All categories

3 years ago

Am soo confused what is conclusions in physics

Physics

1 answer:

ANTONII [103]3 years ago

7 0

Conclusion summarizes the results that either support or contradict your original hypothesis.

You might be interested in

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, $V=IR=1.56\times3.2=4.99 V$

4 0

3 years ago

I have a voltage source of 12V but a light that only burns at 5V. The lamp works on 18 mA. Calculate the resistance that you EXT

Gala2k [10]

Answer:

The resistance that will provide this potential drop is 388.89 ohms.

Explanation:

Given;

Voltage source, E = 12 V

Voltage rating of the lamp, V = 5 V

Current through the lamp, I = 18 mA

Extra voltage or potential drop = 12 V - 5 V = 7 V

The resistance that will provide this potential drop (7 V) is calculated as follows:

$R = \frac{V}{I} = \frac{7 \ V}{18 \times 10^{-3} A} \ = 388.89 \ ohms$

Therefore, the resistance that will provide this potential drop is 388.89 ohms.

5 0

2 years ago

At what point does the comet experience the strongest force of gravity?

spin [16.1K]

I think the answer may be the letter B

5 0

2 years ago

Where would you find convergent and divergent plate boundaries relative to

Sergio [31]

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

8 0

2 years ago

A 0.15 g honeybee acquires a charge of 22 pC while flying. The electric field near the surface of the earth is typically 100 N/C

Rus_ich [418]

Answer:

$1.50\ *10^{-6} }$

Explanation:

Given

e=100 N/C

M=0.15 g

$q=\ 22\ pC\\=\ 22\ *10^{-2}$

The ratio of the electric force on the bee to the bee's weight can be determined by the following formula

$\frac{fe}{M*9.81}$

$\frac{22*10^{-12\ *\ 100} }{0.15*\ 10^{-3} *\ 9.81}$

$=\ 1.50\ *10^{-6}$

4 0

3 years ago