Julie and Ashwin were doing an experiment in the classroom. They

1 answer:

4 0

B, he has to apply less force to lift the box because the mass has decreased.

The equation for force is
F=ma

If m is decreased by half, then f decreases as well. It takes less force to push a lighter object across the floor.

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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

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.62 m/s2 for t1 = 20 s. At that point the

arsen [322]

Total displacement of the car: 405 m

Explanation:

The first part of the motion of the car is a uniformly accelerated motion, so we can use the suvat equation

$s_1=ut_1+\frac{1}{2}a_1 t_1^2$

where:

u = 0 is the initial velocity (the car starts from rest)

$t_1 = 20 s$ is the time elapsed in the 1st part

$a_1=1.62 m/s^2$ is the acceleration of the car in the 1st part

$s_1$ is the displacement of the car in the 1st part

Solving for $s_1$ ,

$s_1=0+\frac{1}{2}(1.62)(20)^2=324 m$

We can also find the velocity of the car after these 20 seconds using the equation:

$v_1 = u +a_1 t_1 = 0 + (1.62)(20)=32.4 m/s$

Now we can find the distance covered by the car in the 2nd part, where it decelerates after having seen the tree limb on the road. We can do it by using the suvat equation:

$s_2 = (\frac{v_1 + v_2}{2})t_2$