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3 years ago

The image of an object placed 30cm from a diverging lens is formed 10cm in front of the lens.

Physics

1 answer:

luda_lava [24]3 years ago

6 0

Answer:

15cm

Explanation:

Since the lens is a diverging lens, the image distance is negative (virtual)

v = -30cm

u = 10cm

Required

focal length f

Using the lens formula;

1/u + 1/v = 1/u

1/10 - 1/30 = 1/f

(3-1)/30 = 1/f

2/30 = 1/f

f = 30/2

f = 15cm

Hence the focal length of the lens is 15cm

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All contraceptives prevent STIs true or false

NNADVOKAT [17]

False. Not all contraceptives prevent STIs.

Contraceptives are items used or taken to prevent conception. These are birth control pill, patch, ring, IUD, birth control shots, and condoms. Among these contraceptives, condoms are the only type of birth control that prevents STIs or Sexually Transmitted Infections.

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4 years ago

A horizontal force acts on an object on a fric- tionless horizontal surface. If the force is halved and the mass of the object i

guapka [62]

(a) If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b) if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

The force on an object is determined by applying Newton's second law of motion;

F = ma

$\frac{F_1}{m_1a_1} = \frac{F_2}{m_2a_2}$

(a)

when the force is halved, F₂ = 0.5F₁,

mass is doubled, m₂ = 2m₁

$\frac{F_1}{m_1a_1} = \frac{0.5F_1}{2m_1a_2} \\\\2m_1a_2F_1 = 0.5F_1 m_1a_1\\\\2a_2 = 0.5a_1\\\\a_2 = \frac{0.5a_1}{2} = \frac{a_1}{2 \times 2} = \frac{a_1}{4} \\\\a_2 = \frac{1}{4} (a_1)$

Thus, If the force is halved and the mass of the object is doubled, the acceleration will be one-fourth as great.

(b)

when the force is doubled, F₂ = 2F₁,

mass is halved, m₂ = 0.5m₁

$\frac{F_1}{m_1 a_1} = \frac{2F_1}{0.5m_1 a_2} \\\\0.5m_1a_2 F_1 = 2F_1m_1a_1\\\\0.5a_2 = 2a_1\\\\a_2 = \frac{2a_1}{0.5} \\\\a_2 = 4(a_1)$

Thus, if the force on it is doubled and its mass is halved, the new acceleration will be four times as great.

Learn more here:brainly.com/question/19887955

3 0

3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface: