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3 years ago

While eating lunch high up in a skyscraper, two construction workers calculate their gravitational potential

Physics

1 answer:

Maurinko [17]3 years ago

8 0

Answer:

The mass of the other worker is 45 kg

Explanation:

The given parameters are;

The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker

The mass of the lighter construction worker, m₁ = 90 kg

The height level of the lighter construction worker's location = h₁

The height level of the other construction worker's location = h₂ = 2·h₁

The gravitational potential energy, P.E., is given as follows;

P.E. = m·g·h

Where;

m = The mass of the object at height

g = The acceleration due to gravity

h = The height at which is located

Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker

We have;

P.E.₁ = P.E.₂

Therefore;

m₁·g·h₁ = m₂·g·h₂

h₂ = 2·h₁

We have;

m₁·g·h₁ = m₂·g·2·h₁

m₁ = 2·m₂

90 kg = 2 × m₂

m₂ = (90 kg)/2 = 45 kg

The mass of the other construction worker is 45 kg.

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The heat generated by a stove element varies directly as the square of the voltage and inversely as the resistance. If the volta

madam [21]

Answer:

The resistance needs to be increased by a factor of 25 to quadruple the amount of heat generated

Explanation:

Let H be the heat generated by the stove element, V its voltage and R its resistance.

So, H ∝ V²/R

H = kV²/R

HR/V² = k

H₁R₁/V₁² = H₂R₂/V₂²

If H₂ = 4H₁ and V₂ = 10V₁

H₁R₁/V₁² = 4H₁R₂/(10V₁)²

H₁R₁/V₁² = 4H₁R₂/100V₁²

R₁ = R₂/25

R₂ = 25R₁

The resistance needs to be increased by a factor of 25 to quadruple the amount of heat generated.

3 0

3 years ago

A 110 kg tackler moving at 2.5m/s meets head on (and holds on to) an 82 kg halfback moving at 5.0m/s. what will be their mutual

Leona [35]

Answer:

0.7 m/s

Explanation:

We can solve the problem by using conservation of momentum.

Before the collision, the momentum of the first tackler is:

$p_1 = m_1 v_1 =(110 kg)(2.5 m/s)=275 kg m/s$

while the momentum of the second tackler is

$p_2 = m_2 v_2 =(82 kg)(-5.0 m/s)=-410 kg m/s$

Note that we used a negative sign because the direction of the second tackler is opposite to that of the first tackler.

Therefore, the total momentum before the collision is

$p_i = p_1 +p_2 =275 kg m/s -410 kg m/s=-135 kg m/s$

Since the total momentum is conserved, this is also equal to the final total momentum :

$p_f = -135 kg m/s$

Which is also equal to

$p_f = (m_1 +m_2 )v_f$

since the two tacklers continue their motion together with final velocity vf. Re-arranging the previous equation, we can find the the new velocity of the two tacklers:

$v_f = \frac{p_f}{m_1 +m_2}=\frac{-135 kg m/s}{110 kg + 82 kg}=-0.7 m/s$