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lord [1]
3 years ago
6

I have to submit my homework soon and I am not familiar with these concepts. can anybody do then for me?

Physics
1 answer:
AlladinOne [14]3 years ago
6 0

#4).  The concepts are:  A). acceleration is always in the direction of the force, and B). friction always acts in the direction opposite to motion.  (that's B)

#5).  The concepts are:  A). the NET force is the sum of all the individual forces acting (on this car, it's 1600N forward).  and B). Force = (mass) x (acceleration).  So Acceleration = (force) / (mass).  For this car, that's (1600N forward) / (800 kg) .

#6).  The concept is:  As long as you don't exceed the "proportionality limit" of a spring, its extension is proportional to the load on it.  That means that the change in extension is always proportional to the change in the load.  So now, look at the table:  As long as the load is 10N or less, the spring stretched 3cm longer for every 2N more of load.  But if the load is somewhere between 10N and 12N, that relationship disappears.  Something changes between 10N and 12N of load.  The spring's "elastic limit" is somewhere in that slot.


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A baseball of mass 1.23 kg is thrown at a speed of 65.8 mi/h. What is its kinetic energy?
frosja888 [35]

Given:

The mass of the ball is

m=1.23\text{ kg}

The speed of the ball is

\begin{gathered} v=65.8\text{ mi/h} \\  \end{gathered}

Required: calculate the kinetic energy of the baseball

Explanation: to calculate the kinetic energy of a body we will use the formula as

K.E=\frac{1}{2}mv^2

first, we convert velocity from mi/h into m/s.

we know that

1\text{ mi=1609.34 m}

and

1\text{ h=3600 sec}

then the velocity is

\begin{gathered} v=\frac{65.8\times1602.34\text{ m}}{3600\text{ s}} \\ v=29.29\text{ m/s} \end{gathered}

now plugging all the values in the above formula, we get

\begin{gathered} K.E=\frac{1}{2}mv^2 \\ K.E=\frac{1}{2}\times1.23\text{ kg}\times(29.29\text{ m/s})^2 \\ K.E=527.61\text{ J} \end{gathered}

Thus, the kinetic energy of the baseball is

527.61\text{ J}

4 0
2 years ago
A softball with a mass of 0.11 kg goes moves at a speed of 12 m/s, then the ball is hit by a bat and rebounds in the opposite di
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7 0
3 years ago
A 50-kg box is being pushed along a horizontal surface. The coefficient of static friction between the box and the ground is 0.6
dsp73

To solve the exercise it is necessary to apply the concepts related to Newton's Second Law, as well as the definition of Weight and Friction Force.

According to the problem there is a movement in the body and it is necessary to make a sum of forces on it, so that

\sum F = ma

There are two forces acting on the body, the Force that is pushing and the opposing force that is that of friction, that is

F - F_f = ma

To find the required force then,

F=F_f+ma

By definition we know that the friction force is equal to the multiplication between the friction coefficient and the weight, that is to say

F = \mu mg +ma

F = 0.35*50*0.8+50*1.2

F=(171.5N)+(50Kg)(1.2m/s^2)

F=231.5N

F\approx 230N

Therefore the horizontal force applied on the block is B) 230N

6 0
4 years ago
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