Hi I’m having some difficulty with this question could really use some help!

A horse canters away from its trainer in a straight line, moving 37 m away in 9.1 s. It then turns abruptly and gallops halfway

vesna_86 [32]

Answer:

Average velocity = 1.69 m/s

Average speed = 5.09 m/s

Explanation:

Given that

Horse cover 37 m in 9.1 sec and 18.5 m in 1.8 sec.

As we know that

Average velocity = Displacement / Total time

Average speed = Total distance / Total time

Average velocity:

Average velocity = Displacement / Total time

Total displacement = 37 - 18.5 = 18.5 m

Total time = 9.1 + 1.8 =10.9 s

Average velocity = Displacement / Total time

Average velocity = 18.5 / 10.9

Average velocity = 1.69 m/s

Average speed:

Average speed = Total distance/ Total time

Total distance = 37 + 18.5 = 55.5 m

Total time = 9.1 + 1.8 =10.9 s

Average speed = Total distance/ Total time

Average speed = 55.5 / 10.9

Average speed = 5.09 m/s

3 0

3 years ago

The universe is mostly made up of what?

Viktor [21]

D. Empty Space

Imagine an apple in the center of a football field. Pretend like that apple is the nucleus of an atom. The electrons would be at the end-zones. The universe is just a butttonne of empty space

5 0

4 years ago

The following questions present a twist on the scenario above to test your understanding. Suppose another stone is thrown horizo

Ipatiy [6.2K]

The first part of the text is missing, you can find on google:

"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."

Let's now solve the different parts.

(a) 3.03 s

The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by

$y(t) = h -\frac{1}{2}gt^2$

where

h = 45 m is the initial height

g = 9.8 m/s^2 is the acceleration of gravity

When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:

$0=h-\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(45)}{9.8}}=3.03 s$

(b) 18.5 m/s

For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.

The horizontal position is given by

$x=v_x t$

where

$v_x$ is the horizontal speed, which is constant

t is the time

At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:

$v_x = \frac{x}{t}=\frac{56}{3.03}=18.5 m/s$

The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.

(c) 35.0 m/s at 58.1 degrees below the horizontal

At the impact, we know that the horizontal speed is still the same:

$v_x = 18.5 m/s$

we need to find the vertical velocity. This can be done by using the equation

$v_y = u_y -gt$

where

$u_y =0$ is the initial vertical velocity

g is the acceleration of gravity

t is the time

Substituting t = 3.03 s, we find the vertical velocity at the time of impact:

$v_y = -(9.8)(3.03)=-29.7 m/s$

So the magnitude of the velocity at the impact (so, the speed at the impact) is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.5^2+(-29.7)^2}=35.0 m/s$

The angle instead can be found as:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-29.7}{18.5})=-58.1^{\circ}$

so, 58.1 degrees below the horizontal.

4 0

3 years ago

Help me its physical science and i don't know what to do tyyy

miskamm [114]

The thing to do here is to balance the eqns that is to balance the stoichometric ratios or coefficients of the cpds.

hope you understand. all the best.

7 0

3 years ago

The potential-energy function u(x) is zero in the interval 0≤x≤l and has the constant value u0 everywhere outside this interval.

VMariaS [17]

Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)

So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀

Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)

Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5

5 0

4 years ago