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Calculate the Work done if the force is 2000 Newtons<br> and the distance is 5 km.

kodGreya [7K]

Answer:

10 million joules or 10,000 KJ

Explanation:

Work= Force x Displacement

convert 5km into meters -5km=5000m

W= 2000N x 5000m

w=10,000,000 Joules

or 10,000KJ

4 0

2 years ago

Identical twins, Angela and Aisha, were both born with cystic fibrosis, a genetic disorder. As they grew older, Aisha was able t

ohaa [14]

Answer:

Their differences are most likely due to non shared environmental influences

4 0

3 years ago

A typical human lens has an index of refraction of 1.430 . The lens has a double convex shape, but its curvature can be varied b

rjkz [21]

Answer:

Maximum Power = 144.3 D

The associated focal length of the lens = $6.92*10^{-3} m$

Explanation:

According to the Lens maker's Formula:

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )$

where;

$n_1$ = the refractive index of the medium

$R_1$ and $R_2$ = radius of curvature on each surface

For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )$

At maximum power

$\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )$

= $0.144 \ mm^{-1}$

This Implies

$f = 6.92 mm\\f = 6.92*10^{-3} \ m$

Therefore; the power is given by the formula:

$P_{max} = \frac{1}{f}$

$P_{max}= \frac{1}{6.92*10^{-3}}$

= 144.3 D

3 0

4 years ago

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

$m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)$

where:

$m_g = 3.0 kg$ is the mass of the gold bar

$C_g=129 J/kg C$ is the specific heat of gold

$T_g=99 C$ is the initial temperature of the gold bar

$m_w = 0.22 kg$ is the mass of the water

$C_w=4186 J/kg C$ is the specific heat of water

$T_w=25 C$ is the initial temperature of the water

$T_f$ is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

$m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C$

5 0

3 years ago

Two light bulbs, A and B, are connected to a 120-V outlet (a constant voltage source). Light bulb A is rated at 60 W and light b

d1i1m1o1n [39]

Answer:

Bulb A has a greater resistance.

Explanation:

Electric power (P) = V²/R

P = V²/R................ Equation 1

Where P = power, V = Voltage, R = Resistance.

Make R the subject of the equation

R = V²/P ................ Equation 2

For Bulb A,

Given: V = 120 V, P = 60 W.

Substitute into equation 2

R = 120²/60

R = 240 Ω

For bulb B

Given: V =120 V, P = 100 W.

Substitute into equation 2

R = 120²/100

R = 14400/100

R = 144 Ω

Hence Bulb A has a greater resistance.

5 0

3 years ago