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4 years ago

Prove mathematically :1. v = u + at2. s = ut+1*2 at

Physics

1 answer:

aleksley [76]4 years ago

6 0

Answer:

a.v=u+v/2

a.v=s/t

combining two equation we get,

u+v/2=s/t

(u+v)t/2=s

{u+(u+at)}t/2=s

(u+u+at)t/2=s

(2u+at)t/2=s

2ut+at^2/2=s

2ut/2+at^2/2=s

UT +1/2at^2=s

proved

a=v-u/t

at=v-u

u+at=v

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Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers

Help! I can't remember how to calculate this.

Mars2501 [29]

You need to draw two lines from the head of the arrow: one parallel to the axis (which will be bent) and the other one passing through the center of the converging lens <span>(which will continue straight) </span>and see where they intersect.

If the arrow is at an infinite distance, the image will be point-like in F.
If the arrow is at a distance greater than 2F, the image will be upside down between F and 2F.
If the arrow is at a distance of 2F, the image will be upside down in 2F.
If the arrow is at a distance between F and 2F, the image will be upside down beyond 2F.
If the arrow is at a distance F, all rays proceed parallel to each other and no image is formed.
If the arrow is at a distance smaller than F, the image will be virtual.

Therefore, the correct matches are 1B, 2C, 3D, and 4A.

5 0

3 years ago

You have a large flashlight that takes 4 D-cell batteries, each with a voltage of 1.5 volts. If the current in the flashlight is

tino4ka555 [31]

Answer:

The answer is "C. 3.0 ohms".

Explanation:

Ohm's law states the following:,

V = IR

where:

Voltage (V) = 1.5 volts × 4 batteries = 6.0 volts

Current (I) = 2.0 amps

Resistance (R) = ? ohms

To solve for Resistance (R) the equation must be rearranged this way:

R = V / I

Then, the variables must be replaced with the known values:

R = 6.0 volts / 2.0 amp

R = 3.0 ohms

The answer is C. 3.0 ohms.

------------

P.D: I got it right on PLATO

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3 years ago

A force of 1,500 N is moved through 0.10 m in order to compress a steel cylinder. The cylinder was compressed 0.010 m by a force

ValentinkaMS [17]

We first calculate the work effectively done by the force when it compressed the steel cylinder. This is solved by multiplying the distance that the cylinder was compressed by and its causing force.
0.010 m * 13,000 N = 130 N-m

Then, we solve for the work done by the 1,500 N force:
0.10 m * 1,500 N = 150 N-m

We then take the ratio of the efficient:experimental
130 N-m / 150 N-m = 0.87 = 87%

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4 years ago

When a virtual image is created in a plane mirrora. the image is upright.b. the image is located behind the mirror.c. reflected

zubka84 [21]

<h2>Answer: all of the above</h2>

A plane mirror is a highly polished flat surface with a very high capacity to reflect incident light.

According to the figure attached:

1. The incident rays coming from the real object reach the mirror and

2.are reflected following the law of Reflection.

3.The prolongation of those reflected rays converge at a point that <u>does not coincide</u> with the actual position of the object. At that point the virtual image of the object is formed.

4.Then, the reflected divergent rays are captured by our eye converging on the retina.

Now, the image is said to be <u>virtual</u> because it is a copy of the object that looks as if the object is behind the mirror and not in front of it or on the surface, but it is not really there. However, it can be seen when we focus it with our eyes.

In addition, the image formed is:

symmetrical, because apparently it is at the same distance from the mirror

the same size as the object.

upright, because it retains the same orientation as the object.

All these characteristics coincide with the first three options proposed in the question, so the correct option is d.

4 0

4 years ago

Prove mathematically :1. v = u + at2. s = ut+1*2 at ​

Prove mathematically :1. v = u + at2. s = ut+1*2 at