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3 years ago

I'm not sure what to do on this work sheet

Physics

2 answers:

Drupady [299]3 years ago

7 0

#1 is 12 u just have to count that one was counting by 5's figure out the pattern on what its counting by and start at the bottom and count your way up till you get to the shaded line

aleksandrvk [35]3 years ago

3 0

You're supposed to write the number on where the line stops. So for example, thr first one goes by 5, 10, 15, 20 and in between there's 4 lines so the lines between 10 and 15 would be 10, 11, 12, 13, 14, 15 and the line stops at the 2nd line so the number would be 12. So the answer is 12 mi

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(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it

faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

$m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

Now, plug in the given values and solve for 'v'. This gives,

$v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s$

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0

3 years ago

Two uncharged metal spheres, spaced 25.0 cm apart, have a capacitance of 26.0 pF. How much work would it take to move 12.0 nC of

viktelen [127]

Answer:

$W=2.76\times 10^{-6}\ J$

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

$U=\dfrac{Q^2}{2C}$

Put all the values,

$U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J$

So, the work done is $2.76\times 10^{-6}\ J$ .

8 0

3 years ago

An igneous rock becomes buried, is subject to high heat and pressure, and recrystallizes. This rock then is eroded, transported,

nevsk [136]

Answer:

Answered

Explanation:

When an igneous rock becomes buried, is subjected to high heat and pressure, and recrystallizes it is formed into Metamorphic rock. Now this rock is eroded, transported, deposited and subsequently lithified to be converted into Sedimentary rock.

The same igneous rock is first converted into Metamorphic and then into sedimentary by the process of weathering.

5 0

3 years ago

A wagon full of manure accidentally rolls down a driveway for 5.0m while a person pushes against the wagon with a force of 420 N

Cerrena [4.2K]

Answer:

2100 J

Explanation:

Parameters given:

Force acting on the object, F = 420 N

Distance moved by object, d = 5m

The change in kinetic energy of an object is equal to the work done by a force acting on the object:

W = F * d

∆KE = F * d

∆KE = 420 * 5

∆KE = 2100 J

8 0

3 years ago

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alexandr402 [8]

Answer:

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4 0

3 years ago