Bradley drops a rock in a well. It falls for 12 seconds. How deep is the well?

Capacitances of 10uF and 20uF are connected in parallel,

jarptica [38.1K]

Answer:

The equivalent capacitance will be $15\mu F$

Explanation:

We have given two capacitance $C_=10\mu F\ and\ C_2=20\mu F$

They are connected in parallel

So equivalent capacitance $C=C_1+C_2=10+20=30\mu F$

This equivalent capacitance is now connected in series with $30\mu F$

In series combination of capacitors the equivalent capacitance is given by $\frac{1}{C}=\frac{1}{30}+\frac{1}{30}$

$C=\frac{30}{2}=15\mu F$

So the equivalent capacitance will be $15\mu F$

5 0

3 years ago

What is an atomic orbital

Serjik [45]

The probability of finding an electron with random motion at a certain energy level (distance away from) the nucleus

8 0

3 years ago

Read 2 more answers

An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os

aniked [119]

Answer:

(a) T = 0.412s

(b) f = 2.42Hz

(c) w = 15.25 rad/s

(d) k = 86.75N/m

(e) vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

$f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz$

The frequency is 2.42 Hz

(c) The angular frequency is:

$\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}$

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

$\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}$

The spring constant is 86.75N/m

(e) The maximum speed is:

$v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}$

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

$F=kA=(86.75N/m)(0.2m)=17.35N$

The maximum force that the spring exerts on the block is 17.35N

3 0

3 years ago

(a) the gamma rays produced by a radioactive nuclide used in medical imaging (b) radiation from an FM radio station at 93.1 MHz

lawyer [7]

Answer:

They can be rank in the following way:

A radio signal from an AM radio station at 680 kHz on the dial

Radiation from an FM radio station at 93.1 MHz on the dial

The red light of a light-emitting diode, such as in a calculator

The yellow light from sodium vapor streetlights

The gamma rays produced by a radioactive nuclide used in medical

Explanation:

The electromagnetic spectrum is the distribution of radiation due to the different frequencies at which it radiates and its different intensities, that radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it.

Radiation is distributed along that electromagnetic spectrum according to the wavelength or frequency.

Highest frequencies

X-rays

Ultraviolet rays

Visible region

Lower frequencies

Infrared

Microwave

Radio waves

Radio waves and the visible region (yellow light, red light) are part of the electromagnetic spectrum, any radiation of that electromagnetic spectrum has a speed of 3.00x10^{8}m/s in vacuum.

However, the following equation relates the velocity, the frequency, and the wavelength:

$c = \nu \cdot \lambda$ (1)

$\nu = \frac{c}{\lambda}$ (2)

It can be see in equation 2 that the frequency and the wavelength are inversely proportional (when the frequency increases the wavelength decreases).

Therefore, for what was already discussed, they can be rank in the next way:

A radio signal from an AM radio station at 680 kHz on the dial

Radiation from an FM radio station at 93.1 MHz on the dial

The red light of a light-emitting diode, such as in a calculator

The yellow light from sodium vapor streetlights

The gamma rays produced by a radioactive nuclide used in medical

Summary:

In the case of the radio waves can be used:

Case for $\nu = 93.1 MHz$ :

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3x10^{8}m/s}{93100000s^{-1}}$

$\lambda = 3.22m$

Case for $\nu = 680 kHz$ :

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3x10^{8}m/s}{680000s^{-1}}$

$\lambda = 441.17m$

7 0

3 years ago

A 2.0-kg cart collides with a 1.0-kg cart that is initially at rest on a low-friction track. After the collision, the 1.0-kg car

nikitadnepr [17]

To solve this problem we will apply the concepts related to the conservation of momentum. The momentum can be defined as the product between the mass of the object and its velocity, and the conservation of the momentum as the equality between the change of the initial momentum versus the final momentum. Mathematically, this relationship can be described as

$m_1u_1+m_2u_2 = m_1v_2+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

According to the statement one of the bodies does not have initial velocity, therefore said term would be zero. And the equation could be rewritten as,

$m_1u_1= m_1v_2+m_2v_2$

Replacing the values respectively (The mass of your body with its respective speed we would have)

$2kg(u_1) = 2kg(0.3m/s)+1kg(0.5m/s)$

$u_1 = \frac{2kg(0.3m/s)+1kg(0.5m/s)}{2kg}$

$u_1 = 0.55m/s$

Therefore the initial velocity of the 2kg cart is 0.55m/s

4 0

3 years ago