The displacement is a straight line from the starting point to the finish in a ___

What did Galileo discover when he dropped two cannon balls of different weights from the Leaning Tower of Pisa?

Maru [420]

The answer is "B".. Galileo discovered that dropping two items of the same mass, they can have different weights but no matter what the force that is acting upon them is the same amount, so this means that both objects will hit the ground at the same time. Galileo tested this theory and proved it right.

6 0

3 years ago

Read 2 more answers

Describe the main distinguishing features of spiral, elliptical, and irregular galaxies.

hram777 [196]

Answer:

Spiral galaxies consist of a flat, rotating disk of stars, gas and dust, and a central concentration of stars known as the bulge. These are surrounded by a much fainter halo of stars, many of which reside in globular clusters.

Elliptical galaxies have smooth, featureless light-profiles and range in shape from nearly spherical to highly flattened, and in size from hundreds of millions to over one trillion stars. In the outer regions, many stars are grouped into globular clusters. Most elliptical galaxies are composed of older, low-mass stars, with a sparse interstellar medium and minimal star formation activity They are often chaotic in appearance, with neither a nuclear bulge nor any trace of spiral arm structure. Collectively they are thought to make up about a quarter of all galaxies.

irregular galaxies were once spiral or elliptical galaxies but were deformed by gravitational action. they are shapeless.

5 0

3 years ago

What is KBr (periodic table) def

adelina 88 [10]

Answer: Answered already

Explanation:

5 0

2 years ago

The

guapka [62]

The NUCLEUS is the center of the atom. it contains protons and NEUTRONS

protons have a POSITIVE charge

neutrons have a NEUTRAL charge

electrons have a NEGATIVE charge

6 0

3 years ago

A child on a sled slides (starting from rest) down an icy slope that makes an angle of 15◦ with the horizontal. After sliding 20

statuscvo [17]

Answer:

A) v₁ = 10.1 m/s t₁= 4.0 s

B) x₂= 17.2 m

C) v₂=7.1 m/s

D) x₂=7.5 m

Explanation:

A)

Assuming no friction, total mechanical energy must keep constant, so the following is always true:

$\Delta K + \Delta U = (K_{f} - K_{o}) +( U_{f} - U_{o}) = 0 (1)$

Choosing the ground level as our zero reference level, Uf =0.
Since the child starts from rest, K₀ = 0.
From (1), ΔU becomes:
$\Delta U = 0- m*g*h = -m*g*h (2)$
In the same way, ΔK becomes:
$\Delta K = \frac{1}{2}*m*v_{1}^{2} (3)$
Replacing (2) and (3) in (1), and simplifying, we get:

$\frac{1}{2}*v_{1}^{2} = g*h (4)$

In order to find v₁, we need first to find h, the height of the slide.
From the definition of sine of an angle, taking the slide as a right triangle, we can find the height h, knowing the distance that the child slides down the slope, x₁, as follows:

$h = x_{1} * sin \theta_{1} = 20.0 m * sin 15 = 5.2 m (5)$

Replacing (5) in (4) and solving for v₁, we get:

$v_{1} = \sqrt{2*g*h} = \sqrt{2*9.8m/s2*5.2m} = 10.1 m/s (6)$

As this speed is achieved when all the energy is kinetic, i.e. at the bottom of the first slide, this is the answer we were looking for.
Now, in order to finish A) we need to find the time that the child used to reach to that point, since she started to slide at the its top.
We can do this in more than one way, but a very simple one is using kinematic equations.
If we assume that the acceleration is constant (which is true due the child is only accelerated by gravity), we can use the following equation:

$v_{1}^{2} - v_{o}^{2} = 2*a* x_{1} (7)$

Since v₀ = 0 (the child starts from rest) we can solve for a:

$a = \frac{v_{1}^{2}}{2*x_{1} } = \frac{(10.1m/s)^{2}}{2* 20.0m} = 2.6 m/s2 (8)$

Since v₀ = 0, applying the definition of acceleration, if we choose t₀=0, we can find t as follows:

$t_{1} =\frac{v_{1} }{a} =\frac{10.1m/s}{2.6m/s2} = 4.0 s (9)$

B)

Since we know the initial speed for this part, the acceleration, and the time, we can use the kinematic equation for displacement, as follows:

$x_{2} = v_{1} * t_{2} + \frac{1}{2} *a_{2}*t_{2}^{2} (10)$

Replacing the values of v₁ = 10.1 m/s, t₂= 2.0s and a₂=-1.5m/s2 in (10):

$x_{2} = 10.1m/s * 2.0s + \frac{1}{2} *(-1.5m/s2)*(2.0s)^{2} = 17.2 m (11)$

C)

From (6) and (8), applying the definition for acceleration, we can find the speed of the child whem she started up the second slope, as follows:

$v_{2} = v_{1} + a_{2} *t_{2} = 10.1m/s - 1.5m/s2*2.0s = 7.1 m/s (12)$

D)

Assuming no friction, all the kinetic energy when she started to go up the second slope, becomes gravitational potential energy when she reaches to the maximum height (her speed becomes zero at that point), so we can write the following equation:

$\frac{1}{2}*v_{2}^{2} = g*h_{2} (13)$

Replacing from (12) in (13), we can solve for h₂:

$h_{2} =\frac{v_{2} ^{2}}{2*g} = \frac{(7.1m/s) ^{2}}{2*9.8m/s2} = 2.57 m (14)$

Since we know that the slide makes an angle of 20º with the horizontal, we can find the distance traveled up the slope applying the definition of sine of an angle, as follows:

$x_{3} = \frac{h_{2} }{sin 20} = \frac{2.57m}{0.342} = 7.5 m (15)$

4 0

2 years ago

The displacement is a straight line from the starting point to the finish in a ____ direction