2 years ago

Impulse equals? A) momentum x velocity B) momentum x time C) mass x velocity

Physics

2 answers:

ladessa [460]2 years ago

8 0

Mass x velocity is the answer i think

mina [271]2 years ago

7 0

Answer:

Explanation:

The impulse experienced by an object is the force•time.

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This is a compound made from a group of covalently bonded atoms.

ra1l [238]

The answer is molecule

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3 years ago

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A long wire carries a current density proportional to the distance from its center, J=(Jo/ro)•r, where Jo and ro are constants a

IgorC [24]

Answer:

$B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)$

Explanation:

As the current density is given as

$J = \frac{J_0}{r_0}r$

now we have current inside wire given as

$i = \int J(2\pi r)dr$

$i = \int \frac{J_0}{r_0} r(2\pi r)dr$

$i = 2\pi \frac{J_0}{r_0} \int r^2 dr$

$i = \frac{2}{3} \pi \frac{J_0}{r_0} r^3$

Now by Ampere's law we will have

$\int B. dl = \mu_0 i$

$B. (2\pi r) = \mu_0(\frac{2}{3} \pi \frac{J_0}{r_0} r^3)$

$B = \mu_0(\frac{1}{3} \frac{J_0}{r_0} r^2)$

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2 years ago

An arrow of mass 415 g is shot at a target with a speed of 68.5m/s. The target, which has a mass of 3.3 kg, is moving toward the

Murrr4er [49]

The velocity of the target and arrow after collision is 6.67m/s

<u>Explanation:</u>

Given:

Mass of arrow, mₐ = 415g

Speed of arrow, vₐ = 68.5m/s

Mass of the target, mₓ = 3.3kg = 3300g

speed of the target, vₓ = -1.1m/s (Because the target moves in opposite direction

Velocity of the target and arrow after collision, vₙ = ?

Applying the conservation of momentum,

mₐvₐ + mₓvₓ = (mₐ+mₓ) vₙ

415 X 68.5 + 3300 X -1.1 = (415+3300) X vₙ

28427.5 - 3630 = 3715 X vₙ

24797.5 = 3715 X vₙ

vₙ = 6.67m/s

Therefore, the velocity of the target and arrow after collision is 6.67m/s

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3 years ago

Suppose a square wave signal has a 65 percent duty cycle and an on-state voltage of 40 volts DC. What is the average DC voltage

Ierofanga [76]

Answer:

The voltage is $\= DC _v = 2.6 \ V$

Explanation:

From the question we are told that

The duty cycle is p = 65% = 0.65

The on - state voltage is $V = 40 \ volt$

Generally the average DC voltage is mathematically represented as

$\= DC _v = p * V$

=> $\= DC _v = 40 * 0.65$

=> $\= DC _v = 2.6 \ V$

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