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2 years ago

if i got an C+ 77% in my science class but i got an 3 F's 0% recently on a science fair project, What will my grade go down to?

Physics

2 answers:

Crank2 years ago

7 0

Your grade will probably go down to a D 68% or little higher than that

vova2212 [387]2 years ago

5 0

An F or a low D like a 60% or 59%

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I need help with questions 4 and 5 because my teacher is not good with explaining them to a better understanding (Geometry)

Helga [31]

Hello! I can help you with this!

4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.

5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.

If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.

4 0

3 years ago

If you hear thunder 5 seconds after you see a lightning bolt how far is the storm

NikAS [45]

I think the answer is 3 miles because its storming now where I live

4 0

3 years ago

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A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a tur

Firlakuza [10]

Answer:

α =18.75 rad/s²

Explanation:

Given that

Acceleration a = 0.15 g

We know that g =10 m/s²

a= 0.15 x 10 = 1.5 m/s²

d= 16 cm

Radius r= 8 cm

Lets take angular acceleration =α rad/s²

As we know that

a= α r

Now by putting the values

1.5 = α x 0.08

α =18.75 rad/s²

6 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

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

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .