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alisha [4.7K]
2 years ago
7

You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

in potential energy of a system containing the book and Earth?
Physics
1 answer:
Sophie [7]2 years ago
8 0
The change in potential energy of an object is given by
U=mg \Delta h
where
m is the mass of the object
g is the gravitational acceleration
\delta h is the increase in altitude of the object

In our problem, m=2.5 kg is the mass of the book, g=9.81 m/s^2 and 
\Delta h=2.6 m -1.2 m=1.4 m is the increase in altitude of the book, so its variation of potential energy is
U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J
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(a) On the axes below, sketch the graphs of the horizontal and vertical components of the sphere’s velocity as a function of tim
jeka94

Answer:

Two identical spheres are released from a device at time t = 0 from the same ... Sphere A has no initial velocity and falls straight down. ... (b) On the axes below, sketch and label a graph of the horizontal component of the velocity of sphere A and of sphere B as a function of time. ... Which ball has the greater vertical velocity

Explanation:

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2 years ago
Choose all the answers that apply. Ecosystems _____.
lana [24]

The answers are; B, C & E.

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3 0
1 year ago
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You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for
mart [117]

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

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= -4.1884k Nm

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8 0
2 years ago
A magnetic dipole with a dipole moment of magnitude 0.0243 J/T is released from rest in a uniform magnetic field of magnitude 57
ololo11 [35]

Answer:

47.76°

Explanation:

Magnitude of dipole moment = 0.0243J/T

Magnetic Field = 57.5mT

kinetic energy = 0.458mJ

∇U = -∇K

Uf - Ui = -0.458mJ

Ui - Uf = 0.458mJ

(-μBcosθi) - (-μBcosθf) = 0.458mJ

rearranging the equation,

(μBcosθf) - (μBcosθi) = 0.458mJ

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μB * (cos 0 - cos θi) = 0.458mJ

but cos 0 = 1

(0.0243 * 0.0575) (1 - cos θi) = 0.458*10⁻³

1 - cos θi = 0.458*10⁻³ / 1.397*10⁻³

1 - cos θi = 0.3278

collect like terms

cosθi = 0.6722

θ = cos⁻ 0.6722

θ = 47.76°

7 0
2 years ago
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Answer:

The answer is B) 3 seconds

Explanation:

I just took the test on 2020 edge and got it right

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