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3 years ago

What best determines whether a borrower’s interest rate goes up or down?

1 answer:

6 0

~Hello there! ^_^

Your question: What best determines whether a borrower’s interest rate goes up or down?

Your answer: A market's condition best determines whether a borrower's interest rate goes down or up.

Hope this helps~

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An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).

sertanlavr [38]

V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm

4 0

3 years ago

A hamster eats a carrot before using its hamster wheel. The hamster wheel is connected to a generator which powers a light bulb.

Andrei [34K]

Answer: Chemical → Mechanical → Electrical → Radiant

Explanation:

First, the Hamster eats the carrot, then the hamster is getting chemical energy.

Now the hamster starts using his wheel, then he "transforms" the chemical energy into mechanical energy.

Now the mechanical energy is connected to a generator, this means that the mechanical energy (the rotation of the wheel) is being converted into electrical energy.

And we know that there is a light bulb powered by this electrical energy, then we have electrical energy being transformed into radiant energy.

Then the correct option is:

Chemical → Mechanical → Electrical → Radiant

6 0

3 years ago

As a gas or liquid increases in heat, what direction will it naturally move?

miss Akunina [59]

Hello!

Answer:

When a gas gets hot it should go up because of the pressure.

Explanation:

Hope this helps!

3 0

3 years ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

An imaginary line perpendicular to a reflecting surface is called _________.

n200080 [17]

<span>An imaginary line perpendicular to a reflecting surface is called "a normal" (principle line)

So, Your Answer would be Option B

Hope this helps!</span>

5 0

3 years ago

Read 2 more answers