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3 years ago

I have this sheet and I don’t understand any of it at all

Physics

1 answer:

katen-ka-za [31]3 years ago

3 0

I know this probably is not what you are looking for, but this is what I would suggest.

Write down on the side what values they are giving you and the corresponding variable used for that value.

Like for the first question, they give you mass is 3000kg.

On the side write

m = 3000kg

m is the variable used for mass

Then list any other variables they give you in the question and then what you are trying to look for.

a = 2m/s^2

F = ?

a is the variable for acceleration and F is the variable for force.

So now you have:

m = 3000kg

a = 2m/s^2

F = ?

You have a and m, and you are looking for F (force), you know that because it asks "with what force".

Also note, that you have to know what the variable is based on the units and you likely have some type of formula sheet with all of them listed.

You can use the variables and what is unknown to match it to a formula that will solve for the unknown variables. Also note, that it might not always be simply a=b+c, you may know what "a" and "b" are, but they may ask for c so you will have to rearrange the formula with some simple algebra.

This question however is simple. You can see you are going to use the formula

F=ma.

So now just substitute in your variables to solve for F.

(3000kg)(2m/s^2) = 6000N

<u>Make sure to write the units</u> which are probably on your formula sheet as well. Force is measured in Newtons, so you use the unit N.

Also note, if they gave you mass in grams (another unit), you cannot use that in the formula (your formula sheet would tell you the unit it must be in), so you would have to convert it to kg.

Hope this helps as a starting point. I don't intend to do your homework for you, but explain a process you can use to help you do it on your own.

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2 years ago

A ball with mass 0.11 kg is thrown upward with initial velocity 10 m/s from the roof of a building 20 m high. Neglect air resist

inessss [21]

Answer:

a) $H=25.1020\ m$

b) $t=3.2838\ s$

Explanation:

Given:

mass of the ball thrown up, $m=0.11\ kg$

initial velocity of the ball thrown up, $u=10\ m.s^{-1}$

height above the ground from where the ball is thrown up, $h=20\ m$

Maximum height attained by the ball above the roof level can be given by the equation of motion.

As,

$v^2=u^2-2g.h'$

where:

$v=$ final velocity at the top height of the upward motion $=0\ m.s^{-1}$

$g=$ acceleration due to gravity

$h'=$ height of the ball above the roof

Now,

$0^2=10^2-2\times 9.8\times h'$

$h'=5.10\ m$

Therefore total height above the ground:

$H=h+h'$

$H=20+5.1020$

$H=25.1020\ m$

Now we find the time taken in raching the height $h'$ :

$v=u-gt'$

$v=$ final velocity at the top of the motion $=0\ m.s^{-1}$

So,

$0=10-9.8\times t'$

$t'=1.0204\ s$

Now the time taken in coming down to the ground from the top height:

$H=u'.t_d+\frac{1}{2} g.t_d^2$

where:

$u'=$ is the initial velocity of the ball in course of coming down to ground from the top $=0\ m.s^{-1}$

Here the direction acceleration due to gravity is same as that of motion so we are taking them positively.

$25.1020=0+0.5\times 9.8\times t_d^2$

$t_d=2.2634\ s$

Therefore the total time taken in by the ball to hit the ground after it begins its motion:

$t=t'+t_d$

$t=1.0204+2.2634$

$t=3.2838\ s$

4 0

3 years ago

What force causes a 1-kg mass to accelerate at a rate of 1 meter per second each second?

Zigmanuir [339]

Answer:

$F=1N$

Explanation:

Conceptual analysis

To solve this problem we apply Newton's second law:

The acceleration of an object is proportional to the force F acting on it and inversely proportional to its mass m.

a = F / m

Where,

F = m * a Formula (1)

F: Force in Newtons (N)

m: mass in kg

a: acceleration in m/(s^2)

a = v / t Formula (2)

v: speed in m/s

t: time in seconds (s)

Known information

We know the following data:

m = 1kg

v = 1 m/s

t = 1s

Development of the problem:

In the Formula (2): $a = \frac{\frac{1m}{s}}{1s} = 1 \frac{m}{s^2}$

In the Formula (1): $F=1kg* 1 \frac{m}{s^2}=1N$

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3 years ago

PERIOD OF THE LEG The period of the leg can be approximated by treating the leg as a physical pendulum, with a period of Equatio

Lesechka [4]

Answer:

Explanation:

We can see from the question that

$T 2\pi\sqrt{\frac{I}{mgh} }$

and $I = \frac{ml^2}{3}$

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=> $m =10kg$

from the question $l =48$ % of $1.83m$

Substituting this into the equation

$I = \frac{10.72 * (0.8784)^2}{3}$

=> $I = 2.7571 \ kg m^2$

$h = 0.5 * L$

$h = 0.5 * 0.8784$

$h = 0.4392m$

From the equation above

$T = 2 \pi \sqrt{\frac{2.7571}{10 .72 * 9.81 * 0.4392} }$

$T = 1,534\ sec$

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3 years ago