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3 years ago

If a car rolls gently off a vertical cliff, how long does it take to reach 90km/hr.

1 answer:

5 0

90 km/h : 3.6 = 25 m/s. If you know that on earth g = 9.81 m/s^2, then all you have to do is divide the speed by g. 25/9.81 = 2.548 seconds

At least, if by 'gently rolls off a vertical cliff' means that your starting velocity equals zero.

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Is work required to pull a nucleon out of an atomic nucleus? Does the nucleon, once outside the nucleus, hove more mass than it

olga2289 [7]

<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>

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4 years ago

A hollow spherical planet is inhabited by people who live inside it, where the gravitational is zero. When a very massive space

Ulleksa [173]

Answer:

c) nonzero, directed toward the spaceship

Explanation:

As we know that net gravitational force due to spherical shell inside all its points will always be zero

So if planet is a spherical shell then inside the planet net gravitational force is zero on the people living in it

So when massive spaceship land on the surface of planet then the gravitational force of the spaceship is experienced by the people inside the shell

So here the gravitational force on the people is nonzero and it is towards the spaceship which landed on the surface of planet

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3 years ago

In which of the following scenarios is the total momentum of the system conserved?

leonid [27]

Answer:

The total momentum of a system is conserved only when the system is closed.

Explanation:

7 0

2 years ago

The uncertainty in the position of an electron along an x axis is given as 68 pm. What is the least uncertainty in any simultane

umka21 [38]

Answer:

$\Delta p_x=7.75\times 10^{-25}\ kg-m/s$

Explanation:

Given that,

The uncertainty in the position of an electron along the x-axis is, $\Delta x=68\ pm=68\times 10^{-12}\ m$

We need to find the east uncertainty in any simultaneous measurement of the momentum component of this electron.

We know that the Heisenberg's uncertainty principle gives the relation between the uncertainty in position and the momentum of electron as :

$\Delta p_x{\cdot}\Delta x\ge \dfrac{h}{4\pi }$

Putting all the values, we get :

$\Delta p_x{\cdot}\ge \dfrac{h}{4\pi \Delta x}\\\\\Delta p_x \ge \dfrac{6.63\times 10^{-34}}{4\pi \times 68\times 10^{-12}}\\\\\Delta p_x\ge 7.75\times 10^{-25}\ kg-m/s$

So, the momentum component of this electrons is greater than $7.75\times 10^{-25}\ kg-m/s$ .

6 0

4 years ago

2. Two projectiles thrown from the same point at angles 0,=60° and 02=30° with the horizontal,

Liono4ka [1.6K]

Answer:

The answer is below

Explanation:

The maximum height (h) of a projectile with an initial velocity of u, acceleration due to gravity g and at an angle θ with the horizontal is given as:

$h=\frac{u^2sin^2\theta}{2g}$

Given that the two projectile has the same height.

$For\ the\ first\ projectile\ with\ an\ angle\ \60^0 \ with\ the\ horizontal\ and initial\ velocity\ u_1:\\\\h=\frac{u_1^2sin^260}{2g} =\frac{0.75u_1^2}{2g} \\\\For\ the\ second\ projectile\ with\ an\ angle\ \30 \ with\ the\ horizontal\ and initial\ velocity\ u_2:\\\\h=\frac{u_2^2sin^230}{2g} =\frac{0.25u_2^2}{2g}\\\\\frac{0.25u_2^2}{2g}=\frac{0.75u_1^2}{2g}\\\\0.25u_2^2=0.75u_1^2\\\\\frac{u_1^2}{u_2^2} =\frac{0.25}{0.75} \\\\\frac{u_1^2}{u_2^2}=\frac{1}{3} \\$ $\\\frac{u_1}{u_2}=\sqrt\frac{1}{3}$

8 0

3 years ago