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3 years ago

12

Which best compares a nondigital and a digital camera?

2 answers:

Fed [463]3 years ago

5 0

<h2>Answer:</h2>

<u>The correct statement is </u><u>A nondigital camera uses film to record images, and a digital camera uses sensors.</u>

<h2>Explanation:</h2>

A non digital camera uses a film and it works on the basis of light which is either produced by the flash light or sunlight. On the other hand a digital camera has its own sensors which has the ability to provide an option of low light compensation and it also allows the user to adjust the light controls digitally with the help of existing sensors.

masya89 [10]3 years ago

3 0

D. Digital cameras use sensors which detect the brightness and color of the light, while non-digital cameras project the light onto film, which develops over time.

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A train travels 98 kilometers in 4 hours, and then 61 kilometers in 3 hours. What is its average speed?

IrinaK [193]

Speed (velocity)=distance/time

V1=98km/4hr=24.5km/hr
V2=61km/3hr=20.3km/hr

Average speed (velocity)=total velocity/ number

Average speed (velocity)=44.8km/hr/2=22.4

So the average speed is 22.4km/hr

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3 years ago

You are given a semiconductor resistor made from silicon with an impurity concentration of resistivity 1.00×10−3ωm. the resistor

Vlad [161]

Answer: Resistance = $3.2 \Omega$ , Current = 1.56 A, Voltage =4.99 V

The resistance,

$R=\frac {\rho l}{A}$

where, $\rho$ is resistivity, A is the area and l is the length of the resistor.

It is given that:

$\rho=1.0\times10^{-3}\Omega m$

Length, l=2 mm

Area, $A= width \times height=1.25 mm\times 0.5 mm=0.625 mm^2$

Hence, $R=\frac{1.0\times10^{-3}\Omega m \times 2\times10^{-3}m}{0.625\times10^{-6}m^2}=3.2\Omega$

We know, Power, $P=I^2R$

$\Rightarrow I=\sqrt{\frac{P}{R}}$

$P=7.81 W$

$I=\sqrt{\frac {7.81 W}{3.2\Omega}}=\sqrt{2.44}A=1.56A$

We know, Voltage, $V=IR=1.56\times3.2=4.99 V$

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2 years ago

How do I solve this question the answer I should get is C

prisoha [69]

Your right. It's C. This question is difficult

7 0

3 years ago

Read 2 more answers

Write the formulas of three important acids and three important bases. Describe their uses.

algol [13]

Explanation:

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2 years ago

Two charged objects are 1 meter apart. Calculate the magnitude of the electric force between them if the two charges are +1.0 μC

Solnce55 [7]

Answer:

0.00899 N

Explanation:

The magnitude of the electrostatic force between two charges is given by the equation:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the charges

r is the distance between the two charges

And the force is:

- Repulsive if the two charges have same sign

- Attractive if the two charges have opposite sign

In this problem we have:

$q_1=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 1)

$q_2=1.0\mu C = 1.0\cdot 10^{6}C$ (charge of object 2)

r = 1 m (separation between the objects)

So, the electric force is

$F=(8.99\cdot 10^9)\frac{(1.0\cdot 10^{-6})^2}{1^2}=0.00899 N$

5 0

3 years ago