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vagabundo [1.1K]

4 years ago

6

Which change decreases the resistance ofva piece of copper wire?

1 answer:

zepelin [54]4 years ago

8 0

Wires would catch fire so i i got that answer bu subtracting 8 then but the fire in there

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3241004551 [841]

Answer:

1.so that we won't do any mistakes and won't have injuries.

Explanation:

yes.

because it is a right thing to use correct tools for correct work.

4 0

3 years ago

A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to

Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass $m_1$ just before the collision

$v_1 = \sqrt{2gh}$

$v_1 = \sqrt{2(9.8)(3)}$

$v_1 = 7.67m/s$

Therefore the momentum just before collision would be

$p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s$

Momentum after the collision

$p_1 = 20*u_1+40u_1\\p_1 = 60u_1$

Since the momentum is conserved we have that

$153.36= 60u_1$

$u_1 = \frac{153.36}{60}$

$u_1 = 2.56m/s$

The velocity of mass $m_2$ after the collision is given by

$v_2 = \frac{2m_1}{m_1+m_2} u_1$

$v_2 = \frac{2(20)}{20+40}(2.56)$

$v_2 = 1.71m/s$

Therefore the change in momentum of mass 2 is

$p_2 = m_2v_2$

$p_2 = 40*1.71$

$p_2 = 68.4kg\cdot m/s$

Therefore the impulse acting on m2 during the collision between the two boxes is $p = 68.4kg\cdot m/s$

8 0

4 years ago

Why is external fertilization not possible in organisms that live on land

Softa [21]

because the sperm and the egg will dry out

4 0

3 years ago

Need help with this question

lakkis [162]

Answer:

1.39m

Explanation:

5 0

2 years ago

Read 2 more answers

A charge of −8.3 μC is traveling at a speed of 7.4 × 106 m/s in a region of space where there is a magnetic field. The angle bet

solniwko [45]

Answer:

The magnitude of the magnetic field is $1.1\times 10^{-4}\, T$

Explanation:

Given charge q = −8.3 μC

speed of charge $v=7.4\times 10^{6}\, \frac{m}{s}$

Angle between magnetic field and the velocity of charge $\Theta =52^{\circ}$

Strength of magnetic force on charge $F=5.4\times 10^{6}\, N$

Let magnitude of magnetic field be B

Since the magnetic force on a free moving charge in the magnetic field given by

$F=\left |q\vec{v}\times \vec{B}\right |=\left | qvBsin\Theta\right |$

=> $5.4\times 10^{-3}=\left | -8.3\times 10^{-6}\times 7.4\times 10^{6}\times sin(52^{\circ})\times B \right |$

=> $B=1.1\times 10^{-4}\, T$

Thus the magnitude of the magnetic field is $1.1\times 10^{-4}\, T$

4 0

3 years ago