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3 years ago

Which of the following is not a necessary condition for seeing a magnified image with a

1 answer:

5 0

The option stating that the observer must be placed within the focal length is false or not a necessary condition for magnification.

Explanation:

Lens are used for magnification provided the nature of the lens is known and by positioning the object at correct position. So the object should be placed in the focal point of the lens for perfect magnification, it can also be kept within the focal point or focal length. But the object should not be kept beyond the focal length of the lens.

So among the given options, the object and image should be on the same side of the lens for magnification is true. Then the nature of lens should be converging for magnification is also true. And the most important point that the object should be placed within the focal length of the lens is the most mandatory for obtaining magnification. So the option stating that the observer must be placed within the focal length is false or not a necessary condition for magnification.

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Why does a 10 gram piece if copper take up more space than a 10 gram piece of gold

algol [13]

Answer:

because the mass of the copper is higher than the mass of the gold.

Explanation:

7 0

3 years ago

Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200

andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

ΔK.E = ΔP.E

K_2 - K_1 = P_1- P_2

Where, K_2: Kinetic energy of hammer head as it hits the I-beam

K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

P_1: Initial gravitational potential energy of hammer head

- The expression simplifies to:

K_2 = P_1

Where, 0.5*m*v2^2 = m*g*s12

v2 = √(2*g*s12) = √(2*9.81*3)

v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

Wnet = ( P_3 - P_1 ) + W_friction

Wnet = m*g*s13 + F*s23

n*s23 = m*g*s13 + F*s23

Where, n: average force the hammerhead exerts on the I-beam.

s13 = s12 + s23

Hence,

n = m*g*( s12/s23 + 1) + F

n = 200*9.81*(3/0.074 + 1) + 60

n = 81562 N

6 0

3 years ago

Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p

PSYCHO15rus [73]

Answer:

Energy density will be 14.73 $J/m^3$

Explanation:

We have given capacitance $C=225\mu F=225\times 10^{-6}F$

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm $0.2\times 10^{-3}m$

We know that there is relation between electric field and potential

$E=\frac{V}{d}$ , here E is electric field, V is potential and d is separation between the plates

So $E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C$

Energy density is given by $E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3$

5 0

3 years ago

You calculate the density of a block of aluminum to be 2.68 g/cm3. You look up the density of a block of aluminum at room temper

Hunter-Best [27]

Answer:

Systematic errors.

Explanation:

The density of the aluminium was calculated by a human and this is not natural but can be due to errors in the calibration of the scale for measuring the weight or taking readings from the measuring cylinder.

Random errors are natural errors. Random errors in experimental measurements are caused by unknown and unpredictable changes in the experiment. Systematic errors are due to imprecision or problems with instruments.

3 0

3 years ago

What do electric and magnetic fields have in common?

Nastasia [14]

Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

Magnetic field are a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic fields are produced by moving electric charges. Everything is made up of atoms, and each atom has a nucleus made of neutrons and protons with electrons that orbit around the nucleus. Since the orbiting electrons are tiny moving charges, a small magnetic field is created around each atom.

Similarities between magnetic fields and electric fields: Magnetic fields are associated with two magnetic poles, north and south, although they are also produced by charges (but moving charges). Like pole repel unlike poles attract. Electric field points in the direction of the force experienced by a positive charge.

7 0

3 years ago