Convert 5 x 105 into normal notation.

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

sweet-ann [11.9K]

Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

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

equating the two equations and cancelling a, we have:

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

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

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

F = ma

equating the above equations and making a subject of formula, we get:

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

a = 25/10 = 2.5 $ms^{-2}$

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

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

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

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

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

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

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

if r is doubled,

F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

7 0

3 years ago

A speedy rabbit is hopping to the right with a velocity of 4.0 \,\dfrac{\text m}{\text s}4.0 s m 4, point, 0, start fraction,

Mrrafil [7]

Answer: 38.25 m

Explanation:

In this situation we need to find the distance $d$ between the rabbit and the carrot, and we can use the following equation, since the rabbit's acceleration is constant:

$V^{2}=V_{o}^{2} + 2ad$ (1)

Where:

$V=13 m/s$ is the rabbit's maximum velocity (final velocity)

$V_{o}=4 m/s$ is the rabbit's initial velocity

$a=2 m/s^{2}$ is the rabbit's acceleration

$d$ is the distance between the rabbit and the carrot

Isolating $d$ :

$d=\frac{V^{2}-V_{o}^{2}}{2a}$ (2)

$d=\frac{(13 m/s)^{2}-(4 m/s)^{2}}{2(2 m/s^{2})}$ (3)

Finally:

$d=38.25 m$

8 0

4 years ago

Read 2 more answers

The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

vaieri [72.5K]

Answer:

v = 44,16 m/s

Explanation:

We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:

$v_y^2 =v_0^2 +2ay$

Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:

$v_y^2 =2ay \rightarrow v_y = \sqrt{2ay}$

Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.

v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s $v_y =- \sqrt{2*(-9,81 m/s^2)*(-99,4 m)} = 44,1613 m$

So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.

I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D

7 0

4 years ago

a professional athlete who takes dance or gymnastics to improve their skills in their professional sport.

scoundrel [369]

Yes, many professional athletes will take dance classes to improve their skills. Such as football players to improve their footwork and use other muscles they wouldn't use while doing just push ups.

4 0

4 years ago

The normal force n with arrow, the gravitational force mg with arrow, and the upward component of the applied force do no work o

lord [1]

Answer and Explanation:

Should let force applied throughout horizontal direction become "F", as well as the displacement of that same sled, become "d." This same force's work seems to be "Fd."
Presently, unless the mass would be doubled as well as the force is applied continues to remain the very same, the work used by the force to keep moving the sled by such the distance "d" would then remain the same.

So that the above seems to be the right answer.

4 0

4 years ago