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3 years ago

Prove the correctness of this equation s=vt + 1/2at

Physics

1 answer:

Kaylis [27]3 years ago

5 0

Answer:

The formula is dimensionally correct.

Explanation:

Given

$s = ut + \frac{1}{2}at^2$

Required

Prove its correctness

Write out the dimension of each:

$s = M^0LT^0$ --- displacement

$ut = M^0LT^{-1} * T$ --- velocity * time

$\frac{1}{2}at^2 = M^0LT^{-2} * T^2$ --- acceleration * square of time

The expression becomes:

$s = ut + \frac{1}{2}at^2$

$M^0LT^0 = M^0LT^{-1} * T + M^0LT^{-2} * T^2$

Apply law of indices

$M^0LT^0 = M^0LT^{-1+1} + M^0LT^{-2+2}$

$M^0LT^0 = M^0LT^{0} + M^0LT^{0}$

$M^0LT^0 = M^0LT^{0}$

Both sides of the equation are equal

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A roller coaster car is going over the top of a 18-mm-radius circular rise. At the top of the hill, the passengers "feel light,"

Andre45 [30]

Answer:

0.29713 m/s

Explanation:

m = Mass of person

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity

r = Radius = 18 mm

By balancing the forces in the system we have

$mg-N=\dfrac{mv^2}{r}\\\Rightarrow mg-\dfrac{mg}{2}=\dfrac{mv^2}{r}\\\Rightarrow v=\sqrt{r(g-\dfrac{g}{2})}\\\Rightarrow v=\sqrt{0.018\times (9.81-\dfrac{9.81}{2})}\\\Rightarrow v=0.29713\ m/s$

The velocity of the coaster is 0.29713 m/s

7 0

3 years ago

So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th

tia_tia [17]

Answer:

a) $a=33.73mm/s^{2}$

b) mg>N

c) $\%_{change}=0.343\%$

d) $a=24.07mm/s^{2}$

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

$a_{c}=\omega ^{2}r$

where:

$\omega=\frac{2\pi}{T}$

we know the period of rotation of the earth is about 24 hours, so:

$T=24hr*\frac{3600s}{1hr}=86400s$

so we can now find the angular speed:

$\omega=\frac{2\pi}{86400s}$

$\omega=72.72x10^{-6} rad/s^{2}$

So the centripetal acceleration will be:

$a_{c} =(72.72x10^{-6} rad/s^{2})^{2}(6478x10^{3}m)$

which yields:

$a_{c}=33.73mm/s^{2}$

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

$\sum F=0$

so we get that:

$N-mg+ma_{c} = 0$

and solve for the normal force:

$N=mg-ma_{c}$

In this case, we can clearly see that:

$mg>mg-ma_{c}$

therefore mg>N

This is because the centripetal acceleration is pulling us upwards, that will make the magnitude of the normal force smaller than the product of the mass times the acceleration of gravity.

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

$mg=(60kg)(9.81m/s^{2})=588.6N$

and let's calculate the normal force:

$N=m(g-a_{c})$

$N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})$

N=586.58N

so now we can calculate the percentage change:

$\%_{change} = \frac{mg-N}{mg}x100\%$

so we get:

$\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%$

$\%_{change}=0.343\%$

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

$cos \theta = \frac{AS}{h}$

In this case:

$cos \theta = \frac{r}{R_{E}}$

so we can solve for r, so we get:

$r= R_{E}cos \theta$

in this case we'll use the average radius of earch which is 6,371 km, so we get:

$r = (6371x10^{3}m)cos (44.4^{o})$

which yields:

r=4,551.91 km

and now we can calculate the acceleration at that point:

$a=\omega ^{2}r$

$a=(72.72x10^{-6} rad/s)^{2}(4,551.91x10^{3}m$

$a=24.07 mm/s^{2}$

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3 years ago

In one scene in the movie The Godfather II, a solid gold phone is passed around a large table for everyone to see. Suppose the v

dangina [55]

Volume of gold in the phone = 10 cm^3
= 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
= 0.193 kg
So
0.193 kg = 0.193 * 2.2 pounds
= 0.43 pounds
I think there is something wrong with the options given in the question.</span>

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3 years ago

An airplane has a momentum of 8.55 x 107 kg.m/s[S] and a velocity of 900 km/h[S]. Determine the mass of the airplane.

kow [346]

Answer:

342,000kg

Explanation:

p=mv

8.55*10^7 kg*m/s=m(900 km/h)

85,500,000 kg*m/s=m(900 km/h)

(85,500,000 kg*m/s)/(900 km/h)=m

Get same units.... 900km/h = 250m/s

m/s cancel in the division, you are left with just kg!!

85,500,000/250=342,000kg! That's it!

6 0

2 years ago

An object with a mass of 375g is moving with a constant velocity. It has a force of -20 N applied to it. Determine the accelerat

olga55 [171]

Answer:

Explanation:

Convert the mass to kg:

375g = 375/1000kg = 0.375kg

F = ma

-20 = 0.375a

a = -20/0.375

a = -53

The object is accelerating at 53m/s/s backwards assuming that the forward motion is positive.

8 0

3 years ago

Prove the correctness of this equation s=vt + 1/2at ​

Prove the correctness of this equation s=vt + 1/2at