15 points

skad [1K]

When an object turns around an internal axis (like the Earth turns around its axis) it is called a rotation. When an object circles an external axis (like the Earth circles the sun) it is called a revolution.

5 0

3 years ago

What is a boyfriend to you? And what age is appropriate to have one?

Sauron [17]

A boyfriend is someone who you find more than a friend. Someone who you have a deeper connection with. An age to have one really depends on the person and is mostly decided by parents. If the person is mature they may have one around 15 to 16+ years old. If the person is still "immature" or is not ready to have one, the age limit may be extended to 17+

7 0

3 years ago

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Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

Part A)

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

Part B)

We can use the equation of the horizontal position here.

First stone

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

Second stone

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago

Using the same cost and time estimates, consider any trade-offs that SciTech may have to make to complete the project.

ch4aika [34]

Answer:

Oracio is the most cost-effective choice because he would cost the least to complete the project. However, he would also take the longest amount of time. Camilla could complete the job the fastest, but she costs more than Oracio. SciTech will have to decide if it is more important to save money or complete the work quickly to meet the deadline.

Hope this helps :)

4 0

3 years ago

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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0

3 years ago

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