Ask question

Ask question

All categories

2 years ago

13

( 8c5p79) A certain force gives mass m1 an acceleration of 13.5 m/s2 and mass m2 an acceleration of 3.5 m/s2. What acceleration

would the force give to an object with a mass of (m2-m1)

1 answer:

Talja [164]2 years ago

4 0

Answer:

$4.725 m/s^{2}$

Explanation:

We know that from Newton's second law of motion, F=ma hence making acceleration the subject then $a=\frac {F}{m}$ where a is acceleration, F is force and m is mass

Also making mass the subject of the formula $m=\frac {F}{a}$

For $m1= \frac {F}{13.5}$ and $m2=\frac {F}{3.5}$ hence $F=(m2-m1)a= (\frac {F}{3.5}-\frac {F}{13.5})a=0.2116402116\\\frac {1}{a}=0.2116402116\\a=4.725 m/s^{2}$

You might be interested in

An overtone that's a whole number multiple of the fundamental frequency of a string is called a

alina1380 [7]

The correct answer in this case is B. Harmonic.

The remaining answers refer to other parts of a musical composition, and some such as the pitch can even be found in the study of human voice.

6 0

3 years ago

How are magnetic poles and electrical charges similar?

exis [7]

I believe the answer is a

3 0

2 years ago

Read 2 more answers

What is a learned behavior of a puppy

Ludmilka [50]

Its when the animal repeats what it is taught similar to a child you never really teach them to lie it just happens because they have seen other people do it

3 0

3 years ago

The nucleus contains protons and neutrons. Being positively charged, the protons repel each other. The nucleus should fly apart

Arte-miy333 [17]

this is due to the existence of other forces called the strong nuclear forces that overcomes the repulsion forces between the protons and keeps the nucleons holding to each other also there is a type of energy that is called the nuclear binding energy and this energy also works on binding the components of the nucleus together

6 0

2 years ago

Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

3 0

2 years ago