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2 years ago

Circle the letter that corresponds to the best answer.

Physics

1 answer:

IrinaK [193]2 years ago

6 0

Answer:

C) China, or "Our largest trade deficit is with China"

Explanation:

China is currently our largest goods trading partner. 85% of goods are imported from China to America. The most prominent goods among the finished products exported from China were consumer electronics, data processing technologies, clothing, other textiles, optical gear, and medical equipment.

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Standing at the same location, which organism would have the GREATEST gravitational attraction to the earth?

Nastasia [14]

To claculate the gravitational attraction between two bodies with mass 1(m1) and mass 2 (m2) you need to use the equation:

F= G ((m1*m2)/r^2)

Where;
G is the gravitational constant (6.67E-11 m^3 s-2 Kg-1) and
r is the distance between the two objects.

7 0

3 years ago

Read 2 more answers

A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo

viva [34]

The fundamental frequency on a vibrating string is given by:

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension

$\mu$ is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}$

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor $\sqrt{2}$

In this case, the mass per unit length is doubled:

$\mu'=2\mu$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}$

So, the fundamental frequency will decrease by a factor $\sqrt{2}$ .

(c) the fundamental frequency will increase by a factor $\sqrt{2}$

In this case, the tension is doubled:

$T'=2T$

Substituting into the expression of the fundamental frequency, we find the new frequency:

$f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f$

So, the fundamental frequency will increase by a factor $\sqrt{2}$ .

8 0

4 years ago

An object is placed at 10.2 cm in front of a diverging lens with a focal length of -10.6 cm. What is the magnification

adoni [48]

Answer:

M= -0.51

Explanation:

After i calculated my v to be -5.2cm from the formula 1/f=1/v+1/u

Then m=v/u which is -0.51

7 0

4 years ago

Beryllium, magnesium and calcium all are located in the alkaline earth group. Which of these characteristics do these elements h

Alona [7]

Shiny, ductility, malleability are some

7 0

4 years ago

A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a massive star at the end ofits

Westkost [7]

Answer:

(a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Explanation:

Given that,

Mass of neutron $M_{n}= 13M_{s}$

Density of neutron $\rho=4.8\times10^{17}\ kg/m^3$

(a). We need to calculate the rotational inertia

Using formula of rotational inertia for sphere

$I=\dfrac{2}{5}MR^2$ ...(I)

We know that,

$\rho=\dfrac{M}{V}$

Put the value of volume

$\rho=\dfrac{3M_{n}}{4\pi R^3}$

$R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value of R in equation (I)

$I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value into the formula

$I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}$

$I=5.72\times10^{39}\ kg m^2$

The rotational inertia is $5.72\times10^{39}\ kg m^2$ .

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

$\tau=I\times \alpha$

Put the value into the formula

$\tau=5.72\times10^{39}\times5.6\times10^{-5}$

$\tau=3.20\times10^{35}\ N-m$

The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Hence, (a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

4 0

3 years ago