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antiseptic1488 [7]
3 years ago
6

The effect of itching power what’s the Independent variable and dependent and constants?

Physics
1 answer:
kenny6666 [7]3 years ago
7 0
The independent variable is what type of powder is used, the dependent variable is how long itching lasts and the constant is the amount of powder used (50 grams)
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When a cannon is fired, how does the size of the force of the cannon on the cannonball compared with the force of the cannonball
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The force applied to the cannonball and cannon is equal. The explosion inside the cannon will generate a pressure which will turn into a force on both cannonball and cannon. The cannon being heavier and fixed to the ground will move a bit, but the cannonball will be thrown away, fired.
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Activity 1 more or less, pls answer this ASAP​
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Which one of the following types of electromagnetic wave travels through space the fastest?
Setler [38]

Answer:

As a result, light travels fastest in empty space, and travels slowest in solids. In glass, for example, light travels about 197,000 km/s.

Explanation:

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2 years ago
On the moon, the acceleration of gravity is 1.6 m/s2. If an object has a
Amanda [17]

Answer:

5.1 kg

Explanation:

Its mass on the moon is 5.1 kg because mass is an intrinsic property of a material and does not change with location. Although, its weight might vary because its acceleration of gravity g is dependent on the mass M and radius r of the planet(in this case, moon) involved g = GM/r². Since weight W = mg is dependent o g, weight varies but mass remains constant.

4 0
3 years ago
In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
lara31 [8.8K]

Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

8 0
3 years ago
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