If 4 resistors are wired in series, what is the equivalent resistance?

olasank [31]

Nonononononononononononononononononono

7 0

3 years ago

A certain gasoline engine has an efficiency of 35.3%. What would the hot reservoir temperature be for a Carnot engine having tha

levacccp [35]

Answer:

The temperature of hot reservoir is 669.24 K.

Explanation:

It is given that,

Efficiency of gasoline engine, $\eta=35.3\%=0.353$

Temperature of cold reservoir, $T_C=160^{\circ}C=433\ K$

We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :

$\eta=1-\dfrac{T_C}{T_H}$

$T_H=\dfrac{T_C}{1-\eta}$

$T_H=\dfrac{433}{1-0.353}$

$T_H=669.24\ K$

So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.

4 0

3 years ago

A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio

Angelina_Jolie [31]

a) -0.259 rad/s/y

b) 1732.8 years

c) 0.0069698 s

Explanation:

a)

The angular acceleration of a rotating object is equal to the rate of change of angular velocity of the object.

Mathematically, it is given by

$\alpha=\frac{\Delta \omega}{\Delta t}$

where

$\Delta \omega$ is the change in angular velocity

$\Delta t$ is the time elapsed

The angular velocity can be written as

$\omega=\frac{2\pi}{T}$

where T is the period of rotation of the object.

Therefore, the change in angular velocity can be written as

$\Delta \omega = \frac{2\pi}{T'}-\frac{2\pi}{T}=2\pi (\frac{1}{T'}-\frac{1}{T})$

In this problem:

T = 0.0140 s is the initial period of the pulsar

The period increases at a rate of 8.09 x 10-6 s/y, so after 1 year, the new period is

$T'=T+8.09\cdot 10^{-6} =0.01400809 s$

Therefore, the change in angular velocity after 1 year is

$\Delta \omega =2\pi (\frac{1}{0.01400809}-\frac{1}{0.0140})=-0.259 rad/s$

So, the angular acceleration of the pulsar is

$\alpha = \frac{-0.259 rad/s}{1 y}=-0.259 rad/s/y$

b)

To solve this part, we can use the following equation of motion:

$\omega'=\omega + \alpha t$

where

$\omega'$ is the final angular velocity

$\omega$ is the initial angular velocity

$\alpha$ is the angular acceleration

t is the time

For the pulsar in this problem:

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.0140}=448.8 rad/s$ is the initial angular velocity

$\omega'=0$ , since we want to find the time t after which the pulsar stops rotating

$\alpha = -0.259 rad/s/y$ is the angular acceleration

Therefore solving for t, we find the time after which the pulsar stops rotating:

$t'=-\frac{\omega}{\alpha}=-\frac{448.8}{-0.259}=1732.8 y$

c)

As we said in the previous part of the problem, the rate of change of the period of the pulsar is

$\frac{\Delta T}{\Delta t}=8.09\cdot 10^{-6} s/y$

which means that the period of the pulsar increases by

$\Delta T=8.09\cdot 10^{-6} s$

For every year:

$\Delta t=1 y$

From part A), we also know that the current period of the pulsar is

T = 0.0140 s

The current period is related to the initial period of the supernova by

$T=T_0+\frac{\Delta T}{\Delta t}\Delta t$

where $T_0$ is the original period and

$\Delta t=869 y$

is the time that has passed; solving for T0,

$T_0=T-\frac{\Delta T}{\Delta t}\Delta t=0.0140 - (8.09\cdot 10^{-6})(869)=0.0069698 s$

6 0

3 years ago

A landing craft with mass 1.21×104 kg is in a circular orbit a distance 5.90×105 m above the surface of a planet. The period of

creativ13 [48]

Answer:

W = 661.6 N

Explanation:

The weight of a body is the force of attraction of the plant on the body, so we must use the law of gravitational attraction

F = G m M / r²

Where G is the gravitational attraction constant that values 6.67 10-11 N m² / kg², M is the mass of the planet and r is the distance from the center of the planet.

Let's look for the mass of the planet, for this we write Newton's second law for the landing craft

F = m a

Acceleration is centripetal a = v² / r

G m M / r² = m (v² / r)

The ship rotates rapidly (constant velocity module), let's use uniform kinematic relationships

v = d / t

The distance of a circle is

d = 2π r

v = 2π r / t

We replace

G m M / r² = m (4π² r² / t² r)

G M = 4 π² r³ / t²

M = 4π² r³ / G t²

The measured distance r from the center of the plant is

r = R orbit + R planet

r = 5.90 10⁵ + ½ 9.80 10⁶

r = 5.49 10⁶ m

M = 4 π² (5.49 10⁶)³ / (6.67 10⁻¹¹ (5.900 10³)²)

M = 6,532 10²¹ / 2,321 10⁺³

M = 2.814 10²⁴ kg

With this data we calculate the astronaut's weight

W = (G M / R²) m

W = (6.67 10⁻¹¹ 2,816 10²⁴ /(4.90 10⁶)2) 84.6

W = 7.82 84.6

W = 661.57 N

3 0

3 years ago

In order to analyze a circuit correctly, what values should be known?

Ann [662]

Answer: A

resistances of resistors and voltages of batteries

Explanation:

Electric circuit comprises of resistors or load, batteries or power supplied e.t.c

Current can only flow in a circuit when there are resistors and voltage supply which is a battery.

In order to analyze a circuit correctly, the values of resistances of resistors and voltages of batteries should be known.

3 0

3 years ago