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3 years ago

What happens in the crushing can experiment?

Physics

1 answer:

Dafna1 [17]3 years ago

7 0

Explanation:

When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped

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In attempting to pull a 1500 kg car out of a ditch, Al exerts a force of 200 N for 5 s, Bill exerts a force of 500 N for two s,

MArishka [77]

Answer:

Clyde will provide greater impulse

Explanation:

We have given that Al exerts a force of 200 N for 5 sec

We know that impulse is given by, impulse = force ×time = 200×5 =1000 N-s

Bill exerts a force of 500 N for 2 sec

So impulse = 500×2 = 1000 N-s

Now the force exerted by Clyde 300 N for 4 sec

So impulse = 300×4 = 1200 N-s

From above calculation we can see that Clyde provide greater impulse than any other

4 0

3 years ago

A car travels at 2000 meters east in 40 seconds. What is the cars velocity

DerKrebs [107]

Answer:

50 ms⁻¹

Explanation:

Given that,

Distance Travelled =2000 meters

Time Taken =40 seconds

velocity = ?

we know,

velocity = Distance / Time

=(2000 / 40) ms⁻¹

=50 ms⁻¹

5 0

3 years ago

A small 22 kilogram canoe is floating downriver at a speed of 5 m/s. What's the canoes kinetic energy? _______ Joules

netineya [11]

The non-relativistic formula for kinetic energy for low speeds is :

K.E = 0.5mv^2 = 0.5 * 22 * (5)^2 = 275 J

8 0

3 years ago

Read 2 more answers

A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

6 0

3 years ago

The age of the universe is around 100,000,000,000,000,000s. A top quark has a lifetime of roughly 0.000000000000000000000001s. W

Maslowich

Answer:

10⁴¹ s quark top lives have been in the history of the universe.

Explanation:

You need to determine how many quark top lives there have been in the history of the universe, that is, what is the age of the universe divided by the lifetime of a top quark. Expressed in a formula, this is:

$t\frac{Age of the universe}{Lifetime of a top quark}$

Yo know that the "Age of the universe" is 100,000,000,000,000,000 which can also be expressed as 10¹⁷ s .

You also know that the "Lifetime of a top quark" is 0.000000000000000000000001 which can also be expressed as 10⁻²⁴ s.

Then $t=\frac{10^{17} }{10^{-24} }$

Recalling that the result of dividing two powers of the same base is another power with the same base where the exponent is the subtraction of the initial exponents, it is possible to calculate this division as follows:

$t=10^{17-(-24)}$

$t=10^{17+24}$

t=10⁴¹ s

So 10⁴¹ s quark top lives have been in the history of the universe.

5 0

3 years ago