All categories

2 years ago

What are some sources of error in a refraction lab using pins and a glass block?

Physics

1 answer:

ser-zykov [4K]2 years ago

6 0

DONT WORRY

ANSWER KO YAN NG MADALI

PERO BRAINLIEST MUNA

MUAAAAAH

TENCHUUU

You might be interested in

What is the definition of specific heat capacity

mamaluj [8]

Answer:

heat energy that is needed to raise tempeture

3 0

3 years ago

An object 1,000 km above earths surface would accelerate

Verizon [17]

The acceleration which is gained by an object because of the gravitational force is called its acceleration due to gravity. Its SI unit is m/s2. Acceleration due to gravity is a vector, which means it has both a magnitude and a direction. The formula is ‘the change in velocity= gravity x time’ The acceleration due to gravity at the surface of Earth is represented as g. It has a standard value defined as 9.80665 m/s2.[1]

5 0

3 years ago

An astronaut on a space walk floats a little too far away from the space station. Without air to push against, he cannot paddle

Artyom0805 [142]

He can throw the hammer in the direction opposite to the direction he wants to travel in. The hammer will exert an equal and opposite force on him, as per Newton's third law, and this will help him move towards the space station.

5 0

3 years ago

Read 2 more answers

A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

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

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

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

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

Why are some substances dissolve both polar and nonpolar substances?

dlinn [17]

Answer:

Because water is polar and oil is nonpolar, their molecules are not attracted to each other. polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.

please mark me as the brainliest :)

use your brain :)

8 0

3 years ago