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Sergio039 [100]
2 years ago
13

A 2 kg box of taffy candy has 40j of potential energy relative to the ground. Its height above the ground is​

Physics
2 answers:
lord [1]2 years ago
8 0

Answer:

2.04m

Explanation:

PE=mgh

h= PE/mg

h= 40/2(9.82)

h=40/19.64

h=2.04

Alona [7]2 years ago
7 0

Answer:

20m

Explanation:

Formula to find the potential energy is,

Potential Energy = mgh

Here,

m = mass

g = gravitational field

h = height

According to the question, we have to find the height.

Usually, we get g is 10ms^{-2}

Let us solve for the answer now.

P.E. =mgh\\40J =2kg*10ms^{-2} *h\\40 = 20h\\\frac{40}{20}= \frac{20h}{20} \\20m=h

Hope this helps you.

Let me know if you have any other questions :-)

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A charge alters the space around it. What is this alteration of space called?
iogann1982 [59]

Answer: electric field

Explanation: when a charge is placed in space, it alters the space around it by creating an electric field.

This electric field has the ability to exert a force (f) on any test charge(q) placed within this vicinity.

This is the reason why a charge can either attract or repel another charge.

6 0
3 years ago
Which substance is a combination of different atoms?
Lady_Fox [76]

Answer:

The answer is compound

Explanation:

heterogeneous mixture is wronggggg

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3 years ago
A star’s parallax angle is 1.0. How far away is the star in light years?
larisa [96]
Distance = 2AU / tan1.0

If you mean 1.0 is in degrees, then Distance = 114.58 AU
7 0
2 years ago
Read 2 more answers
What is the electric force acting between two charges of -0.0045 C and -0.0025 C that are 0.0060 m apart? Use Fe=kq1q2/r^2 and k
oksano4ka [1.4K]

Answer:

D. 2.8 × 10⁹ N

Explanation:

The force between two charges is directly proportional to the amount of charges at the two points and inversely proportional to the square of distance between the two points.

Fe= k Q₁Q₂/r²

Q₁= -0.0045 C

Q₂= -0.0025 C

r= 0.0060 m

k= 9.00 × 10 ⁹ Nm²/C²

Fe= (9.00 × 10 ⁹ Nm²/C²×-0.0045 C×-0.0025 C)/0.0060²

=2.8 × 10⁹ N

4 0
3 years ago
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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
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