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2 years ago

The electric field 30cm from a van de Graaff generator is measured to be 28,300N/C. What is the charge of the van de Graaf?

Physics

1 answer:

gizmo_the_mogwai [7]2 years ago

8 0

Answer:

Explanation:

EWAN KO LANG DIN BASTA YAN ALAM KO

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A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

2 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

2 years ago

An airplane accelerates from rest to its required takeoff speed with an acceleration 5.0 m/s^2 after travelling on the runway a

Alex Ar [27]

Answer:

Explanation:

.75 km = 750 m

acceleration a = 5 m/s²

distance s = 750 m

initial velocity u = 0

Final velocity v = ?

v² = u² + 2as

v² = 0 + 2 x 5 x 750

v = 86.60 m /s.

5 0

3 years ago

While vacationing in the mountains you do some hiking. In the morning, your displacement is S⃗ morning= (2200 m , east) + (4000

Minchanka [31]

Answer:

The hiker (you ) is 200 m below his/her(your) starting point

The resultant displacement in the north east direction is

$a = 6562.0 \ m$

The resultant displacement in vertical direction (i.e the altitude change )

$b =6503.1 \ m$

Explanation:

From the question we are told that

The displacement in the morning is $S_{morning} = (2200 \m , east) + (4000\ m\ north) + (100 \ m ,\ vertical)$

The displacement in the afternoon is $S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)$

Generally the direction west is negative , the direction east is positive

the direction south is negative , the direction north is positive

resultant displacement is mathematically evaluated as

$(2200 \m , east) +( - 1300\ m ,\ west) = 900 \ m \ east$

$(4000\ m\ north) + (2500 \ m ,\ north) = 6500 \ m ,\ north$

$(100 \ m ,\ vertical) - (300\ m ,\ vertical) = -200 \ m$

From the above calculation we see that at the end of the hiking the hiker (you) is 200 m below his/her(your) initial position

Generally from Pythagoras theorem , the resultant displacement in the north east direction is

$a = \sqrt{900^2 + 6500^2}$

=> $a = 6562.0 \ m$

Generally from Pythagoras theorem , the resultant displacement in vertical direction (i.e the altitude change )

$b = \sqrt{6500^2 +(-200)^2 }$

=> $b =6503.1 \ m$

5 0

3 years ago

What is the difference between obtaining knowledge from a controlled experiment versus other forms of scientific investigation?

solong [7]

Answer:

Any scientific investigation in its simplest form involves collecting information, while a controlled experiment is a tightly controlled investigation that tests a hypothesis.

Explanation:

All the scientific investigations start with a particular question. The question decides what type of investigation should research take. A controlled experiment is an investigation which allows a particular variable to change and all other variable controlled to remain constant to see the effect on dependent variable to test a particular hypothesis while other types of investigations are used to increase the knowledge or information about the particular research subject.

6 0

2 years ago