Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up to

Which of the following about Dalton's theory of matter is true?

Lemur [1.5K]

D.Atoms can be divided into smaller parts.

Dalton's atomic theory was the first complete attempt to describe allmatter in terms of atoms and their properties. Dalton based his theory on the law of conservation of mass and the law of constant composition. The first part of his theory states that allmatter is made of atoms, which are indivisible.

6 0

3 years ago

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the

Diano4ka-milaya [45]

Answer:

<em>The object could fall from six times the original height and still be safe</em>

Explanation:

<u>Free Falling</u>

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

$U=m.g.h$

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

$\displaystyle K=\frac{1}{2}m.v^2$

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

$\displaystyle m.g.h=\frac{1}{2}m.v^2$

Solving for h

$\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}$

If the drop had occurred in the Moon, then

$\displaystyle h_M=\frac{v_M^2}{2g_M}$

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

$\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}$

$\displaystyle h_M=6\frac{v^2}{2g}=6h$

This means the object could fall from six times the original height and still be safe

6 0

3 years ago

If you throw an object straight up into the air with an initial velocity of 42m/s. What is it’s velocity at the peak of its fl

solniwko [45]

0 m/s. It’s trivial

7 0

3 years ago

Explain the rules for mirror and lens in order to form ray diagram

Vinil7 [7]

See below.

The following are the guidelines for creating ray diagrams for the generation of an image by a concave mirror:

Rule 1: If the incident ray is parallel to the major axis, the reflected ray will pass through the concave mirror's focus.

The reflected ray is parallel to the principal axis if the incident ray passes through the focus of the concave mirror.

Rule 3: If the incident ray passes through the concave mirror's center of curvature, the reflected ray follows the same path.

Hope this helps!

4 0

2 years ago

James threw a ball vertically upward with a velocity of 41.67ms-1 and after 2 second David threw a ball vertically upward with a

Reptile [31]

Answer:

When have passed 3.9[s], since James threw the ball.

Explanation:

First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.

We'll use the kinematics equations to find these two unknowns.

$y=y_{0} +v_{0} *t+\frac{1}{2} *g*t^{2} \\where:\\y= elevation [m]\\y_{0}=initial height [m]\\v_{0}= initial velocity [m/s] =41.67[m/s]\\t = time passed [s]\\g= gravity [m/s^2]=9.81[m/s^2]\\Now replacing:\\y=0+41.67 *(2)-\frac{1}{2} *(9.81)*(2)^{2} \\\\y=63.72[m]\\$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can find the velocity after 2 seconds.

$v_{f} =v_{o} +g*t\\replacing:\\v_{f} =41.67-(9.81)*(2)\\\\v_{f}=22.05[m/s]$

Note: The sign for the gravity is minus because it is acting against the movement.

Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.

$y=y_{o} +v_{o} *t-\frac{1}{2} *g*t^{2} \\y=63.72 +22.05 *t-\frac{1}{2} *(9.81)*t^{2} \\\\y=63.72 +22.05 *t-4.905*t^{2} \\$

As we can see the equation is based on Time (t).

Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t

$y=y_{o} +v_{o} *t+\frac{1}{2} *g*t^{2} \\where:\\v_{o} =55.56[m/s] = initial velocity\\y_{o} =0[m]\\now replacing\\63.72 +22.05 *t-(4.905)*t^{2} =0 +55.56 *t-(4.905)*t^{2} \\63.72 +22.05 *t =0 +55.56 *t\\63.72 = 33.51*t\\t=1.9[s]$

Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.

Time = 2 + 1.9 = 3.9[s]

4 0

3 years ago