Planet with the most extreme temperature range

A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right

V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

g: Gravitational Acceleration

θ: Angle with the vertical

N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

m*g*cos(θ) - 0 = m*vf^2 / R

g*cos(θ) = vf^2 / R

vf^2 = R*g*cos(θ)

vf^2 = 1.45*32.2*cos(34)

vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

ΔK.E = ΔP.E

0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))

vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

vi^2 = 22.744

vi = 4.77 ft/s

4 0

4 years ago

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

Explain why two different elements cannot occupy the same box in the periodic table

bezimeni [28]

They cant occupy the same box in the periodic table because each atom has a certain number of protons which is the atomic number and no two atoms have the same amount of protons.

8 0

3 years ago

A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance

seropon [69]

Answer: Option B. R = (1/2)gt^2

Explanation:

S = R (horizontal distance)

V^2 = 2gS

V^2 = 2gR

R = V^2 / 2g

But V = gt

R = (gt)^2 / 2g

R = (g^2 x t^2) / 2g

R = gt^2 / 2

But t^2 = 2h/g

R = ( g x 2h/g) / 2

R = h

But h = (1/2)gt^2

R = h = (1/2)gt^2

4 0

3 years ago

What is nuclear fission? (Points : 1)

mixas84 [53]

The splitting of the atomic nucleus into parts

4 0

3 years ago

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