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3 years ago

If 2.0 j of work Is done raising a 180 g Apple how far was it lifted

Physics

1 answer:

andrew11 [14]3 years ago

6 0

The apple is lifted by 1.1 m

<u>Explanation</u>:

Given that, mgh = 2 J and mass of apple = 180g
Step 1: convert grams to kg as work is given in SI units.

180 g = 0.18 kg

Step 2: substitute in the formula for work done.

0.18 * 10 * h = 2

Step 3: calculate the value of h from this formula.

h = 2/1.8

h = 1.11 m

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ou have designed and constructed a solenoid to produce a magnetic field equal in magnitude to that of the Earth (5.0 10-5 T). If

Nikitich [7]

Answer:

$I = 0.0256 A$

Explanation:

Given,

Magnetic field, B = 5 x 10⁻⁵ T

Number of turns, N = 450

length = 29 cm

Current,I = ?

Using formula of magnetic field due to solenoid

$B = \mu_0 NI$

$I = \dfrac{B}{\mu_0N}$

$I = \dfrac{5\times 10^{-5}\times 0.29}{450\times 4\pi \times 10^{-7}}$

$I = 0.0256 A$

Hence, the current obtained is equal to 0.0256 A.

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3 years ago

With what maximum precision can its position be ascertained? [hint: ?p=m?v.]

Nana76 [90]

The formula for the momentum is
p = mv
As a consequence of the conservation of energy, there is also the law of conservation of momentum.
So,
Δp = Δmv = mΔv
The maximum precision that the position can be ascertained is less than 100% because of dissipation.

5 0

3 years ago

Will give brainiest to best answer

Gala2k [10]

Answer:

Explanation:

6 0

2 years ago

Convert 23 miles into feet.<br> **There are 5280 feet in 1 mile** *

zhenek [66]

Answer:

121440 feet is 23 miles.

5280 x 23= 121440

To check your answer just divide 121440 by 5280, which gives you 23.

Hopefully this helps you!

8 0

3 years ago

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

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2 years ago