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3 years ago

12

1. when estimating the area covered by an object, what type of errors might you make and what are the sources that might have ca

used it.

1 answer:

defon3 years ago

6 0

Answer:

Rounding errors, measuring errors

Explanation:

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A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

goblinko [34]

Oh my lord lol I was do ready to help then I saw numbers

4 0

2 years ago

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago

I need help it is due today

siniylev [52]

Answer:

Option 3. The tennis ball began from rest and rolls at a rate of 14.7 m/s safer 1.5 seconds.

Explanation:

To know the the correct answer to the question, it is important that we know the definition of acceleration.

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as:

a = (v – u) /t

Where

a => acceleration

v => final velocity

u => Initial velocity

t => time

With the above information in mind, let us consider the options given in the question above to know which conform to the difinition of acceleration.

For Option 1,

We were told that the tennis ball has the following:

Distance = 4 m

Time = 1.5 s

This talks about the speed and not the acceleration.

Speed = distance / time

For Option 2,

We were only told about the average speed and nothing else.

For Option 3,

We were told that the tennis ball have the following:

Initial velocity (u) = 0 m/s

Final velocity (v) = 14.7 m/s

Time = 1.5 s

This talks about the acceleration.

a = (v – u) /t

For Option 4,

We were only told that the tennis rolls to the right at an average speed. This talks about the average velocity. We need more information like time to justify the acceleration.

From the above illustrations, option 3 gives the correct answer to the question.

8 0

2 years ago

A 500 g ball swings in a vertical circle at the end of a 1.4-m-long string. when the ball is at the bottom of the circle, the te

sergij07 [2.7K]

A 500 g ball swings in a vertical circle at the end of a 1.4-m-long string. when the ball is at the bottom of the circle, the tension in the string is 18 n.

6 0

3 years ago

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air

fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

$s=u_y t+\frac{1}{2}at^2$

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

$u_y=0$ is the initial vertical velocity

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

$v_x = 10 m/s$

Therefore, the horizontal distance covered after a time t is

$d=v_x t$

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

$d=(10)(2.02)=20 m$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0

2 years ago