4. A car travels 3 km due North, then 5 km East Represent

A spring is used as part of a lift system and follows Hooke's law. If the spring is

salantis [7]

Answer:

1.05m or 105cm

Explanation:

Using the hooke's law equation as follows;

F = –k.x

Where;

F = force (N)

x = extension length (m)

k = constant of proportionality (N/m)

According to the information given in this question;

Displacement (x) = 85cm = 85/100 = 0.85m

Force = 12500N

Using F = kx, we find the proportionality constant

k = F/x

K = 12500/0.85

K = 14705.8N/m.

Also, since K = 14705.8N/m, the displacement (x), when the force increases to 15500N is;

F = kx

x = F/k

x = 15500/14705.8

x = 1.05m or 105cm

6 0

2 years ago

Please choose the answer that describes the scientific notation for 5,098,000.

galben [10]

D. 5.098 x 106 is the correct answer when reduced to the proper notation.

3 0

3 years ago

The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:

kow [346]

Answer:

$\frac{d_{1}}{d_{2}}=0.36$

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

$\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]}$ (1)

So, the temperature of the first and the second star will be:

$T_{1}=3866.7 K$

$T_{2}=6444.4 K$

Now the relation between the absolute luminosity and apparent brightness is given:

$L=l\cdot 4\pi r^{2}$ (2)

Where:

L is the absolute luminosity
l is the apparent brightness
r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

$\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}$

So the relative distance between both stars will be:

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}}$ (3)

The Boltzmann Law says, $L=A\sigma T^{4}$ (4)

σ is the Boltzmann constant
A is the area
T is the temperature
L is the absolute luminosity

Let's put (4) in (3) for each star.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}$

As we know both stars have the same size we can canceled out the areas.

$\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}$

$\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}$

$\frac{d_{1}}{d_{2}}=0.36$

I hope it helps!

5 0

2 years ago

The orbit of a certain a satellite has a semimajor axis of 4.0 x 107 m and an eccentricity of 0.15. Its perigee (minimum distanc

Mashutka [201]

Answer:

100KM and 1kkm

Explanation:

5 0

3 years ago

How many electrons should be removed from an initially uncharged spherical conductor of radius 0.300 m to produce a potential of

Simora [160]

1.56 x 10^12 electrons electrons should be removed.

What are electrons ?

The subatomic particles that revolve around an atom's nucleus are called electrons. They are smaller than protons or neutrons and have a negative charge. They are actually 1,800 times smaller. They also convey electric current.

Radius = 0. 300 m

Potential = 7.50 kV = 7.5 x 10^ 3 V

We kniw,

(Electric Potential) V = ke q/r

and q= ne

Therefore, n= Vr/ ke e

now substituting the values we get,

n= 1.56 x 10^12 electrons

To learn more about electrons click on the link below:

brainly.com/question/13822373

#SPJ4

3 0

1 year ago