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3 years ago

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A horizontal spring with spring constant 290 N/m is compressed by 10 cm and then used to launch a 300 g box across the floor. Th

e coefficient of kinetic friction between the box and the floor is 0.23. What is the box's launch speed?

1 answer:

Marta_Voda [28]3 years ago

5 0

Answer:

Explanation:

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Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0

3 years ago

Air pressure may be represented as a function of height above the surface of the Earth as shown below.. P(h) = P_0 e^-.00012h. .

Rufina [12.5K]

Answer. Second Option: .85p_o=p_o e^-.00012h

Solution:

P(h)=Po e^(-0.00012h)

Air pressure: P(h)

Height above the surface of the Earth (in meters): h

Air pressure at the sea level: Po

Height at which air pressure is 85% of the air pressure at sea level:

h=?, P(h)=85% Po

P(h)=(85/100) Po

P(h)=0.85 Po

Replacing P(h) by 0.85 Po in the formula above:

P(h)=Po e^(-0.00012h)

0.85 Po = Po e^(-0.00012h)

5 0

3 years ago

Read 2 more answers

Is adequate health care are available to all

tankabanditka [31]

Explanation:

I think for anyone to answer this we need more info on what you want answered. The Sentence Itself doesn't Make Since To Me

6 0

2 years ago

According to Coulomb's law, what happens to the attraction of two oppositely charged objects as their distance of separation inc

NARA [144]

Answer:

Option B. Decreases

Explanation:

Coulomb's law states that:

F = Kq₁q₂ / r²

Where:

F => is the force of attraction between two charges

K => is the electrical constant.

q₁ and q₂ => are the two charges

r => is the distance apart.

From the formula:

F = Kq₁q₂ / r²

The force of attraction (F) is inversely proportional to the square of their separating distance (r).

This implies that as the distance between them increase, the force of attraction between the two charges will decrease and as the distance between two charges decrease, the force of attraction between them will increase.

Considering the question given above and the illustration given above, the force of attraction will decrease as their distance of separation increases.

Option B gives the right answer to the question.

7 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.

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

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

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

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

I hope it helps!

5 0

3 years ago