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3 years ago

Write a paragraph containing 8 rnery transformation

1 answer:

4 0

Energy Transformations - Energy Types - This product is a bundle of two of my energy card sort of lab station activities. Included in the bundle: Energy Transformations Card Sort Energy Types Card Sort My students love both

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Match these items.

horrorfan [7]

Iron oxide = small region within a magnet
drop or hammer =man-made magnet
strength decreases rapidly with distance lines of force

4 0

3 years ago

Atoms containing radioactive nuclei are called

miss Akunina [59]

The correct answer is radioisotopes.

3 0

3 years ago

A spool of thread has an average radius of 1.00 cm. If the spool contains 62.8 m of thread, how many turns of thread are on the

Simora [160]

To solve this problem we will start from the given concept in which the number of turns is equivalent to the length of the thread per circumference of spool. That is:

$N = \frac{l}{\phi}$

Where,

l = length of the thread

$\phi$ = circumference of spool

For \phi we have that,

$\phi = 2\pi r \rightarrow 2\pi (0.01)$

For l we have that

l = 62.8m

Finally the number of Turns would be,

$N = \frac{l}{\phi}$

$N = \frac{62.8}{2\pi (0.01)}$

$N = 1000turns$

Therefore the number of turns of thread on the spool are 1000turns.

7 0

3 years ago

Apakah yang berlaku kepada kekuatan medan magnet jika satu lagi sel kering 1.5 v ditambahkan

ad-work [718]

Answer:

The magnetic field is doubled.

Explanation:

What happens to the strength of the magnetic field if one more 1.5 v dry cell is added?

The magnetic field due to a current carrying conductor is directly proportional to the current flowing in the wire.

If we connect one more battery of 1.5 V so the voltage is doubled and according to the Ohm's law, as the resistance of wire is constant, so the current in the wire is also doubled.

When the current doubles, the magnetic field produced by the wire is also doubled.

8 0

3 years ago

The frictionless system shown is released from rest. After the right-hand mass has risen 75 cm, the object of mass 0.50m falls l

Andreas93 [3]

Let $a$ be the acceleration of the masses. By Newton's second law, we have

• for the masses on the left,

$1.3mg - T = 1.3ma$

where $T$ is the magnitude of tension in the pulley cord, and

• for the mass on the right,

$T - mg = ma$

Eliminate $T$ to get

$(1.3mg - T) + (T - mg) = 1.3ma + ma$

$0.3mg = 2.3ma$

$\implies a = \dfrac{0.3}{2.3}g \approx 0.13g \approx 1.3 \dfrac{\rm m}{\mathrm s^2}$

Starting from rest and accelerating uniformly, the right-hand mass moves up 75 cm = 0.75 m and attains an upward velocity $v$ such that

$v^2 = 2a(0.75\,\mathrm m) \\\\ \implies v \approx \sqrt{2\left(1.3\frac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx 1.4\dfrac{\rm m}{\rm s}$

When the 0.5m mass is released, the new net force equations change to

• for the mass on the right,

$mg - T' = ma'$

where $T'$ and $a'$ are still tension and acceleration, but not having the same magnitude as before the mass was removed; and

• for the mass on the left,

$T' - 0.8mg = 0.8ma'$

Eliminate $T'$ .

$(mg - T') + (T' - 0.8mg) = ma' + 0.8ma'$

$0.2mg = 1.8 ma'$

$\implies a' = \dfrac{0.2}{1.8}g = \dfrac19 g \approx 1.1\dfrac{\rm m}{\mathrm s^2}$

Now, the right-hand mass has an initial upward velocity of $v$ , but we're now treating down as the positive direction. As it returns to its starting position, its speed $v'$ at that point is such that

${v'}^2 - v^2 = 2a'(0.75\,\mathrm m) \\\\ \implies v' \approx \sqrt{\left(1.4\dfrac{\rm m}{\rm s}\right)^2 + 2\left(1.1\dfrac{\rm m}{\mathrm s^2}\right)(0.75\,\mathrm m)} \approx \boxed{1.9\dfrac{\rm m}{\rm s}}$

3 0

2 years ago