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3 years ago

In a lever, the effort arm is two times as long as the load arm. The resultant force will be

1 answer:

Taya2010 [7]3 years ago

5 0

So we want to know what the resultant force would be if on the lever the effort arm is as twice as long as the load arm. So in order for the lever to be in a state of balance it must be: F1*r1=F2*r2 where F1 is the effort arm and F2 is the load arm, r1 is the perpendicular distance from the force F1 to the pivot and r2 is the perpendicular distance from the force F2 to the pivot. For both sides of the equation to be equal, since r1=2*r2, F1*2*r2=F2*r2. Now we can divide by r2 and we get: 2F1=F2. When we divide by 2: F1=(1/2)*F2. So the force of the hand that is trying to lift the load is twice as small as the force of the load we are trying to lift.

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Which of the following is made up of vibrating electric and magnetic fields

hichkok12 [17]

A i think
hope it helps!!:)))

4 0

3 years ago

Read 2 more answers

A 70.0 kg ice hockey goalie, originally at rest, has a 0.110 kg hockey puck slapped at him at a velocity of 31.5 m/s. Suppose th

NISA [10]

Answer

given,

mass of the goalie(m₁) = 70 kg

mass of the puck (m₂)= 0.11 kg

velocity of the puck = 31.5 m/s

elastic collision

$v_1=\dfrac{m_2-m_1}{m_1+m_2}v_1+\dfrac{2m_2}{m_1+m_2}v_2$

$v_{pf}=\dfrac{0.11-70}{0.11+70}31.5+\dfrac{2m_2}{m_1+m_2}\times (0)$

$v_{pf}=-31.4\ m/s$

$v'_2 = \dfrac{2m_1v_1}{m_1+m_2}-\dfrac{(m_2-m_1)v_2}{m_2+m_1}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}-\dfrac{(0.11-70)\times 0}{m_1+m_2}$

$v_{gf} = \dfrac{2\times 0.11\times 31.5}{0.11+70}$

$v_{gf} = 0.0988\ m/s$

4 0

3 years ago

In a two-body collision, if the momentum of the system is conserved, then which of the

lakkis [162]

Answer:

c) may also be conserved

Explanation:

Momentum is conserved in both elastic and inelastic type of collisions.

But the differences is that:

In an ELASTIC type of collisions, KINETIC ENERGY IS ALSO CONSERVED.

whereas, In an INELASTIC type of collision, KINETIC ENERGY IS NOT CONSERVED.

So unless until type of collision is specified, we can not say anything about the conservation of kinetic energy after collision.

Hence, may also be conserved is the appropriate option here.

5 0

2 years ago

Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.

jasenka [17]

(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

<h3>Force per unit length</h3>

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d - I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L = (μI/2π) x (I/2d + I/d)

F₃/L = (μI/2π) x (3I/2d)

F₃/L = 3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

Learn more about magnetic force here: brainly.com/question/13277365

#SPJ1

6 0

2 years ago

As seen in the figure, a bullet with mass of 15.0-g is fired vertically and penetrates a block with mass of 2.5-kg and the block

rodikova [14]

Answer:

KE = 2.03 J

Explanation:

After impact, the kinetic energy of the bullet+block will convert to potential energy

½mv² = mgh

v = √(2gh) = √(2(9.81)(0.00500) = 0.0981 m/s

conservation of momentum during the collision

0.015u + 2.50(0) = (2.50 + 0.015)(0.0981)

u = 16.4481 m/s

KE = ½mv² = ½(0.015)16.4481² = 2.0290499...

KE = 2.03 J

4 0

3 years ago