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2 years ago

Do you think that the universe has a starting point and ending point? explain.

Physics

1 answer:

Eddi Din [679]2 years ago

3 0

Answer:

Explanation:

I don't think the universe has a starting and ending point because the universe is vast and has not been fully discovered. I think that the universe is endless because all of the stars, planets, and galaxy's have yet to be found and the end or start of the universe is also yet to be discovered so I think that the universe could go on forever.

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A roller coaster's velocity at the top of the hill is 10 m/s. Two seconds later it reaches the bottom of the hill with a velocit

svet-max [94.6K]

your answer is 8 m / sec²

5 0

2 years ago

You are traveling at 100 km/hr to make it to work on time. You have to be to work in .25 hr or you will be late. You have 16 km

alexgriva [62]

Answer: Yes

Explanation:

Velocity $V$ is defined as the distance traveled $d$ in a specific time $t$ :

$V=\frac{d}{t}$

If you are traveling at $V=100 km/h$ a distance $d=16 m$ , then the time it will take you to be at work is:

$t=\frac{d}{V}=\frac{16 km}{100 km/h}$

$t=0.16 h$

This means you will make it on time, because this time is less than 0.25 h.

4 0

3 years ago

Suppose a plane accelerates from rest for 30 s, achieving a takeoff speed of 80 m/s after traveling a distance of 1200 m down th

Margaret [11]

Answer:

300 m

Explanation:

The train accelerate from the rest so u = 0 m/sec

Final speed that is v = 80 m/sec

Time t = 30 sec

The distance traveled by first plane = 1200 m

We know the equation of motion $S=ut+\frac{1}{2}at^2$ where s is distance a is acceleration and u is initial velocity

Using this equation for first plane $1200=0\times 30+\frac{1}{2}a30^2$

$a=2.67\frac{m}{sec^2}$

As the acceleration is same for both the plane so a for second plane will be 2.67 $\frac{m}{sec^2}$

The another equation of motion is $v^2=u^2+2as$ using this equation for second plane $40^2=0+2\times 2.67\times s$

s = 300 m

5 0

2 years ago

How much time will it take to do 33j of work with 11 w of power

Yuri [45]

Time required : 3 s

<h3>Further explanation </h3>

Power is the work done/second.

$\tt P=\dfrac{W}{t}\\\\P=power,j/s,watt\\\\W=work, J\\\\t=times=s$

To do 33 J of work with 11 W of power

P = 11 W

W = 33 J

$\tt t=\dfrac{W}{P}\\\\t=\dfrac{33}{11}\\\\t=\boxed{\bold{3~s}}$

8 0

3 years ago

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 32.50 N/m, is 0.640 J. Find by how

liraira [26]

Answer:

$x = 0.198456 \ m$

$h = 1.3061 \ m$

$v = 5.06 \ m/s$

$d = 4.0273 \ m$

Explanation:

Considering the first question

From the question we are told that

The spring constant is $k = 32.50 N/m$

The potential energy is $PE = 0.640 \ J$

Generally the potential energy stored in spring is mathematically represented as $PE = \frac{1}{2} * k * x^2$

=> $0.640= \frac{1}{2} * 32.50 * x^2$

=> $x = \sqrt{0.03938}$

=> $x = 0.198456 \ m$

Considering the second question

From the question we are told that

The mass of the dart is m = 0.050 kg

Generally from the law of energy conservation

$PE = mgh$

=> $0.640 = 0.050 * 9.8 * h$

=> $h = 1.3061 \ m$

Considering the third question

The height at which the dart was fired horizontally is $H = 3.90\ m$

Generally from the law of energy conservation

$PE = KE$

Here KE is kinetic energy of the dart which is mathematical represented as

$KE = \frac{1}{2} * mv^2$

=> $0.640 = \frac{1}{2} * 0.050 * v^2$

=> $v^2 = 25.6$

=> $v = 5.06 \ m/s$

Considering the fourth question

Generally the total time of flight of the dart is mathematically represented as

$t = \frac{ 2 * H }{g}$

=> $t = \frac{ 2 * 3.90 }{9.8 }$

=> $t = 0.7959 \ s$

Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as

$d = v * t$

=> $d = 5.06 * 0.7959$

=> $d = 4.0273 \ m$

5 0

2 years ago