \frac{v^2}{R}

a_c = \frac{4^2}{2} = 8 m/s^2

now we know that both centripetal acceleration and tangential acceleration is perpendicular to each other

so total acceleration is vector sum of both and given as

a_{net} = \sqrt{8^2 + 3^2} = 8.54 m/s^2

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3 years ago
Two grocery carts have the same initial total momentum. The two carts collide, and now cart B has a greater final velocity.
Angelina_Jolie [31]

Answer:

Cart B has a smaller mass

Explanation:

I may be wrong but if mass is decreased then acceleration increases, and if velocity increases then acceleration increases so my logic is that if mass decreases then velocity increases.

7 0
3 years ago
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n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

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q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

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M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

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b). Period of revolution.

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T = 1.19*10⁻⁷s

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K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

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