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3 years ago

Tina just threw a 0.15 kg ball straight up in the air. If the ball reaches

Physics

1 answer:

marta [7]3 years ago

4 0

Answer:

13 j

Explanation:

I took the test and got it wrong lol

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An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion

anyanavicka [17]

Answer:

attached below is the free body diagram of the missing illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

$E_{e} = E_{3} - E_{1}$

E $_{e}$ = initial kinetic energy of the electron

E $_{1}$ = -4 eV

E $_{3}$ = -1 eV

insert the values into the equation above

$E_{e}$ = -1 -(-4) eV

= -1 + 4 = 3 eV

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3 years ago

shelly starts from rest on her bicycle at the top of a hill. After 6.0s she has reached a final velocity of 14m/s. what is shell

earnstyle [38]

Divide 14 by 6 and there is your answer with the unit of m

8 0

3 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLEST

mart [117]

1. Meiosis has 2 cell divisions but mitosis has only one cell division. Meiosis I is same as mitosis because parent cell produces diploid cells. But the difference between meiosis and mitosis is after the complete process meiosis produces genetically different haploid daughter cells and mitosis produces genetically same diploid daughter cells.

2. The main advantage of the sexual reproduction is It helps to produce offspring which can be well adapted to the environmental conditions due to the genetically differentiations. Hence, there are various types of offspring in the population. But the disadvantage is it requires two parents and it requires more time than asexual reproduction.

3. Advantages of asexual reproduction are it requires only parent, time for the process is relatively low, it produces genetically identical offspring. The main disadvantage is due to the production of genetically identical offspring, the survival rate is low when there is a disaster.

4. The organism can do either sexual reproduction or asexual reproduction to produce offspring. If it is difficult to find a partner to mate, the living being can do asexual reproduction to make offspring in a short period of time. When the organism needs genetically different offspring for survival, they can do sexual reproduction.

7 0

3 years ago

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Groups of students are making measurements of small cars rolling across a table. the class has measuring tapes with markings in

TiliK225 [7]

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Centimeters Measurements:

A centimeter is a metric unit of measurement used for measuring the length of an object, It is written as cm
Centimeter is one hundredth of a meter, 1 meter is 0.01 cm.

Inches measurements:

An inch can be defined as a unit of length in the customary system of measurement. Length in inches is either represented by in or ''.
1 meter is equal to 39.37 inches

here, the cars are small objects.

The number of centimeters is always bigger,

because a centimeter unit is smaller than an inch unit, and it takes more of them when we are measuring.

Hence,

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

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4 0

2 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

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3 years ago